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Introduction to calculus

Table of Contents

Introduction

Welcome to the Introduction to calculus module. This module provides an overview of some key concepts in calculus, including limits, differentiation, integration, differential equations and vectors.

You may wish to work your way through the entirety of the module in the order provided, or you may wish to jump to certain pages or sections of the module.

Your feedback on this module is very welcome and can be provided at any time on the feedback page, or alternatively for any questions about the module please contact Library-UniSkills@curtin.edu.au

What you will learn

In calculus, limits are used to define continuity, derivatives and integrals. This page introduces limits and continuity, while later pages explain derivatives and integration.

In brief, it covers the following:

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Limits

Suppose we have a function \(f(x)\). We want to look at the value this function gets close to as the input value \(x\) gets close to a particular value. We will call this value \(a.\)

In this case we say we are finding the limit of \(f(x)\) as \(x\) approaches \(a,\) and we use the following notation to represent this:

\[\lim_{x \to a} f(x)\]

Sometimes we need to show whether we are approaching \(a\) from the left or from the right. This is becasue the limit does not always give the same value from each side. When approaching from the left, we use the notation \(\lim_{x \to a^-} f(x)\). This is called a left-hand limit. When approaching from the right, we use the notation \(\lim_{x \to a^+} f(x)\). This is called a right-hand limit.

For \(\lim_{x \to a} f(x)\) to exist, \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\) need to be the same as eeach other.

We can use tables of values to calculate the left hand limit and the right hand limit of a function. This lets us work out if the limit of that function exists as \(x\) approaches a given value.

Example 1:

Use tables of values to investigate \(\lim_{x \to 3} (x^2 + 4x + 5)\)

Solution:

Limit as \(x\) approaches \(3\) from the left:

\(x\) \(x^2 + 4x + 5\)
2.9 25.01
2.99 25.9001
2.999 25.990001
2.9999 25.99900001

Limit as \(x\) approaches \(3\) from the right:

\(x\) \(x^2 + 4x + 5\)
3.1 27.01
3.01 26.1001
3.001 26.010001
3.0001 26.00100001

These tables indicate that:
\(\lim_{x \to 3^-} (x^2 + 4x + 5) =\) \(\lim_{x \to 3^+} (x^2 + 4x + 5) =\) \(26\)

Therefore:
\(\lim_{x \to 3} (x^2 + 4x + 5) = 26\)

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The fact that \(\lim_{x \to 3} (x^2 + 4x + 5) = 26\) should not be surprising, as this is the value of \(x^2 + 4x + 5\) when \(x = 3\). In other words, \(\lim_{x \to 3} (x^2 + 4x + 5) = f(3)\)

In fact, the following is true for any function that is continuous at \(x = a\):

\(\lim_{x \to a} f(x) = f(a)\)

More information on continuity, and the conditions required for a function to be continuous at \(x = a,\) are covered in the following section.

Example 2:

Use tables of values to investigate \(\lim_{x \to 2} (\frac{1}{x-2})\)

Solution:

Limit as \(x\) approaches \(2\) from the left:

\(x\) \(x^2 + 4x + 5\)
1.9 -10
1.99 -100
1.999 -1000
1.9999 -10000
1.99999 -100000

Limit as \(x\) approaches \(2\) from the right:

\(x\) \(x^2 + 4x + 5\)
2.1 10
2.01 100
2.001 1000
2.0001 10000
2.00001 100000

These tables indicate that:
\(\lim_{x \to 2^-} (\frac{1}{x-2}) = -\infty\)
\(\lim_{x \to 2^+} (\frac{1}{x-2}) = \infty\)

Therefore:
\(\lim_{x \to 2} (\frac{1}{x-2})\) does not exist

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It is important to understand something about infinity here. Technically, a limit cannot equal positive infinity or negative infinity. This means that if a limit tends towards positive or negative infinity, the limit does not exist.

However, saying that the limit tends towards positive or negative infinity (as shown above) still gives us useful information. It tells us more about how the function behaves close to the specified point.

The fact that this limit does not exist also tells us something else. It means that this function is not continuous. More information about continuity is provided in the following section.

For now, test your understanding of limits by working through the questions below.

1. Determine \(\lim_{x \to 3} (x^2 + 4)\) if it exists

Using tables of values, or by substituting \(x = 3\) into the function (since it is continuous at this point), we can determine that \(\lim_{x \to 3} (x^2 + 4) = 13\)

2. Determine \(\lim_{x \to 1} (\frac{x^2-3x+2}{x-1})\) if it exists

Using tables of values, we can determine that \(\lim_{x \to 1} (\frac{x^2-3x+2}{x-1}) = -1\)

3. Determine \(\lim_{y \to -5} (\frac{y^2+4}{y+5})\) if it exists

Using tables of values, we can determine that this limit does not exist. If we want to provide more information about the limit, however, we could state that:

4. Determine \(\lim_{x \to 2} f(x)\) if it exists, where \(f(x) = 5 - x^2\) when \(x \leqslant 2\) and \(f(x) = x - 1\) when \(x > 2\)

Using tables of values, or by substituting \(x = 2\) into \(f(x) = 5 - x^2\), we can determine that:
\(\lim_{x \to 2^-} f(x) =1\)

Similarly, using tables of values or by substituting \(x = 2\) into \(f(x) = x - 1\), we can determine that:
\(\lim_{x \to 2^+} f(x) =1\)

Therefore, \(\lim_{x \to 2} f(x) =1\)

Continuity

A function is called continuous if its graph has no breaks or jumps. This means you could draw the whole graph without ever lifting your pencil off the paper. If a function is continuous everywhere on its graph, we say it is continuous.

Sometimes, however, a function is not continuous everywhere. In many cases, the function might have just one point where it is not continuous. This means it is continuous at every other value of \(x\), except for that one point.

For example:

Three graphs. The first two graphs are two curves with no breaks or jumps, and with the label continuous everywhere. The third graph is broken when x = 0, and has the label not continuous at x = 0.

The idea above is a helpful way to think about continuity, but there is also a formal mathematical definition. This definition says that a function \(f(x)\) is continuous at \(x = a\) only if all of the following are true:

Example:

Is the function \(f(x) = \frac{x+4}{x^2 + 3x - 40}\) continuous? If not, for which value or values is it discontinuous?

Solution:

Noting that dividing by \(0\) is undefined, we can factorise the denominator of this function to find the values of \(x\) such that it is equal to \(0.\) Factorising gives the following: \(f(x) = \frac{x+4}{(x+8)(x-5)}\)
Finding the values of \(x\) such that \((x+8)(x-5) = 0\) gives \(x = -8\) and \(x = 5,\) which means the function is discontinuous at these values.

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All polynomials are continuous everywhere. This means that if you have two functions \(f(x)\) and \(g(x)\), and both of them are continuous at \(x = a\) then the following are also true:

Test your understanding of continuity by working through the questions below:

1. Is \(f(x) = x^3 + 2x - 4\) continuous? If not, for which values is it discontinuous?

Yes, this function is continuous everywhere as it is a polynomial.

2. Is \(\frac{x^2-4x+5}{x-3}\) continuous? If not, for which value or values is it discontinuous?

The denominator of this function is equal to \(0\) when \(x = 3,\) hence the function is discontinuous when \(x = 3\)

3. Is \(f(x) = \frac{x - 2}{x^2 + 10x + 16}\) continuous? If not, for which value or values is it discontinuous?

Factorising the denominator of this function gives:
\(f(x) = \frac{x - 2}{(x+8)(x+2)}\)
Since you can’t divide by \(0\) we need to find the values for \(x\) such that \((x+8)(x+2) = 0,\) which gives \(x = -8\) and \(x = -2\)
This means the function is discontinuous at these values.

4. The functions \(f(x) = \frac{x + 2}{x + 4}\) and \(g(x) = \frac{2x + 5}{x + 4}\) are both continuous at \(x = 1\). Explain how you can use this information to determine that the function \(h(x) = \frac{3x + 7}{x + 4}\) is also continuous at \(x = 1\)

\(h(x) = f(x) + g(x),\) and since \(f(x)\) and \(g(x)\) are both continuous at \(x = 1\) it follows that \(h(x)\) must also be continuous at \(x = 1\)

Derivatives

The derivative of a function \(f(x)\) is the rate of change of that function with respect to its input value, \(x\). To find the derivative we use a process called differentiation. The rules for how to do this are explained on this page.

In brief, it covers the following:

</div>

Before we look at the rules, let’s first look more closely at what a derivative actually is.

Understanding derivatives

In order to understand derivatives we need to start with the graph of a function, and the slope of that graph.

Consider the graph:

A graph of a line, y=2x-1.

We can find the slope, or gradient, of this graph by choosing two points along the graph and dividing the change in \(y\) by the change in \(x,\) as shown:

A graph of a line, y=2x-1, with two points marked and the change in y and the change in x labelled.

\[Slope\,=\,\frac{change\;in\;y}{change\;in\;x}\]

Note that the function plotted in the graph above is a linear function. Because of this, the slope is the same at every point on the graph. However, if the function is not linear this is not true. Instead the slope will be different at different points on the graph.

For any given point on a non-linear graph, the slope at that point is the slope of the tangent to the curve at that point. The tangent is a straight line that touches the curve at that one point.

For example, the diagram below shows the tangent to the function \(f(x) = x^2 + 1\) at the point where \(x = 0\):

A graph of a function with a tangent line at x = 0

As this tangent is horizontal it has a slope of \(0,\) meaning the slope of \(f(x) = x^2 + 1\) at \(x = 0\) is \(0.\) However, while this is true at this particular point, to calculate the slope at other points we need to calculate the slope of the line between that point and another point as the second point gets closer and closer. In other words, it requires us to find the slope of the line as the difference between the two \(x\) values approaches zero.

To do this, we can use the equation for the slope of a line. Writing change in using the Greek letter delta \(\Delta\) instead, it becomes:

\[Slope\,=\,\frac{\Delta\,y}{\Delta\,x}\]

Then as the change in the \(x\) values approaches zero, it can be written as the following limit:

\[\lim_{\Delta x \to 0} \frac{\Delta\,y}{\Delta\,x}\]

This is called the derivative of the function. Rather than writing it as above though, we write it as \(\frac{dy}{dx}\)

Alternatively, if we use the function name \(f(x)\) instead we can write the derivative as the following limit, where \((x, f(x))\) is the point of interest and \(h\) is the difference between this and the second point, \((x+h, f(x+h)):\)

\[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]

Example:

Find the derivative of \(f(x) = x^2 + 1\)

Solution:

Substituting this function into the limit above gives:

\[\begin{aligned} \lim_{h \to 0} \frac{((x+h)^2 + 1) - (x^2 + 1)}{h} &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 1 - x^2 - 1}{h}\\\ &= \lim_{h \to 0} \frac{2xh + h^2}{h}\\\ &= \lim_{h \to 0} 2x + h\\\ &= 2x\\\ \end{aligned}\]

So the derivative of this function is \(2x\) </div>

In this case, the derivative of the function is called the prime of the function and is written as \(f'(x)\). This is read as f-prime of \(x\) and means the derivative of \(f(x)\) with respect to \(x\).

Therefore, if \(y\,=\,f(x)\), the derivative of that function will be written as:

\[\frac{dy}{dx}\;\]

or

\[f'(x)\]

(It may also be written as \(\frac{d}{dx}\,y\) or \(\frac{d}{dx}\,f(x),\) but this notation won’t be used here.) </div>

Note that the derivatives discussed so far are referred to as first derivatives, as we are only differentiating once. While these are the only kinds of derivatives covered on this page, it is important to know that you can differentiate again to give the second derivative \(\frac{d^2y}{dx^2},\) and again to give the third derivative \(\frac{d^3y}{dx^3},\) and so on.

Differentiation rules

Rather than calculating the limit each time as above, there are a set of general rules to follow in order to find the derivative of any given function. You may need to use any number of these rules, depending on the elements within the function, in order to find the derivative.

Constant rule

The derivative of a constant number, \(c\), is always zero.

If

\[f(x)\,=\,c\]

then

\[f'(x)\,=\,0\]

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Example:

Find the derivative of \(f(x)\,=\,3\)

Solution:

Using the constant rule, \(f'(x)\,=\,0\)

</div>

Line rule

When \(x\) has a coefficient, the derivative will just be the coefficient.

If

\[f(x)\,=\,nx\]

then

\[f'(x)\,=\,n\]

</div>

Example:

Find the derivative of:

\[f(x)\,=\,7x\]

Solution:

\[f'(x)\,=\,7\]

</div>

Power rule

When \(x\) has an indice, multiply \(x\) by that indice, and then reduce the indice by one.

If

\[f(x)\,=\,x^n\]

then

\[f'(x)\,=\,n\,x^{n-1}\]

</div>

Example 1:

What is \(f'(x)\) given \(f(x)\,=\,x^4\)

Solution:

\[f'(x)\,=\,4x^3\]

Example 2:

What is the derivative of \(y\,=\,3\,x^2\)

Solution:

\(\begin{aligned} \frac{dy}{dx} &= 2 \times 3x^{2-1} \\\ &= 6x \\\ \end{aligned}\) </div>

Trigonometric function rules

We can also find the derivatives of the trigonometric functions \(sin\), \(cos\), and \(tan\). It’s important to note that when working with the derivatives of trigonometric functions, \(x\) will be in radians.

The derivative of \(sin\) is \(cos\). So if

\[f(x)\,=\,sin\,x\]

then

\[f'(x)\,=\,cos\,x\]

The derivative of \(cos\) is negative \(sin\). So if

\[f(x)\,=\,cos\,x\]

then

\[f'(x)\,=\,-\,sin\,x\]

And the derivative of \(tan\) is \(sec^2\), where \(sec\) is another trigonometric function. So if

\[f(x)\,=\,tan\,x\]

then

\[f'(x)\,=\,sec^2\,x\]

</div>

Example:

If \(f(x)\,=\,2sin\,x\;+\;cos\,x\), find \(f'(x)\)

Solution:

\[f'(x)\,=\,2cos\,x\;-\;sin\,x\]

</div>

Taking it further, when \(x\) has a coefficient in a trigonometric function, the derivative is multiplied by this coefficient.

Example:

If \(f(x)\,=\,2tan\,2x\), find \(f'(x)\)

Solution:

Since the \(x\) in the trigonometric function has a coefficient of \(2\), our solution will be:

\[\begin{aligned} f'(x) &= 2 \times 2\,sec^2\,2x \\\ &= 4\,sec^2\,2x \\\ \end{aligned}\]

</div>

Euler’s number

To find the derivative of a power function of Euler’s number, \(e\), simply multiply the function by the derivative of the power.

If

\[y = e^{f(x)}\]

then

\[\frac{dy}{dx} = f'(x)\,e^{f(x)}\]

</div>

Example:

Find the derivative of \(f(x) = e^x\)

Solution:

\[\begin{aligned} f'(x) &= 1 \times e^x \\\ &= e^x \\\ \end{aligned}\]

Example 2:

Find the derivative of \(f(x) = e^{3x^5 + 1}\)

Solution:

\[f'(x) = 15x^4\,e^{3x^5 + 1}\]

</div>

Logarithms

You may also have to find the derivatives of functions containing logarithms.

If

\[f(x) = log_a\,x\]

then

\[f'(x) = \frac{1}{x\,ln\,a}\]

</div>

Example 1:

Find the derivative of \(f(x) = log_2\,x\)

Solution:
\(\begin{aligned} f'(x) &= \frac{1}{x\,ln\,a} \\\ &= \frac{1}{x\,ln\,2} \\\ \end{aligned}\)

Example 2:

Find the derivative of \(f(x) = log_{10}\,{5x}\)

Solution:

Using the logarithm laws we have:

\[f(x) = log_{10}\,5 + log_{10}\,x\]

Since \(log_{10}\,5\) is a constant, it’s derivative will be \(0\). Therefore \(f(x) = log_{10}\,x\) is the only part that needs to be differentiated:

\[f'(x) = \frac{1}{x\,ln\,{10}}\]

</div>

If your logarithm is the natural logarithm \(ln,\) the derivative is simplified.

If

\[f(x) = ln\,x\]

then

\[f'(x) = \frac{1}{x}\]

</div>

Example 1:

Find the derivative of \(f(x) = ln\,{3x}\)

Solution:

Using log laws, we know that:

\[f(x) = ln\,3 + ln\,x\]

Since \(ln\,3\) is a constant, we have:

\[f'(x) = \frac 1x\]

Example 2:

Find the derivative of \(f(x) = ln(x^2)\)

Solution:

Using log laws, we know that:

\[f(x) = 2\,ln\,x\]

Therefore, we have:

\[\begin{aligned} f'(x) &= 2 \times \frac 1x \\\ &= \frac 2x \\\ \end{aligned}\]

</div>

Note: To differentiate logarithms of functions, you will need to use the chain rule.

Test your knowledge of the above differentiation rules by having a go at finding the derivatives of the functions below:

1. \(y = 3x^5 - 7x + 4\)

\[\frac{dy}{dx} = 15x^4 - 7\]

2. \(f(x) = x - 3\,cos\,x\)

\[f'(x) = 1 + 3\,sin\,x\]

3. \(f(x) = 3\,sin\,4x + 5 - x^7\)

\[f'(x) = 12\,cos\,4x - 7x^6\]

4. \(f(x) = e^{tan5x}\)

\[f'(x) = 5sec^2\,5x\,e^{tan5x}\]

5. \(f(x) = log_4 (x^3)\)

Using log laws, we know that

\[f(x) = 3 log_4 x\]

So the derivative will be

\[\begin{aligned} f'(x) &= 3(\frac{1}{xln4}) \\\ &= \frac{3}{xln4} \\\ \end{aligned}\]

Two functions

You may have to use differentiation to find the derivatives of two related functions, for example \(f(x)\) and \(g(x).\) How these functions are related will determine which rule you use to find the derivative.

Sum and difference

When two functions are added together or subtracted from each other, we find the derivative of each function and keep the same operation.

If

\[y\,=\,f(x)\,\pm\,g(x)\]

then

\[\frac{dy}{dx}\,=\,f'(x)\,\pm\,g'(x)\]

</div>

Example:

Find the derivative of \(y\) given that \(y\,=\,f(x)\,+\,g(x),\) \(f(x)\,=\,4x\,+2\) and \(g(x)\,=\,3x^2\)

Solution:

\(f'(x)\,=\,4\) and \(g'(x)\,=\,6x,\) therefore:

\[\frac{dy}{dx}\,=\,4\,+\,6x\]

</div>

Test your knowledge of the sum and difference rule by using it to find the derivatives of the functions below:

1. \(f(x) = 3x^2 + tan\,x\)
\[f'(x) = 6x + sec^2x\]
2. \(f(x) =\) \((5x^2 + 2x) - (4x^3 - 3x - 7)\)
\[f'(x) = 10x + 2 - (12x^2 - 3)\]

Which is then simplified to be:

\[f'(x) = -12x^2 + 10x + 5\]
3. \(y = 2x^3 + 5\,cos\,2x\)
\[\frac{dy}{dx} = 6x^2 - 10\,sin\,2x\]

Product rule

The product rule is used when two functions are being multiplied together.

If we have

\[y\,=\,f(x)\,\times\,g(x)\]

then the derivative will be

\[\frac{dy}{dx}\,=\,g(x)\,\times\,f'(x)\;+\;f(x)\,\times\,g'(x)\]

</div>

Example:

Given \(y\,=\,(3x\,+\,4) \times (2x^3\,+\,x\,-\,7)\), use the product rule to find \(\frac{dy}{dx}\)

Solution:

Let \(f(x)=3x\,+\,4\) and \(g(x)=2x^3\,+\,x\,-\,7\)

We can now use the product rule to find \(\frac{dy}{dx}\)

\[\begin{aligned} \frac{dy}{dx} &= g(x)\,\times\,f'(x)\;+\;f(x)\,\times\,g'(x) \\\ &= (2x^3\,+\,x\,-\,7) \times (3)\;+\;(3x\,+\,4) \times (6x\,+\,1) \\\ \end{aligned}\]

We then simplify the expression:

\[\begin{aligned} \frac{dy}{dx} &= (2x^3\,+\,x\,-\,7) \times (3)\;+\;(3x\,+\,4) \times (6x\,+\,1) \\\ &= 6x^3\,+\,3x\,-\,21\,+\,18x^2\,+\,3x\,+\,24x\,+\,4 \\\ &= 6x^3\,+\,3x\,-\,21\,+\,18x^2\,+\,27x\,+\,4 \\\ &= 6x^3\,+\,18x^2\,+\,30x\,-\,17 \\\ \end{aligned}\]

</div>

Test your knowledge of the product rule by using it to find the derivatives of the functions below:

1. \(y = x^3\,sin\,x\)
\[\frac{dy}{dx} = 3x^2\;sin\,x + x^3\;cos\,x\]
2. \(f(x) = (2x^3 + 5x)(4x^2 - 3)\)
\[f'(x) = (6x^2 + 5)(4x^2 - 3) + (2x^3 + 5x)(8x)\]

Which can then be expanded and simplified to:

\[f'(x) = 40x^4 + 42x^2 - 15\]
3. \(f(x) =\) \((x^2 + 2x + 1)(3x^2 - 4x + 5)\)
\[f'(x) = (2x + 2)(3x^2 - 4x + 5) + (x^2 + 2x + 1)(6x - 4)\]

Which can then be expanded and simplified into:

\[f'(x) = 12x^3 - 6x^2 + 6\]

Quotient rule

The quotient rule is used when one function is being divided by another.

If we have

\[y\,=\,\frac{f(x)}{g(x)}\]

then the derivative will be

\[\frac{dy}{dx}\,=\,\frac{g(x)\,\times\,f'(x)\;-\;f(x)\,\times\,g'(x)}{(g(x))^2}\]

</div>

Example:

Find \(\frac{dy}{dx}\) given \(y=\frac{3x+5}{2x-2}\)

Solution:

Let \(f(x)=3x+5\) and \(g(x)=2x-2\)

We can now find \(\frac{dy}{dx}\) using the quotient rule.

\[\begin{aligned} \frac{dy}{dx} &= \frac{g(x)\,\times\,f'(x)\;-\;f(x)\,\times\,g'(x)}{(g(x))^2} \\\ &= \frac{(2x-2) \times (3)\;-\;(3x+5) \times (2)}{(2x-2)^2} \\\ \end{aligned}\]

We then simplify the expression:

\[\begin{aligned} \frac{dy}{dx} &= \frac{(2x-2) \times (3)\;-\;(3x+5) \times (2)}{(2x-2)^2} \\\ &= \frac{(6x-6)\,-\,(6x+10)}{(2x-2)^2} \\\ &= -\,\frac{16}{(2x-2)^2} \\\ \end{aligned}\]

</div>

Test your knowledge of the quotient rule by using it to find the derivatives of the functions below:

1. \(y = \frac{3x^2 - 4x + 1}{x^2 + 2}\)
\[\frac{dy}{dx} = \frac{(6x - 4)(x^2 + 2) - (3x^2 - 4x + 1)(2x)}{(x^2 + 2)^2}\]

Which we can then expand and simplify into:

\[\frac{dy}{dx} = \frac{4x^2 + 10x - 8}{(x^2 + 2)^2}\]
2. \(y = \frac{x^2 + 3x}{sin\,x}\)
\[\frac{dy}{dx} = \frac{(2x + 3)\,sin\,x - (x^2 + 3)\,cos\,x}{(sin\,x)^2}\]
3. \(y = \frac{5x^2 + 3}{cos\,3x}\)
\[\frac{dy}{dx} = \frac{10x\,cos\,3x - (5x^2 + 3)(-3sin\,3x)}{(cos\,3x)^2}\]

Which can then be simplified into:

\[\frac{dy}{dx} = \frac{10x\,cos\,3x + 15x^2sin\,3x + 9sin\,3x}{(cos\,3x)^2}\]

Chain rule

We use the chain rule when one function relies on the answer from another function, or when one function contains another function.

For example, if we have a function \(y\) that contains the variable \(u\), and we have a function \(u\) that contains the variable \(x\), to find the derivative \(\frac{dy}{dx}\) we first need to find \(\frac{dy}{du}\) and then \(\frac{du}{dx}\). The chain rule says we can then multiply these together to get \(\frac{dy}{dx}\).

If we want to find \(\frac{dy}{dx}\) and we know \(\frac{dy}{du}\) and \(\frac{du}{dx}\) then

\[\frac{dy}{dx}\,=\,\frac{dy}{du}\,\times\,\frac{du}{dx}\]

</div>

Example 1:

Find \(\frac{dy}{dx}\) given \(y\,=\,10u\) and \(u\,=\,-4x^2\,+\,2x\)

Solution:
The first thing we have to do is find \(\frac{dy}{du}\):

\[\frac{dy}{du}=10\]

Next, we need to find \(\frac{du}{dx}\):

\[\begin{aligned} \frac{du}{dx} &= 2 \times (-4)x^{2-1}\,+\,2 \\\ &= -8x\,+\,2 \\\ \end{aligned}\]

Now, using the chain rule, we can find \(\frac{dy}{dx}\):

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du}\,\times\,\frac{du}{dx} \\\ &= 10 \times (-8x\,+\,2) \\\ &= -80\,x\,+\,20 \\\ \end{aligned}\]

Example 2:

We can use the chain rule to differentiate the logs of functions. As an example, find the derivative of \(y = ln(2x^2 + 3)\)

Solution:

Let \(u = 2x^2 + 3,\) so \(y = ln\,u\) and we have \(\frac{du}{dx} = 4x\) and \(\frac{dy}{du} = \frac{1}{u}.\) Our integral can then be evaluated as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\\ &= \frac{1}{u} \times 4x \\\ &= \frac{4x}{u} \\\ &= \frac{4x}{2x^2 + 3} \\\ \end{aligned}\]

</div>

Test your knowledge of the chain rule by using it to find the derivatives of the functions below:

1. \(y = cos(2x^2 + 1)\)

Let \(u = 2x^2 + 1,\) so \(y = cos\,u\) and we have \(\frac{du}{dx} = 4x\) and \(\frac{dy}{du} = -sin\,u.\) Our derivative can then be evaluated as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\\ &= -sin\,u \times 4x \\\ &= -4x\,sin\,u \\\ &= -4x\,sin(2x^2 + 1) \\\ \end{aligned}\]
2. \(y = (3x^2 + 2)^4\)

Let \(u=3x^2 + 2,\) so \(y = u^4\) and we have\(\frac{du}{dx} = 6x\) and \(\frac{dy}{du} = 4u^3.\) Our derivative can then be evaluated as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\\ &= 4u^3 \times 6x \\\ &= 4(3x^2 + 2)^3 \times 6x \\\ &= 24x(3x^2 + 2)^3 \\\ \end{aligned}\]
3. \(y = log_5 (3x^2 + 1)\)

Let \(u = 3x^2 + 1,\) so \(y = log_5\,u\) and we have \(\frac{du}{dx} = 6x\) and \(\frac{dy}{du} = \frac{1}{u\,ln\,5}.\) Our derivative can then be evaluated as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\\ &= \frac{1}{u\,ln\,5} \times 6x \\\ &= \frac{6x}{u\,ln\,5} \\\ &= \frac{6x}{(3x^2 + 1)\,ln\,5} \\\ \end{aligned}\]
4. Find \(\frac{dy}{dx}\) given \(u = sin\,x\) and \(y = 5u^2 + 2u\)

\(\frac{du}{dx} = cos\,x\) and \(\frac{dy}{du} = 10u + 2,\) so our derivative can be evaluated as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\\ &= (10u + 2) \times cos\,x \\\ &= (10\,sin\,x + 2)(cos\,x) \\\ \end{aligned}\]

Integration

The integral of a function is used to find the area under the graph of that function. There are two types of integrals: definite integrals and indefinite integrals. To find the integral of a function, we use a set of rules called integration.

Indefinite integrals can also be called antiderivatives, and integration can also be called antidifferentiation. These names are used because integration is the inverse of differentiation. This means that integration “undoes” differentiation.

In brief, this page covers the following:

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Understanding integration

Consider the graph:

A graph of a line, y=2sinx+3.

If we wanted to find the area between the graph and the \(x\)-axis we could calculate the function at a few points and add these up, essentially breaking the area up into smaller parts.

A graph of a line, y=2sinx+3. The are 10 blue rectangles sitting next to each other, touching the x-axis and going up to the line of the graph.

The smaller the parts we use, or the smaller the change in \(x\), the more accurate we can be in our answer. However, this also means more work as there will be an infinite number of parts that could be used to find the area. Instead of doing a lot of individual calculations and adding up all these individual parts, we can use integration to find the integral of the function and use that to calculate the area under the graph.

For example, consider the graph below:

A graph of a line, y=2x. There is a blue triangle sitting under the graph from x=0 to x=4.

We know that the formula to calculate the area of a triangle is \(\frac 12 base \times height\). In this example, our base is going to be the value of \(x\), and our height will be \(2x\). So we have:

\[\begin{aligned} Area\;under\;the\;graph &= \frac 12 (x)(2x) \\\ &= x^2 \\\ \end{aligned}\]

This means that the integral of the function \(y = 2x\) is \(x^2\). If we were asked to find the derivative of that integral, the answer would be \(2x\), which is the function we started with.

Integrals don’t always appear as simple as this example, but that’s why we have rules we can follow to help us integrate any given function.

If we have \(y = f(x)\), the notation for the integral of that function will look like:

\[\int f(x)\,dx\]

The \(\int\) is the mathematical integral symbol, and \(dx\) represents the change in \(x\).

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This notation is used when finding an indefinite integral.

Definite vs indefinite

An indefinite integral does not have any defined limits. Finding the indefinite integral of a function is the same as finding the antiderivative of that function. Your answer will still look like a function rather than a value, and will be the inverse of the derivative.

A definite integral has defined upper and lower limits. These are the values of \(x\) between which you are being asked to find the area under the graph.

The notation of a definite integral will have values, or limits, attached to the integral symbol \(\int\):

\[\int_a^b f(x)\,dx\]

This is asking us to find the integral of the function \(f(x)\) between \(x=a\) and \(x=b\), that is, the value of the area under the graph between \(x=a\) and \(x=b\):

A graph of a line, y=2sinx+3. There is a blue shaded area sitting under the graph from x=a to x=b.

This is performed by finding the value of the integral at \(x=a\) and subtracting it from the value of the integral at \(x=b\):

\[\int_a^b f(x)\,dx = \int f(b) - \int f(a)\]

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Note: if you have a negative definite integral, you can make it positive by swapping the limits.

\[- \int_b^a f(x)\,dx = \int_a^b f(x)\,dx\]

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In general, if you are required to find the integral of a function, you are probably being asked to find the definite integral. If you are required to find the indefinite integral of a function, you may be asked to find the antiderivative and will not be given any limits.

Integration rules

The rules of integration are the inverse of the rules of differentiation. If you’d like to know more about differentiation and how it compares to integration, have a look at our guide to derivatives.

Each of the rules below have the notation expected for indefinite integrals. However, there are also examples for finding definite integrals using each rule.

You’ll notice that each indefinite integral includes a \(+ C\). This is known as the constant of integration. This is added onto the end of indefinite integrals as the derivative of any real constant is zero, and we need to include the possibility that our integral contains a constant. The constant of integration is not required for definite integrals.

Constant rule

The integral of a constant will always be that same constant multiplied by a variable, usually \(x\).

The integral of a constant will be:

\[\int a\;dx\;=\;ax+C\]

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Example 1:

Find the antiderivative of \(y = 8\)

Solution:

\[\int 8\;dx = 8x + C\]

Example 2:

Evaluate the definite integral \(\int_2^5 7\,dx\)

Solution:

\[\begin{aligned} \int_2^5 \,7\,dx &= 7(5) - 7(2) \\\ &= 35 - 14 \\\ &= 21 \\\ \end{aligned}\]

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Power rule

When finding the integral of a variable with a power, add one to the power and then divide it by that same new power.

The integral of a variable with a power will be:

\[\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\]

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Example 1:

Find the indefinite integral of \(y=x\)

Solution:

\[\int x\,dx = \frac{x^2}{2} + C\]

Example 2:

Find the indefinite integral of \(y=2x^5\)

Solution:

\[\begin{aligned} \int 2x^5\,dx &= \frac{2x^6}{6} + C \\\ &= \frac{x^6}{3} + C \\\ \end{aligned}\]

Example 3:

Evaluate \(\int_5^8 2x^2 dx\)

Solution:

\[\begin{aligned} \int_5^8 \,2x^2\,dx &= \frac{2(8)^3}{3} - \frac{2(5)^3}{3} \\\ &= \frac{1024}{3} - \frac{250}{3} \\\ &= 258 \\\ \end{aligned}\]

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Multiplier constant

If the function is being multiplied by a constant, we can move the constant and place it outside the integration symbol. We can then put it back in front of our integrated function. Let’s have a look at some of the examples from above and see if we get the same results using the multiplier constant rule.

Example 1:

Find the indefinite integral of \(y=2x^5\)

Solution:

We want to find \(\int 2x^5\,dx\), but let’s move the constant first.

\[2 \int x^5\,dx\]

Now, we can find the integral of our function, and multiply it by our constant, \(2\), again.

\[\begin{aligned} 2 \int x^5\,dx &= 2 \frac{x^6}{6} + C \\\ &= \frac{x^6}{3} + C \\\ \end{aligned}\]

Example 2:

Evaluate \(\int_5^8 2x^2 dx\)

Solution:

Again, we can first move the multiplier constant outside of the integration.

\[\begin{aligned} 2\,\int_5^8 \,x^2\,dx &= 2 (\frac{(8)^3}{3} - \frac{(5)^3}{3}) \\\ &= 2 (\frac{512}{3} - \frac{125}{3}) \\\ &= 2(129) \\\ &= 258 \\\ \end{aligned}\]

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As you can see, using the multiplier constant rule gives the same answer for the examples that we saw while trying out the power rule above. These examples were just to demonstrate that it worked and while it may not always be necessary to use the multiplier constant rule, it’s good to know it is an option when integrating functions as it can assist in solving more difficult integrals, such as when having to use integration by substitution.

Trigonometric functions

We can also find the integral of the trigonometric functions \(sin\), \(cos\), and \(tan\). It’s important to note that when working with the integrals of trigonometric functions, \(x\) will be in radians.

The integral of \(sin\) is \(-cos\):

\[\int sin\,x\,dx = -cos\,x + C\]

The integral of \(cos\) is \(sin\):

\[\int cos\,x\,dx = sin\,x + C\]

The integral of \(tan\) is a little more complicated, due to the nature of the graph of \(y = tan\,x\). To find the integral of \(tan\) you need to use integration by substitution. This will give you the following (where \(\vert sec\,x \vert\) represents the absolute value of \(sec\,x\) , which is its non-negative value):

\[\int tan\,x\,dx = ln\,|sec\,x|\, + C\]

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Example 1:

Find \(\int 3\,sin\,x\,dx\)

Solution:

\[\begin{aligned} \int 3\,sin\,x\,dx &= 3\,\int sin\,x\,dx \\\ &= - 3\,cos\,x + C \\\ \end{aligned}\]

Example 2:

Evaluate \(\int_0^{\pi} 2\,sin\,x\,dx\)

Solution:

\[\begin{aligned} \int_0^{\pi} 2\,sin\,x\,dx &= 2 \int_0^{\pi} sin\,x\,dx \\\ &= 2(- cos\,\pi - (-cos\,0)) \\\ &= 2(1 + 1) \\\ &= 4 \\\ \end{aligned}\]

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Euler’s number

We can also find the integral of \(e^x\). We can easily find the integral of \(e\), even when \(x\) has a coefficient. If \(x\) has its own power, we need to start using imaginary numbers, and that is not covered in this module.

The integral of \(e^x\) is \(e^x\):

\[\int e^x\,dx = e^x + C\]

When \(x\) has a coefficient, we divide the answer by this number:

\[\int e^{ax}\,dx = \frac{e^{ax}}{a} + C\]

Finally, we can use exponent laws to expand this one step further as follows:

\[\int e^{ax + b}\,dx = \frac{e^{ax + b}}{a} + C\]

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Example 1:

Determine the integral of

\[\int e^{3x+2}\,dx\]

Solution:

\[\int e^{3x+2}\,dx = \frac{e^{3x+2}}{3} + C\]

Example 2:

Evaluate the integral

\[\int_2^3 e^{3x}\,dx\]

Solution:

\[\begin{aligned} \int_2^3 e^{3x}\,dx &= \frac{e^{3(3)}}{3} - \frac{e^{3(2)}}{3} \\\ &= 2701.03 - 134.48 \\\ &= 2566.55 \end{aligned}\]

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Logarithms

We can find the integral of logarithms with various bases, however as the natural log \(ln\) (the log with base \(e\), so \(log_e\)) is more frequently used in calculus it is the only one covered here.

The integral of the natural log, \(ln\,x,\) is:

\[\int ln\,x\,dx = x\,ln\,x - x + C\]

or

\[\int ln\,x\,dx = x\,(ln\,x - 1) + C\]

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Example 1:

Find the antiderivative of \(f(x) = ln\,(2x)\)

Solution:

Using the logarithm laws we know that \(ln\,(2x) = ln\,x + ln\,2\)

\[\begin{aligned} \therefore \quad \int ln\,2x\,dx &= \int (ln\,x + ln\,2)\,dx \\\ &= \int ln\,x\,dx + \int ln\,2\,dx \\\ &= x\,ln\,x - x + x\,ln\,2 + C \\\ \end{aligned}\]

Example 2:

Calculate the integral below to 3 decimal places.

\[\int_1^3 \frac{1}{2} ln\,x\,dx\]

Solution:

\[\begin{aligned} \int_1^3 \frac{1}{2} ln\,x\,dx &= \frac{1}{2} \int_1^3 ln\,x\,dx \\\ &= \frac{1}{2}((3\,ln\,3 - 3) - (ln\,1 - 1)) \\\ &= \frac{1}{2}((3\,ln\,3 - 3) - (-1)) \\\ &= \frac{1}{2}(3\,ln\,3 - 2) \\\ &= 0.648 \\\ \end{aligned}\]

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Reciprocals

Now that we’ve had a look at logarithms, it’s important to have a look at the integral of reciprocals. A reciprocal is also known as a (multiplicative) inverse. For example, the inverse of \(2\) is \(\frac{1}{2},\) and the inverse of \(x\) is \(\frac{1}{x}\) (this is known as the reciprocal function).

The integral of the reciprocal function is as follows (where \(\vert x \vert\) represents the absolute value of \(x\) , which is its non-negative value):

\[\int \frac{1}{x}\,dx = ln\,|x|\,+ C\]

Furthermore, we can use the multiplier constant rule to establish that:

\[\int \frac{a}{x}\,dx = a\,ln\,|x|\,+ C\]

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Example 1:

Find the antiderivative of \(f(x) = \frac{2}{x}\)

Solution:

\[\int \frac{2}{x}\,dx = 2\,ln\,|x|\,+ C\]

Example 2:

Calculate the integral below. Give your answer to two decimal places.

\[\int_1^2 \frac{1}{4x}\,dx\]

Solution:

\[\begin{aligned} \int_1^2 \frac{1}{4x}\,dx &= \frac{ln\,(2)}{4} - \frac{ln\,(1)}{4} \\\ &= \frac{ln\,(2)}{4} - 0 \\\ &= \frac{ln\,(2)}{4} \\\ &= 0.17 \\\ \end{aligned}\]

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Sum and difference

To find the integral of two (or more) functions that are being added together or subtracted from each other, we can find the integral of each one separately first, and keep the relevant operation.

The integral of \(y\,=\,f(x)\,\pm\,g(x)\) will be:

\[\int (f(x)\,\pm\,g(x))\,dx = \int f(x)\,dx \pm \int g(x)\,dx\]

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Example 1:

Find the antiderivative:

\[\int (sin\,x - 3x^2 + 7x - 4)\,dx\]

Solution:

\[\begin{aligned} \int (sin\,x - 3x^2 + 7x - 4)\,dx &= \int sin\,x\,dx - \int 3x^2\,dx + \int 7x\,dx - \int 4\,dx \\\ &= -cos\,x - \frac{3x^3}{3} + \frac{7x^2}{2} - 4x + C \\\ \end{aligned}\]

Example 2:

Evaluate the definite integral below. Give your answer to 3 decimal places.

\[\int_5^{10} (3x - cos\,x)\,dx\]

Solution:

\[\begin{aligned} \int_5^{10} (3x - cos\,x)\,dx &= \int_5^{10} 3x\,dx - \int_5^{10} cos\,x\,dx \\\ &= (\frac{3(10^2)}{2} - \frac{3(5^2)}{2}) - (sin\,10 - sin\,5) \\\ &= (150 - 37.5) - ((-0.544) - (-0.959)) \\\ &= 112.5 - 0.415 \\\ &= 112.085 \end{aligned}\]

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Check your understanding of the rules of integration by working through the questions below:

1. \(\int (7x^3 - x - 1)\,dx\)

\[\int (7x^3 - x - 1)\,dx = \frac{7x^4}{4} - \frac{x^2}{2} - x + C\]

2. \(\int 3\,cos\,x\,dx\)

\[\begin{aligned} \int 3\,cos\,x\,dx &= 3\int\,cos\,x\,dx \\\ &= 3\,sin\,x + C \\\ \end{aligned}\]

3. \(\int 4\,ln\,5x\,dx\)

\[\begin{aligned} \int 4\,ln\,5x\,dx &= 4\int (ln\,x + ln\,5) dx \\\ &= 4(x\,ln\,x - x + x\,ln\,5) + C \\\ \end{aligned}\]

4. Evaluate \(\int_4^5 (\frac{x^3}{2} - 2x^2)\,dx\) giving your answer to 2 decimal places.

\[\begin{aligned} \int_4^5 (\frac{x^3}{2} - 2x^2)\,dx &= \int_4^5 (\frac{1}{2}\,x^3 - 2x^2)\,dx \\\ &= (\frac{1}{2} \times \frac{(5)^4}{4} - \frac{2(5)^3}{3}) - (\frac{1}{2} \times \frac{(4)^4}{4} - \frac{2(4)^3}{3}) \\\ &= (-5.208) - (-10.667) \\\ &= 5.46 \\\ \end{aligned}\]

5. Calculate \(\int_3^5 \frac{2}{3x}\,dx\) giving your answer to 2 decimal places.

\[\begin{aligned} \int_3^5 \frac{2}{3x}\,dx &= \frac{2}{3} \int_3^5 \frac{1}{x}\,dx \\\ &= \frac{2}{3} (ln\,5 - ln\,3) \\\ &= \frac{2}{3}(0.511) \\\ &= 0.34 \\\ \end{aligned}\]

6. Find \(\int (cos\,x + ln\,4x)\,dx\)

\[\int (cos\,x + ln\,4x)\,dx = sin\,x + x(ln\,x - 1) + x\,ln\,4 +C\]

7. Find \(\int (x^2 + e^{2x})\,dx\)

\[\int (x^2 + e^{2x})\,dx = \frac{x^3}{3} + \frac{e^{2x}}{2} + C\]

8. Calculate \(\int_{1}^{3} (\frac{1}{x} + cos\,x)\,dx\) giving your answer to 3 decimal places.

\[\begin{aligned} \int_{1}^{3} (\frac{1}{x} + cos\,x)\,dx &= (ln\,3 - ln\,1) + (sin\,3 - sin\,1) \\\ &= ln\,3 + sin\,3 - sin\,1 \\\ &= 0.398 \\\ \end{aligned}\]

Integration by substitution

In some specific cases, we can use integration by substitution to find the integral of a function. This method is also known as \(u\)-substitution as it involves using the letter \(u\) to represent part of the function we are integrating, or as the reverse chain rule as it requires you to recognise the derivative of a function.

Given two functions \(f(x)\) and \(g(x)\), we can use integration by substitution if our integral looks like this:

\[\int f(g(x))\;g'(x)\,dx\]

Leaving the above as it is would make finding the integral difficult, but we can change it to make it simpler for ourselves using integration by substitution. Specifically, we let \(u = g(x),\) which gives \(g'(x)\,dx = (\frac{du}{dx})\,dx = du.\) The integral is then simplified as shown below.

Integration by substitution says that if we have:

\[\int f(g(x))\;g'(x)\,dx\]

then we can substitute \(u = g(x)\) and \(du = g'(x)\,dx\) to give:

\[\int f(u)\,du\]

We can then integrate with respect to \(u,\) before substituting \(g(x)\) back into the answer to remove \(u.\) Note also that even if the integral is not in the stated format, you can still integrate by substitution by taking out a multiplier constant as required.

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Example 1:

Evaluate \(\int ln(2x+3)\,dx\)

Solution:

For this example it makes sense to use \(u = 2x+3.\) However, the derivative of \(2x+3\) is \(2\) and we don’t have this in our integral. We can fix this by including it in the integral as \(g'(x)\) and then taking out the multiplier constant of \(\frac{1}{2}\) to compensate though, as follows:

\[\int ln(2x+3)\,dx = \frac{1}{2}\int 2\,ln(2x+3)\,dx\]

We can then use integration by substitution with \(u = 2x+3\) to give:

\[\begin{aligned} \frac{1}{2}\int 2\,ln(2x+3)\,dx &= \frac{1}{2}\int ln\,u\,du \\\ &= \frac{1}{2}\,u(ln\,u - 1) + C \\\ \end{aligned}\]

Finally, we can substitute \(u = 2x + 3\) back into this answer to remove the variable \(u\) and therefore evaluate the original integral:

\[\int ln(2x+3)\,dx = \frac{2x+3}{2}\,(ln (2x+3) - 1) + C\]

Example 2:

Calculate the definite integral of \(f(x) = 2x(1+x^2)^3\) from \(x=0\) to \(x=1\)

Solution:

Looking at our integral, \(\int_0^1 2x(1+x^2)^3\,dx\) we can see that \(2x\) is the derivative of \(1 + x^2\), and therefore we can use \(u = 1 + x^2\)

While we don’t need to make any changes to what is being integrated in this example, because we are changing the variables in the integral we do need to change the limits of the integral so that they are related to the new variable \(u\). We do this by finding the value of \(u\) at the given limits, which in this example is at \(x=0\) and \(x=1\)

So, at \(x=0\):

\[\begin{aligned} u &= 1+(0)^2 \\\ &= 1 \\\ \end{aligned}\]

And, at \(x=1\):

\[\begin{aligned} u &= 1+(1)^2 \\\ &= 2 \\\ \end{aligned}\]

Therefore our integral, after substitution, is now:

\[\int_1^2\,u^3\, du\]

Now we can calculate the definite integral:

\[\begin{aligned} \int_1^2\,u^3\, du &= \frac{(2)^4}{4} - \frac{(1)^4}{4} \\\ &= \frac{16}{4} - \frac{1}{4} \\\ &= \frac{15}{4} \\\ \end{aligned}\]

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Check your understanding of integration by substitution by working through the questions below:

1. Find \(\int 2xe^{x^2}\,dx\)

If we let \(u = x^2,\) this has the derivative \(2x\) and hence the integral is in the required form to use integration by substitution as follows:

\[\begin{aligned} \int 2xe^{x^2}\,dx &= \int e^u du \\\ &= e^u + C \\\ &= e^{x^2} + C \\\ \end{aligned}\]

2. Calculate \(\int_0^1 \frac{2x}{(x^2 + 1)^2}\,dx\)

If we let \(u = x^2 + 1,\) this has the derivative \(2x\) and hence the integral is in the required form to use integration by substitution. However, we do need to change the limits from \(x=0\) to \(u=1\) and from \(x=1\) to \(u=2.\) We can then evaluate the integral as follows:

\[\begin{aligned} \int_0^1 \frac{2x}{(x^2 + 1)^2}\,dx &= \int_1^2 \frac{1}{u^2}\,du \\\ &= \int_1^2 u^{(-2)}\,du \\\ &= [\frac{u^{-1}}{-1}]_1^2 \\\ &= [-\frac{1}{u}]_1^2 \\\ &= -(\frac{1}{2} - \frac{1}{1}) \\\ &= - \frac{1}{2} + 1 \\\ &= \frac{1}{2} \end{aligned}\]

3. Calculate \(\int_0^{\frac{\pi}{2}} \frac{sin\,x}{1 + cos\,x} dx\)

We can use \(u = 1 + cos\,x,\) however since this has the derivative \(-sin\,x\) we need to rewrite the integral as:

\[-\int_0^{\frac{\pi}{2}} \frac{-sin\,x}{1 + cos\,x} dx\]

We also need to rewrite the defined limits in terms of \(u\):

\[\begin{aligned} u &= 1 + cos\,0 \\\ &= 2 \end{aligned}\] \[\begin{aligned} u &= 1 + cos\,\frac{\pi}{2} \\\ &= 1 \end{aligned}\]

Therefore, the integral can be evaluated as follows:

\[\begin{aligned} \int_0^{\frac{\pi}{2}} \frac{sin\,x}{1 + cos\,x} dx &= -\int_0^{\frac{\pi}{2}} \frac{-sin\,x}{1 + cos\,x}\,dx \\\ &= - \int_2^1 \frac{1}{u}\,du \\\ &= \int_1^2 \frac{1}{u}\,du \\\ &= ln\,2 - ln 1 \\\ &= ln\,2 \\\ \end{aligned}\]

Integration by parts

When we have the product of two functions, we can use integration by parts to find the integral. In order to understand how integration by parts works, we need to think back to derivatives and specifically to the product rule. This says that if we have:

\[y\,=\,f(x)\,\times\,g(x)\]

then the derivative will be:

\[\frac{dy}{dx}\,=\,g(x)\,\times\,f'(x)\;+\;f(x)\,\times\,g'(x)\]

If we rewrite the above formula using integration notation, we know that this must also be true:

\[f(x)\,\times\,g(x) = \int\,(g(x)\,\times\,f'(x)\;+\;f(x)\,\times\,g'(x))\,dx\]

This can also be written as:

\[f(x)\,\times\,g(x) = \int\,g(x)\,\times\,f'(x)\, dx + \int\,f(x)\,\times\,g'(x)\,dx\]

When we rearrange this we get the formula for integration by parts:

\[\int\,f(x)\,\times\,g'(x)\,dx = f(x)\,\times\,g(x) - \int\,g(x)\,\times\,f'(x)\, dx\]

where \(f(x)\) and \(g(x)\) are two functions, \(f'(x)\) is the derivative of \(f(x)\) and \(g'(x)\) is the derivative of \(g(x)\)

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Example 1:

Find \(\int\,x\,sin\,x\,dx\)

Solution:

To evaluate this we need:

\[\int\,x\,sin\,x\,dx = \int\,f(x)\,\times\,g'(x)\,dx\]

So let \(f(x) = x\) and \(g'(x) = sin\,x,\) which means \(f'(x) = 1\) and \(g(x) = -\,cos\,x.\) Our indefinite integral can then be evaluated in parts as follows:

\[\begin{aligned} \int\,x\,sin\,x\,dx &= x(-cos\,x) - \int(-cos\,x)\,dx \\\ &= -x\,cos\,x + \int cos\,x\,dx \\\ &= -x\,cos\,x + sin\,x + C \\\ \end{aligned}\]

Example 2:

Evaluate \(\int_0^{\pi} x^2\,cos\,x\,dx\)

Solution:

Let \(f(x) = x^2\) and \(g'(x) = cos\,x,\) which means \(f'(x) = 2x\) and \(g(x) = sin\,x.\) Our definite integral can then be rewritten as follows:

\[\int_0^{\pi}\,x^2\,cos\,x\,dx = [x^2\,sin\,x]_0^{\pi} - \int_0^{\pi}\,2x\,sin\,x\,dx\]

Note that since we don’t need to find the integral of \(x^2\,sin\,x\), we use square brackets with the limits to denote that we still need calculate this part of the formula at those values of \(x\). Doing this gives:

\[\begin{aligned} \,[x^2\,sin\,x]_0^{\pi} &= {\pi}^2 \times sin(\pi) - 0^2 \times sin\,(0) \\\ &= 0 \\\ \end{aligned}\]

Next we need to integrate the second half of our formula \(\int_0^{\pi}\,2x\,sin\,x\,dx,\) which we can do using integration by parts again. This time let \(f(x) = 2x\) and \(g'(x) = sin\,x,\) so \(f'(x) = 2\) and \(g(x) = - cos\,x\) and we have:

\[\begin{aligned} \int_0^{\pi}\,2x\,sin\,x\,dx &= [-2x\,cos\,x]_0^{\pi} - \int_0^{\pi} -2\,cos\,x\,dx \\\ &= -[2x\,cos\,x]_0^{\pi} + 2\,\int_0^{\pi}\,cos\,x\, dx \\\ \end{aligned}\]

Now we calculate \(-[2x\,cos\,x]_0^{\pi}\) as follows:

\[\begin{aligned} -[2x\,cos\,x]_0^{\pi} &= -2(\pi)\,cos\,(\pi) - -2(0)\,cos\,(0) \\\ &= 2 \pi \\\ \end{aligned}\]

Next we calculate the second half of this integral:

\[\begin{aligned} 2 \int_0^{\pi}\,cos\,x\,dx &= 2(sin\,(\pi) - sin\,(0)) \\\ &= 0 \\\ \end{aligned}\]

Therefore we have established that the second half of our original integral is:

\[\begin{aligned} \int_0^{\pi}\,2x\,sin\,x\,dx &= 2 \pi + 0 \\\ &= 2 \pi \\\ \end{aligned}\]

Finally, we can use all of this to calculate the answer to our original integral:

\[\begin{aligned} \int_0^{\pi} x^2\,cos\,x\,dx &= [x^2\,sin\,x]_0^{\pi} - \int_0^{\pi}\,2x\,sin\,x\,dx \\\ &= 0 - 2 \pi \\\ &= -2 \pi \\\ \end{aligned}\]

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Check your understanding of integration by parts by working through the questions below:

1. Find \(\int xe^{2x}\,dx\)

Let \(f(x) = x\) and \(g'(x) = e^{2x},\) which means \(f'(x) = 1\) and \(g(x) = \frac{e^{2x}}{2}\)

So using our integration by parts formula, we have:

\[\begin{aligned} \int xe^{2x}\,dx &= \frac{xe^{2x}}{2} - \int \frac{e^{2x}}{2} \\\ &= \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C \\\ &= \frac{e^{2x}}{4} (2x - 1) + C \\\ \end{aligned}\]

2. Calculate \(\int_0^1 xe^x\,dx\)

Let \(f(x) = x\) and \(g'(x) = e^x,\) which means \(f'(x) = 1\) and \(g(x) = e^x\)

So using our integration by parts formula, we have:

\[\begin{aligned} \int_0^1 xe^x\,dx &= [xe^x]_0^1 - \int_0^1 e^x\,dx \\\ &= (1e^1 - 0 \times e^0) - (e^1 - e^0) \\\ &= e - (e - 1) \\\ &= 1 \end{aligned}\]

3. Find \(\int x^2\,ln\,x\,dx\)

Let \(f(x) = ln\,x\) and \(g'(x) = x^2,\) which means \(f'(x) = \frac{1}{x}\) and \(g(x) = \frac{x^3}{3}\)

So using our integration by parts formula, we have:

\[\begin{aligned} \int x^2\,ln\,x\,dx &= \frac{x^3\,ln\,x}{3} - \int \frac{x^2}{3} \\\ &= \frac{x^3\,ln\,x}{3} - \frac{x^3}{9} + C\\\ \end{aligned}\]

Differential equations

Differential equations are an important idea in calculus because they can be used to describe how things change over time or space. For example, differential equations can be used to model:

They can also be used to model many other situations.

Differential equations are very good at describing these changes. However, they are generally difficult to use in their original form. For this reason, it is helpful to understand how to solve them. This page covers one technique for doing this.

In brief, it covers the following:

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First, though, we look at what exactly differential equations are, and the different types.

Differential equations

A differential equation is an equation that involves one or more derivatives. For example, the following are all differential equations:

\[\frac{dy}{dx} = 4x + 5\]
\[\frac{dP}{dt} = rP(1-\frac{P}{k})\]
\[4x^2\cos\,y\,\frac{dy}{dx} = 10\]
\[7\,\frac{d^2x}{dt^2} + 2\,\frac{dx}{dt} + 2x = \frac{4}{t^2}\]

These are all examples of what are called ordinary differential equations. This is because each equation contains derivatives taken with respect to only one variable. In other words, each equation has derivatives with respect to \(x\) or \(t\), but not both.

If a differential equation contains derivatives with respect to more than one variable, these are called partial derivatives, and the equation is called a partial differential equation. Partial differential equations are not covered on this page.

Differential equations can also be described by their order and their degree.

The order of a differential equation is the highest derivative that appears in the equation. If the highest derivative is a first derivative, the equation is called a first order differential equation. If the highest derivative is a second derivative, the equation is called a second order differential equation, and so on. For example, the first three differential equations above are first order differential equations. The fourth one is a second order differential equation.

The degree of a differential equation is the exponent of the highest derivative in the equation. For example, the following differential equation is a second degree differential equation. It is actually a first order, second degree, ordinary differential equation:

\[(\frac{dy}{dx})^2 + 2y = 3x^3\]

The differential equation below, on the other hand, is only a first degree differential equation as its highest derivative does not have an exponent (in fact it is a second order first degree ordinary differential equation).

\[\frac{d^2y}{dx^2} - (\frac{dy}{dx})^3 = 4x\]

In this page we will look at a method of solving first order first degree differential equations only.

Finally, a special kind of first order differential equation is a linear differential equation, which is a first order differential equation in the following form (where \(p(x)\) and \(q(x)\) are functions of \(x\)):

\[\frac{dy}{dx} + p(x)y = q(x)\]

Test your understanding of the different types of differential equations by working through the questions below:

1. What is the order of the differential equation \(\frac{d^2y}{dx^2} + xy = \frac{dy}{dx} + 2\)

This is a second order differential equation, as the highest derivative is a second derivative.

2. What is the degree of the differential equation \((\frac{d^3y}{dx^3})^2 = (\frac{dy}{dx})^4 + 3x^2\)

This is a second degree differential equation, as the highest derivative has an exponent of \(2.\)

3. What is the full name of the differential equation \((\frac{d^2y}{dx^2})^2 - 2xy = sin\,x\)

This is a second order second degree ordinary differential equation.

4. Is the differential equation \(\frac{dy}{dx} + 2xy^3 = 3x^2\) a linear differential equation?

No, because it contains a \(y^3\) instead of just \(y.\)

Solving differential equations

There are many different types of differential equations, and there are also many different ways to solve them. Some differential equations cannot be solved at all. Some of the methods for solving them are very complex, but one of the simplest is a technique called separation of variables. This technique can be used when a differential equation is separable. A differential equation is separable when it can be written in the form:

\[f(y)\,\frac{dy}{dx} = g(x)\]

If it can be written in this form, then the differential equation can be rearranged as follows:

\[f(y)\,dy = g(x) dx\]

We can then integrate both sides of the equation to solve for \(y\):

\[\int f(y)\,dy = \int g(x)\,dx\]

Sometimes you are given an initial condition. This is the value of \(y\) when \(x\) is equal to a certain number. If you have an initial condition, you can substitute these values into the solution. This allows you to find the constant of integration. If you are not given an initial condition, you can leave the constant of integration as part of the solution.

Example 1:

Solve \(2y\,\frac{dy}{dx} = 4x - 5\)

Solution:

This equation is in the form \(f(y)\,\frac{dy}{dx} = g(x),\) so we can rearrange it and then integrate as follows:

\[\begin{aligned} 2y\,\frac{dy}{dx} &= 4x - 5 \\\ \therefore \int 2y\,dy &= \int (4x - 5)\,dx \\\ \therefore y^2 &= 2x^2 - 5x + C \\\ \therefore y &= \sqrt{2x^2 - 5x + C} \\\ \end{aligned}\]

Example 2:

Solve \(\frac{dy}{dx} = \frac{3x^2(4x + 1)}{2y}\) given that when \(x = 1,\) \(y = 6\)

Solution:

This equation is in the form \(f(y)\,\frac{dy}{dx} = g(x),\) so we can rearrange it and then integrate as follows:

\[\begin{aligned} \frac{dy}{dx} &= \frac{3x^2(4x + 1)}{2y} \\\ \therefore \int 2y\,dy &= \int 3x^2(4x + 1)\,dx \\\ \therefore \int 2y\,dy &= \int (12x^3 + 3x^2)\,dx \\\ \therefore y^2 &= 3x^4 + x^3 + C\\\ \therefore y &= \sqrt{3x^4 + x^3 + C} \\\ \end{aligned}\]

Given that when \(x = 1\) we have \(y = 6,\) it follows that:

\[\begin{aligned} 6 &= \sqrt{1 + 1 + C} \\\ \therefore 6 &= \sqrt{2 + C} \\\ \therefore 36 &= 2 + C \\\ \therefore C &= 34 \\\ \end{aligned}\]

So the solution to the differential equation is \(y = \sqrt{3x^4 + x^3 + 34}\)

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Test your knowledge of the separation of variables technique by using it to solve the differential equations below:

1. \(\frac{dy}{dx} = 4xy\)

\[\begin{aligned} \frac{dy}{dx} &= 4xy \\\ \therefore \int \frac{1}{y}\,dy &= \int 4x\,dx \\\ \therefore ln\,|y|\, &= 2x^2 + C \\\ \therefore y &= Ce^{2x^2} \\\ \end{aligned}\]

Note we can write this as \(y = Ce^{2x^2}\) as the exponent laws tell us we can rewrite \(e^{2x^2 + C}\) as \(e^{2x^2}.e^C,\) and \(e^C\) is just another constant.

2. \(\frac{dy}{dx} = (2x + 1)\,y\)

\[\begin{aligned} \frac{dy}{dx} &= (2x + 1)\,y \\\ \therefore \int \frac{1}{y}\,dy &= \int (2x + 1)\,dx \\\ \therefore ln\,|y|\, &= x^2 + x + C \\\ \therefore y &= Ce^{x^2 + x} \\\ \end{aligned}\]

Note we can write this as \(y = Ce^{x^2 + x}\) as the exponent laws tell us we can rewrite \(e^{x^2 + x + C}\) as \(e^{x^2 + x}.e^C,\) and \(e^C\) is just another constant.

3. \(\frac{dy}{dx} = \frac{cos\,x}{2y}\)

\[\begin{aligned} \frac{dy}{dx} &= \frac{cos\,x}{2y} \\\ \therefore \int 2y\,dy &= \int cos\,x\,dx \\\ \therefore y^2 &= sin\,x + C \\\ \therefore y &= \sqrt{sin\,x + C} \\\ \end{aligned}\]

4. \(\frac{dy}{dx} = 8x^3e^{-y}\) given that when \(x = 0,\) \(y = 5\)

\[\begin{aligned} \frac{dy}{dx} &= 8x^3e^{-y} \\\ \therefore \int e^y\,dy &= \int 8x^3\,dx \\\ \therefore e^y &= 2x^4 + C \\\ \therefore y &= ln(2x^4 + C) \\\ \end{aligned}\]

Given that when \(x = 0\) we have \(y = 5,\) it follows that:

\[\begin{aligned} 5 &= ln\,C \\\ \therefore C &= e^5 \\\ \end{aligned}\]

So the solution to the differential equation is \(y = ln(2x^4 + e^5)\)

5. \(2y - x\,\frac{dy}{dx} = 8\) given that when \(x = 3,\) \(y = 22\)

\[\begin{aligned} 2y - x\,\frac{dy}{dx} &= 8 \\\ \therefore 2y - 8 &= x\,\frac{dy}{dx} \\\ \therefore y - 4 &= \frac{x}{2}\,\frac{dy}{dx} \\\ \therefore \int \frac{1}{y-4}\,dy &= \int \frac{2}{x}\,dx\\\ \end{aligned}\]

We can integrate the left hand side of this equation using integration by substitution where we set \(u = y - 4.\) This gives:

\[\begin{aligned} \int \frac{1}{u}\,du &= ln\,|u|\,\\\ &= ln\,|y - 4|\,\\\ \end{aligned}\]

And so, using the logarithm laws, we have:

\[\begin{aligned} \int \frac{1}{y-4}\,dy &= \int \frac{2}{x}\,dx\\\ \therefore ln\,|y - 4|\, &= 2\,ln\,|x|\, + ln\,C \\\ \therefore ln\,|y - 4|\, &= ln\,|x^2|\, + ln\,C \\\ \therefore ln\,|y - 4|\, &= ln\,|Cx^2|\, \\\ \therefore y - 4 &= Cx^2 \\\ \therefore y &= Cx^2 + 4\\\ \end{aligned}\]

Finally, given that when \(x = 3\) we have \(y = 22,\) it follows that:

\[\begin{aligned} 22 &= 9C + 4 \\\ \therefore 9C &= 18 \\\ \therefore C &= 2 \\\ \end{aligned}\]

So the solution to the differential equation is \(y = 2x^2 + 4\)

Vectors

In calculus, we often work with two different types of quantities: scalars and vectors.

Scalar quantities can be described by their magnitude (or size) alone. This means they can be represented by a single number. For example, if we say it is \(30^{\circ} C\) outside, we have a clear idea of how warm the weather is. Other examples of scalar quantities are \(40\textrm{km/hr}\), \(5\textrm{kg}\) and \(15\textrm{minutes.}\) This is because speed, mass and time are all scalar quantities. In other words, a scalar tells us “how much” but does not include any direction.

Vectors are quantities that have both a magnitude and a direction. For example, we might say that a car is travelling at \(60\textrm{km/hr}\) to the north, or that someone has moved \(5\textrm{m}\) to the right. These are vector quantities because each one gives us both a magnitude and a direction.

This page explains how to do different calculations that involve vectors.

In brief, it covers the following:

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Vector diagrams

A vector can be shown in a diagram as an arrow between two points. The length of the arrow shows the magnitude of the vector, and the direction the arrow points shows which way the vector is going.

We can label a vector in two ways. The first way is to use a single lower case letter, which is usually written in bold. The second way is to use the names of the two points at each end of the arrow, with a small arrow written above them.

For example, the vector shown below in green goes from A to B. Point A is called the initial point, and point B is called the terminal point. This vector could be labelled as \(\mathbf{a}\) or as \(\overrightarrow{AB}.\)

A vector represented as an arrow and labelled A B, with the direction and magnitude labelled.

If two vectors have the same magnitude and direction they are called equivalent or equal vectors.

We can add vectors together on a diagram using the nose to tail method (also known as the triangle law).

The nose to tail method says that if two vectors are positioned so that the initial point of one is at the terminal point of another, then we can add the two vectors together to get the displacement vector.

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Example: Suppose a person has walked from point A to point B, then changed direction and walked from point B to another point, C:

Two vectors, one going from point A to point B and another going from point B to point C.

If we label these vectors \(\mathbf{u}\) and \(\mathbf{v}\), where \(\mathbf{u} = \overrightarrow{AB}\) and \(\mathbf{v} = \overrightarrow{BC},\) what is the displacement vector indicating the distance and direction the person is from their starting point?

Two vectors, one going from point A to point B and labelled u, and another going from point B to point C, labelled v.

Solution: Since the initial point of \(\mathbf{v}\) is at the terminal point of \(\mathbf{u}\), our displacement vector will be \(\mathbf{u} + \mathbf{v}\)

Two vectors, one going from point A to point B and labelled u, and another going from point B to point C, labelled v.

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Subtracting vectors on a diagram is done in a similar way. All we need to do is reverse the direction of the vector we want to subtract, and then add as normal. This however means that the addition becomes a subtraction since we would have \(\mathbf{u} + (-\mathbf{v})\) which is the same as \(\mathbf{u} - \mathbf{v}\).

Two vectors, one going from point A to point B and labelled u, and another going from point B to point C, labelled v.

Components

It can be difficult to show vectors accurately in diagrams. Because of this, vectors are usually written using coordinates when we need to perform calculations.

One way to do this is to split a vector into parts. These parts are the vector’s \(x\) and \(y\) components, which are also known as Cartesian coordinates. (For more information about Cartesian coordinates, and about using polar coordinates instead, see converting between coordinates.)

A vector’s \(x\) component tells us how many units ‘long’ the vector is in relation to the \(x\)-axis. A vector’s \(y\) component tells us how many units ‘high’ the vector is in relation to the \(y\)-axis.

If we have vector \(\mathbf{v}\), its components can be written as \(\mathbf{v_x}\) and \(\mathbf{v_y}\). These components are then expressed in the form \(\mathbf{v}=(\mathbf{v}_x, \mathbf{v}_y),\) similar to the coordinates of a graph, or as \(\begin{pmatrix} \mathbf{v}_x \\ \mathbf{v}_y \end{pmatrix}\)

An x-axis and a y-axis with a vector labelled v placed on a blank graph. The x and y components are marked.

Example:

What are the components of the vectors shown below?

Two vectors labelled a and b on numbered graph paper.

Solution:

Vector \(\mathbf{a}\) begins at \((0,0)\) and ends at \((2,4)\) therefore we have \(\mathbf{a} = (2,4)\)

Vector \(\mathbf{b}\) begins at \((1,3)\) and ends at \((4,8)\). To find the components of \(\mathbf{b}\), we first have to find the difference between the coordinates of these points:

\[\begin{aligned} \mathbf{b}_x &= 4 - 1 \\\ &= 3 \\\ \mathbf{b}_y &= 8 - 3 \\\ &= 5 \\\ \end {aligned}\]

Therefore, the components for \(\mathbf{b}\) will be \(\mathbf{b} = (3,5)\)

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Adding and subtracting vectors

Once vectors are written using components, we can add or subtract them quite easily. To do this, we add or subtract the matching \(x\) components, and then we add or subtract the matching \(y\) components. The results give us the \(x\) and \(y\) components of the new vector.

Example:

Add the two vectors below to find the displacement vector:

Two vectors, u=(8,5) and v=(6, -2) sitting nose to tail.

Solution:

The displacement vector’s components will be:

\[\begin{aligned} \mathbf{u} + \mathbf{v} &= (8+6,\,5+(-2)) \\\ &=(14,\,3) \\\ \end{aligned}\]

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Check your understanding of adding and subtracting vectors by working through the questions below:

1. Find \(\mathbf{a} + \mathbf{b}\) given that vector \(\mathbf{a} = (3,\,5)\) and vector \(\mathbf{b} = (4, -7)\)

\(\begin{aligned} \mathbf{a} + \mathbf{b} &= (3 + 4,\,5+(-7))\\\ &= (7, -2) \\\ \end{aligned}\)

2. Find \(\mathbf{c} - \mathbf{d}\) given that vector \(\mathbf{c} = (12, -3)\) and vector \(\mathbf{d} = (7,\,2)\)

\(\begin{aligned} \mathbf{c} - \mathbf{d} &= (12 - 7, -3 - 2)\\\ &= (5, -5) \\\ \end{aligned}\)

3. Find \(\mathbf{y} - \mathbf{x} + \mathbf{z}\) given that vector \(\mathbf{x} = (2,\,3),\) vector \(\mathbf{y} = (8, -3)\) and vector \(\mathbf{z} = (7, 4)\)

\(\begin{aligned} \mathbf{y} - \mathbf{x} + \mathbf{z} &= (8 - 2 + 7, -3 - 3 + 4)\\\ &= (13, -2) \\\ \end{aligned}\)

Magnitude of a vector

Vectors have both a magnitude and a direction. If we have a vector \(\mathbf{u}\), the magnitude of that vector is written as \(\vert \mathbf{u} \vert,\) or sometimes as \(\Vert \mathbf{u} \Vert\) so that it isn’t confused with the symbol for absolute value.

The magnitude of a vector is different to the components of a vector, as the magnitude is the numeric value of the vector. However, we can find the magnitude of a vector given its components using Pythagoras’ theorem.

The magnitude of a vector \(\mathbf{u} = (\mathbf{u}_x , \mathbf{u}_y)\) will be:

\[\vert \mathbf{u} \vert = \sqrt{\mathbf{u_x}^2 + \mathbf{u_y}^2}\]

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Example:

Find the magnitude of \(\mathbf{v} = (6,\,8)\)

Solution:

\[\begin{aligned} \vert \mathbf{v} \vert &= \sqrt{\mathbf{v_x}^2 + \mathbf{v_y}^2} \\\ &= \sqrt{6^2 + 8^2} \\\ &= \sqrt{100} \\\ &= 10 \\\ \end{aligned}\]

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Unit vectors

If a vector has a magnitude of \(1\) it is called a unit vector. There are two commonly known unit vectors that are often used in vector notation. The unit vectors are \(\mathbf{i}\) and \(\mathbf{j}\) and they sit along the \(x\) and \(y\) axes, respectively:

The two unit vectors i and j.

Vectors can also be represented using these unit vectors, for example as \(\mathbf{v} = x\mathbf{i} + y\mathbf{j}\)

Check your understanding of the magnitude of vectors and unit vector notation by working through the questions below:

1. Find the magnitude of \(\mathbf{a} = (4,-3)\)

\(\begin{aligned} \vert \mathbf{a} \vert &= \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2} \\\ &= \sqrt{4^2 + (-3)^2} \\\ &= \sqrt{25} \\\ &= 5 \\\ \end{aligned}\)

2. If \(\mathbf{b} = (-7,\,5),\) find \(\vert \mathbf{b} \vert\) correct to three decimal places.

\(\begin{aligned} \vert \mathbf{b} \vert &= \sqrt{\mathbf{b_x}^2 + \mathbf{b_y}^2} \\\ &= \sqrt{(-7)^2 + 5^2} \\\ &= \sqrt{74} \\\ &= 8.602 \\\ \end{aligned}\)

3. If \(\mathbf{c} = 4\mathbf{i} - 2\mathbf{j},\) find \(\vert \mathbf{c} \vert\) correct to two decimal places.

\(\begin{aligned} \vert \mathbf{c} \vert &= \sqrt{\mathbf{c_x}^2 + \mathbf{c_y}^2} \\\ &= \sqrt{4^2 + (-2)^2} \\\ &= \sqrt{20} \\\ &= 4.47 \\\ \end{aligned}\)

Scalar multiples

Vectors can be multiplied by scalar quantities in order to scale them up or down. To multiply a vector by a scalar, we multiply each of the vector’s components by the scalar.

If we have a vector \(\mathbf{u}\), the scaled vector will be \(k\mathbf{u}\) where \(k\) is the scalar and \(\mathbf{u}\) is the vector.

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Example 1:

Multiply the vector \(\mathbf{f} = (9,3)\) by \(4\)

Solution:

\[\begin{aligned} 4 \mathbf{f} &= (4 \times 9 ,\,4 \times 3) \\\ &= (36, 12) \\\ \end{aligned}\]

Example 2:

Calculate \(2.5 \mathbf{u}\) when \(\mathbf{u} = (5, 8)\)

Solution:

\[\begin{aligned} 2.5 \mathbf{u} &= (2.5 \times 5,\,2.5 \times 8) \\\ &= (12.5, 20) \\\ \end{aligned}\]

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Check your understanding of calculating scalar multiples by working through the questions below:

1. Multiply the vector \(\mathbf{a} = (7,-6)\) by \(3\)

\(\begin{aligned} 3 \mathbf{a} &= (3 \times 7,\,3 \times -6) \\\ &= (21, -18) \\\ \end{aligned}\)

2. If \(\mathbf{b} = -3 \mathbf{i} + 4 \mathbf{j},\) find \(7 \mathbf{b}\)

\(\begin{aligned} 7 \mathbf{b} &= (7 \times -3) \mathbf{i} + (7 \times 4) \mathbf{j} \\\ &= -21 \mathbf{i} + 28 \mathbf{j} \\\ \end{aligned}\)

3. If \(\mathbf{c} = 12 \mathbf{i} + 3 \mathbf{j}\) and \(\mathbf{d} = -3 \mathbf{i} + 5 \mathbf{j},\) find \(3 \mathbf{c} - 4 \mathbf{d}\)

\(\begin{aligned} 3 \mathbf{c} - 4 \mathbf{d} &= (3 \times 12 - 4 \times (-3)) \mathbf{i} + (3 \times 3 - 4 \times 5) \mathbf{j} \\\ &= (36 + 12) \mathbf{i} + (9 - 20) \mathbf{j} \\\ &= 48 \mathbf{i} - 11 \mathbf{j} \\\ \end{aligned}\)

Converting between coordinates

So far, we have used Cartesian coordinates to write vectors in terms of their \(x\) and \(y\) components. However, vectors can also be written using polar coordinates. When we use polar coordinates, we state the magnitude of the vector and its direction instead of its components.

The direction is given as an angle. This angle can be measured in degrees or in radians. We will use degrees on this page. For example, we could write a vector \(\mathbf{v}\) as \((12, 30^\circ)\) using polar coordinates.

To convert between Cartesian and polar coordinates, we need to use trigonometry. To convert from Cartesian coordinates to polar coordinates, we use two steps. First, we use Pythagoras’ theorem to calculate the magnitude. Second, we use the tangent trigonometric ratio to calculate the direction.

To write the vector \(\mathbf{a} = (\mathbf{a}_x, \mathbf{a}_y)\) using polar coordinates \((r, \theta)\) we calculate:

\[r = \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2}\] \[\theta = \tan^{-1}\frac{a_y}{a_x}\]

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Example:

Write \(\mathbf{a} = (10, 4)\) using polar coordinates rounded to two decimal places.

Solution:

\[\begin{aligned} r &= \sqrt{\mathbf{a_x}^2 + \mathbf{a_y}^2} \\\ &= \sqrt{10^2 + 4^2} \\\ &= \sqrt{116} \\\ &= 10.77 \\\ \end{aligned}\] \[\begin{aligned} \theta &= \tan^{-1}\frac{a_y}{a_x} \\\ &= \tan^{-1}\frac{4}{10} \\\ &= 21.80^\circ \end{aligned}\]

Therefore \(\mathbf{a} = (10.77,\,21.80^\circ)\)

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To convert from polar coordinates to Cartesian coordinates we use the sine and cosine trigonometric ratios.

To write the vector \(\mathbf{b} = (r, \theta)\) using Cartesian coordinates \((\mathbf{b}_x, \mathbf{b}_y)\) we calculate:

\[\mathbf{b}_x = r\,cos\,\theta\] \[\mathbf{b}_y = r\,sin\,\theta\]

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Example:

Write \(\mathbf{b} = (25, 30^\circ)\) using Cartesian coordinates rounded to two decimal places.

Solution:

\[\begin{aligned} \mathbf{b}_x &= r\,cos\,\theta \\\ &= 25\,cos\,30^\circ \\\ &= 21.65 \\\ \end{aligned}\] \[\begin{aligned} \mathbf{b}_y &= r\,sin\,\theta \\\ &= 25\,sin\,30^\circ \\\ &= 12.5 \\\ \end{aligned}\]

Therefore \(\mathbf{b} = (21.65, 12.5)\)

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Check your understanding of how to convert between coordinates by working through the questions below:

1. Write \(\mathbf{u} = (12, 3)\) using polar coordinates rounded to two decimal places.

\[\begin{aligned} r &= \sqrt{\mathbf{u_x}^2 + \mathbf{u_y}^2} \\\ &= \sqrt{12^2 + 3^2} \\\ &= \sqrt{144 + 9} \\\ &= \sqrt{153} \\\ &= 12.37 \\\ \end{aligned}\] \[\begin{aligned} \theta &= \tan^{-1}\frac{a_y}{a_x} \\\ &= \tan^{-1}\frac{3}{12} \\\ &= 14.04^\circ \\\ \end{aligned}\]

Therefore \(\mathbf{u} = (12.37,\,14.04^\circ)\)

2. Write \(\mathbf{v} = (14, 63^\circ)\) using Cartesian coordinates rounded to two decimal places.

\[\begin{aligned} \mathbf{v}_x &= r\,cos\,\theta\\\ &= 14\,cos\,63^\circ \\\ &= 6.36 \\\ \end{aligned}\] \[\begin{aligned} \mathbf{v}_y &= r\,sin\,\theta\\\ &= 14\,sin\,63^\circ \\\ &= 12.47 \\\ \end{aligned}\]

Therefore \(\mathbf{v} = (6.36,\,12.47)\)

3. A vector has magnitude \(3\textrm{m}\) and direction \(145^\circ.\) What are its \(x\) and \(y\) components rounded to two decimal places?

\[\begin{aligned} x\textrm{ component} &= r\,cos\,\theta \\\ &= 3\,cos\,145 \\\ &= -2.46\textrm{m} \end{aligned}\] \[\begin{aligned} y\textrm{ component} &= r\,sin\,\theta \\\ &= 3\,sin\,145 \\\ &= 1.72\textrm{m} \end{aligned}\]

Therefore the vector can be expressed as \((-2.46\textrm{m},\,1.72\textrm{m})\)

Multiplying vectors

If we want to multiply two vectors together, instead of just multiplying a vector by a scalar, there are two different methods we can use. The first method is called the dot product, and it gives an answer that is a scalar. The second method is called the cross product, and it gives an answer that is a vector.

The cross product is not covered on this page. However, how to calculate the dot product is explained below.

Dot product

The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is written as:

\[\mathbf{a} . \mathbf{b}\]

The dot product can be calculated in two different ways. The method we use depends on whether the vectors are written using Cartesian coordinates or polar coordinates.

If the vectors are written using Cartesian coordinates, we follow three steps. First, we multiply the two \(x\) components together. Second, we multiply the two \(y\) components together. Third, we add the two results.

The dot product of two vectors written using Cartesian coordinates is:

\[\mathbf{a} . \mathbf{b} = \mathbf{a}_x \times \mathbf{b}_x + \mathbf{a}_y \times \mathbf{b}_y\]

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Example:

Calculate \(\mathbf{a} . \mathbf{b}\) given that \(\mathbf{a} = (-2, 4)\) and \(\mathbf{a} = (3, 5)\)

Solution:

\[\begin{aligned} \mathbf{a} . \mathbf{b} &= \mathbf{a}_x \times \mathbf{b}_x + \mathbf{a}_y \times \mathbf{b}_y \\\ &= -2 \times 3 + 4 \times 5 \\\ &= -6 + 20 \\\ &= 14 \\\ \end{aligned}\]

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If the vectors are written using polar coordinates, we multiply the magnitude of each vector together and then multiply by the cosine of the angle between them.

The dot product of two vectors written using polar coordinates is:

\[\mathbf{a} . \mathbf{b} = \vert \mathbf{a} \vert \times \vert \mathbf{b} \vert \times cos\,\theta\]

where \(\theta\) is the angle between the two vectors.

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Example:

Calculate \(\mathbf{a} . \mathbf{b}\) given that \(\mathbf{a} = (4, 45^\circ)\) and \(\mathbf{b} = (7, 105^\circ)\)

Solution:

The angle between the two vectors is \(105^\circ - 45^\circ = 60^\circ,\) so:

\[\begin{aligned} \mathbf{a} . \mathbf{b} &= \vert \mathbf{a} \vert \times \vert \mathbf{b} \vert \times cos\,\theta \\\ &= 4 \times 7 \times cos\,60^\circ \\\ &= 28 \times 0.5 \\\ &= 14 \\\ \end{aligned}\]

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Check your understanding of using the dot product to multiply vectors by working through the questions below:

1. Calculate \(\mathbf{u} . \mathbf{v},\) rounded to two decimal places, given that \(\mathbf{u} = (3, 75^\circ)\) and \(\mathbf{v} = (8, 22^\circ)\)

The angle between the two vectors is \(75^\circ - 22^\circ = 53^\circ,\) so:

\[\begin{aligned} \mathbf{u} . \mathbf{v} &= \vert \mathbf{u} \vert \times \vert \mathbf{v} \vert \times cos\,\theta \\\ &= 3 \times 8 \times cos\,53^\circ \\\ &= 24 \times 0.602 \\\ &= 14.44 \\\ \end{aligned}\]

2. Calculate \(\mathbf{c} . \mathbf{d},\) rounded to two decimal places, given that \(\mathbf{c} = (6,\,7)\) and \(\mathbf{d} = (9, -3)\)

\[\begin{aligned} \mathbf{c} . \mathbf{d} &= \mathbf{c}_x \times \mathbf{d}_x + \mathbf{c}_y \times \mathbf{d}_y \\\ &= 6 \times 9 + 7 \times -3 \\\ &= 54 - 21 \\\ &= 33 \\\ \end{aligned}\]

3. Given your answer to question 2 above, determine the angle between vectors \(\mathbf{c}\) and \(\mathbf{d}\) rounded to two decimal places.

From question 2 above, we have: \(\mathbf{c} . \mathbf{d} = 33\)

Furthermore, calculating the magnitude of the vectors gives:

\[\begin{aligned} \vert \mathbf{c} \vert &= \sqrt{\mathbf{c_x}^2 + \mathbf{c_y}^2} \\\ &= \sqrt{6^2 + 7^2} \\\ &= \sqrt{85} \\\ &= 9.22 \\\ \end{aligned}\] \[\begin{aligned} \vert \mathbf{d} \vert &= \sqrt{\mathbf{d_x}^2 + \mathbf{d_y}^2} \\\ &= \sqrt{9^2 + (-3)^2} \\\ &= \sqrt{90} \\\ &= 9.49 \\\ \end{aligned}\]

So we have:

\[\begin{aligned} \vert \mathbf{u} \vert \times \vert \mathbf{v} \vert \times cos\,\theta &= 33\\\ \therefore 9.22 \times 9.49 \times cos\,\theta &= 33 \\\ \therefore cos\,\theta &= 0.38 \\\ \therefore \theta &= cos^{-1}0.38 \\\ \therefore \theta &= 67.83^\circ \\\ \end{aligned}\]

So the angle between the two vectors is \(67.83^\circ\)

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