Definite Integration – A Level Maths Revision

Definite Integration - A Level Maths Revision

Definite integration – what’s it all about, you might wonder? Well, you’re on an A Level Maths revision page so I guess you’re no casual mathematician (or are you?).

This post will look at the specifics of the topic, with example content, step-by-step solutions and even some practice questions chucked in for good measure.

So, what exactly is it all about?…

Let’s Get Started

Integration is one of the bigger new topics you will see at A Level.

Integration can roughly be divided into indefinite integration (where your integral includes a constant of integration) and definite integration (where your integral can be expressed as a value).

If you’re not comfortable with indefinite integration, check out this revision blog first.

  • To see the content of this revision blog in PDF and PowerPoint formats, click here.
  • If you’d like to practice some of the prior knowledge required for this topic, try this multiple-choice quiz.
  • If you’re happy you’ve fully grasped the topic, try this activity, or, if you think you’re ready, try these exam-style questions on all integration topics.

Integration can be considered the reverse of differentiation. If integrating π‘₯𝑛, with respect to π‘₯:

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

In other words, you can find the integral of a power term by adding one to the power and dividing by the new power.

You must then add the constant of integration (C) to take into account any constant terms which would have been removed when differentiating.

This is called indefinite integration because, without knowing the value of the constant of integration, you cannot calculate the exact value of the integral.

However, you can also calculate the value of an integral between two limits. You can do this by evaluating the value of the integral at the upper limit then subtracting the value of the integral at the lower limit.

Consider the integral below; 𝑏 is the upper limit and π‘Ž is the lower limit:

\int^b_a{x^n}\,dx

Note that the lower limit is placed at the bottom of the integral sign and the upper limit is placed at the top. Integrate, as normal, writing the integral in square brackets with the limits after the closing bracket:

\left[\frac{x^{n\,+\,1}}{n\,+\,1}+C\right]^b_a

Next, substitute each of the limits into the integral and find the difference:

\left(\frac{b^{n\,+\,1}}{n\,+\,1}+C\right)-\left(\frac{a^{n\,+\,1}}{n\,+\,1}+C\right) = \left(\frac{b^{n\,+\,1}}{n\,+\,1}\right) - \left(\frac{a^{n\,+\,1}}{n\,+\,1}\right)

As you can see, by subtracting C from C you remove the constants of integration. Consequently, you can omit C when integrating with limits.

This is called definite integration and allows you to evaluate an integral between two limits.

This method can be generalised by considering a function f(π‘₯), with derivative f’(π‘₯), over an interval [π‘Ž, 𝑏]. We can then write a definite integral as:

\int_a^b{f'(x)}\,dx=\left[f(x)\right]^b_a=f(b)-f(a)

This is called the Fundamental Theorem of Calculus.


Example Question 1

Evaluate \int^1_0{x^2}\,dx

To integrate a power term, add one to the power, then divide the whole term by that new number.

\begin{aligned}\int^1_0{x^2}\,dx = \left[\frac{x^3}{3}\right]^1_0 \end{aligned}

These square brackets are shorthand to indicate that you need to evaluate the expression between these bounds. To do this, substitute each bound, then find the difference between the expressions (subtracting the lower bound):

\begin{aligned} \left[\frac{x^3}{3}\right]^1_0 &= \left(\frac{1^3}{3}\right)-\left(\frac{0^3}{3}\right)\\&=\frac{1}{3} \end{aligned}


Example Question 2

Given that:
\int_9^2{f'(x)}\,dx = 15
Show that:
\int_2^9{f'(x)}\,dx = -15

Using the fundamental theorem of calculus:

\begin{aligned}&\int^2_9{f'(x)}\,dx=f(2)-f(9) \\ &f(2)-f(9)=15\end{aligned}

Likewise:

\begin{aligned}\int^9_2{f'(x)}\,dx&=f(9)-f(2) \\ &=-(f(2)-f(9))\\&=-15 \end{aligned}

This example shows an interesting property of the definite integral that can be generalised – if you switch the bounds, the sign of the final integral will change.

\int^b_a{f(x)}\,dx=-\int^a_b{f(x)}\,dx 

Remember that there are some rules for integration that can help make the integral easier to work with. These rules apply to definite integrals too:

For integrable function f(π‘₯) and g(π‘₯) and real constants π‘Ž, 𝑏 and 𝑐:
\int_a^b{(f(x) + g(x))}\,dx = \int^b_a{f(x)}\,dx+\int_a^b{g(x)}\,dx 
\int^b_a{cf(x)}\,dx = c\int^b_a{f(x)}\,dx 

This means you can integrate each term separately before substituting the limits.


Example Question 3

Evaluate \int_{-2}^4{(2x+5x^2)}\,dx

\begin{aligned}\int_{-2}^4{(2x+5x^2)}\,dx &= \left[ x^2+\frac{5x^3}{3}\right]^4_{-2}\\&=\left(4^2+\frac{5(4)^3}{3}\right)-\left((-2)^2+\frac{5(-2)^3}{3}\right) \\ &= \frac{368}{3} - -\frac{28}{3} \\&=132 \end{aligned}

Some calculators can evaluate definite integrals using this button:

Evaluate definite integration using this button

For example, to evaluate the integral in this example, start by pressing the button above.

Then input the expression you are integrating, in this case, 2π‘₯ + 5π‘₯2 (make sure you use the β€˜π‘₯’ button to input the variable).

Next, input the limits. Pressing the right arrow will navigate to the lower limit – here, input -2. Pressing the right arrow again will take you to the upper limit – here, input 4.

Finally, press β€˜=’ to get the answer of 132.


Example Question 4

Evaluate \int^4_1{\sqrt{x}}\,dx . Use your calculator to check your answer.

First, write the equation in index form: \sqrt{x} = x^{\frac{1}{2}}

Next, integrate by adding one to the power and dividing by the new power:

\begin{aligned} \int^4_1{x^{\frac{1}{2}}}\,dx &= \left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]^4_1\\&=\left[\frac{2x^{\frac{3}{2}}}{3}\right]^4_1\\&=\left(\frac{2(4)^{\frac{3}{2}}}{3}\right)-\left(\frac{2(1)^{\frac{3}{2}}}{3}\right)\\&=\frac{16}{3}-\frac{2}{3}\\&=\frac{14}{3} \end{aligned}

When checking this on your calculator, input the function in its original rather than index form. That way, you are checking each stage of your calculation.

Exam questions with a simple integration will often be phrased in a way that requires you to show your working; that way, you can’t simply input into your calculator. However, you should always check your answer using your calculator.


Definite Integration – Practice Questions

1. Evaluate the following integrals without using your calculator.

a. \int^3_1{x^4}\,dx

b. \int^2_{-1}{3x^2}\,dx

c. \int^9_4{\sqrt{x}}\,dx

d. \int^{-2}_{-5}{\frac{6}{x^2}}\,dx

2. Evaluate the following integrals. Check your answers on your calculator.

a. \int^6_2{(3x^3-4x)}\,dx

b. \int^3_{1}{\left(\frac{2}{x^3}+3\right)}\,dx

c. \int^0_{-4}{\frac{2x^2+3x}{x}}\,dx

d. \int^9_4{3x-2\sqrt{x}}\,dx

3. Given that \int^b_1{x^{-\frac{1}{2}}}\,dx =8 find the value of 𝑏.

4. Given that 𝑛 > 0, find the value of 𝑛 such that \int^{3n}_n{(4x-1)}\,dx = \frac{1}{2}

5. Given that 𝑝 > 0, find the value of 𝑝 such that \int^p_0{\sqrt{x}}\,dx=2p

Answers

1. Evaluate the following integrals without using your calculator.

a. \int^3_1{x^4}\,dx

\boldsymbol{\begin{aligned}\left[\frac{x^5}{5}\right]^3_1&=\left(\frac{3^5}{5}\right)-\left(\frac{1^5}{5}\right)\\&=\frac{242}{5}\end{aligned}}

b. \int^2_{-1}{3x^2}\,dx

\boldsymbol{\begin{aligned}[x^3]^2_{-1}&=(2^3)-((-1)^3)\\&=9\end{aligned}}

c. \int^9_4{\sqrt{x}}\,dx

\boldsymbol{\begin{aligned}\int^9_4{x^{\frac{1}{2}}}\,dx&=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]^9_4\\&=\left[\frac{2x^{\frac{3}{2}}}{3}\right]^9_4\\&=\left(\frac{2\times 9^{\frac{3}{2}}}{3}\right)-\left(\frac{2\times 4^{\frac{3}{2}}}{3}\right)\\&=\frac{38}{3}\end{aligned}}

d. \int^{-2}_{-5}{\frac{6}{x^2}}\,dx

\boldsymbol{\begin{aligned}\int^{-2}_{-5}{6x^{-2}}\,dx&=\left[-6x^{-1}\right]^{-2}_{-5}\\&=(-6\times(-2)^{-1})-(-6\times (-5)^{-1})\\&=\frac{9}{5}\end{aligned}}

2. Evaluate the following integrals. Check your answers on your calculator.

a. \int^6_2{(3x^3-4x)}\,dx

\boldsymbol{\begin{aligned}\left[\frac{3x^4}{4}-2x^2\right]^6_2&=\left(\frac{3\times6^4}{4}-2\times 6^2\right)-\left(\frac{3\times2^4}{4}-2\times 2^2 \right) \\ &=896\end{aligned}}

b. \int^3_{1}{\left(\frac{2}{x^3}+3\right)}\,dx

\boldsymbol{\begin{aligned}\left[-x^{-2}+3x\right]^3_{1}&=(-3^{-2}+3 \times 3)-(-(1)^{-2}+3 \times (1))\\&=\frac{62}{9}\end{aligned}}

c. \int^0_{-4}{\frac{2x^2+3x}{x}}\,dx

\boldsymbol{\begin{aligned}\int^0_{-4}{(2x+3)}\,dx&=[x^2+3x]^0_{-4}\\&=(0^2+3\times0)-((-4)^2+3\times(-4))\\&=-4\end{aligned}}

d. \int^9_4{3x-2\sqrt{x}}\,dx

\boldsymbol{\begin{aligned}\int^9_4{\left(3x-2x^{\frac{1}{2}}\right)}\,dx&=\left[\frac{3x^2}{2}-\frac{4x^{\frac{3}{2}}}{3}\right]^9_4\\&=\left(\frac{3\times9^2}{2}-\frac{4\times 9^{\frac{3}{2}}}{3}\right)-\left(\frac{3\times 4^2}{2}-\frac{4\times 4^{\frac{3}{2}}}{3}\right)\\&=\frac{433}{6}\end{aligned}}

3. Given that \int^b_1{x^{-\frac{1}{2}}}\,dx =8 find the value of 𝑏.

\boldsymbol{\begin{aligned}&\left[2x^{\frac{1}{2}}\right]^b_1=8\\&2b^{\frac{1}{2}}-2=8\\&2b^{\frac{1}{2}}=10\\&b=25\end{aligned}}

4. Given that 𝑛 > 0, find the value of 𝑛 such that \int^{3n}_n{(4x-1)}\,dx = \frac{1}{2}

\boldsymbol{\begin{aligned}\int^{3n}_n{(4x-1)}\,dx&=[2x^2-x]^{3n}_n\\&=(2(3n)^2-3n)-(2(n)^2-n)\\&=(18n^2-3n)-(2n^2-n)\\&=16n^2-2n\end{aligned}}

\boldsymbol{\begin{aligned}&16n^2-2n=\frac{1}{2}\\&16n^2-2n-\frac{1}{2}=0\\&n=\frac{1}{4} \text{ or } n=-\frac{1}{8} \text { (discard the negative value)}\\&n=\frac{1}{4}\end{aligned}}

5. Given that 𝑝 > 0, find the value of 𝑝 such that \int^p_0{\sqrt{x}}\,dx=2p

\boldsymbol{\begin{aligned}\int^p_0{\sqrt{x}}\,dx&=\left[\frac{2x^{\frac{3}{2}}}{3}\right]^p_0\\&=\left(\frac{2p^{\frac{3}{2}}}{3}\right)-\left(\frac{2(0)^{\frac{3}{2}}}{3}\right)\\&=\frac{2}{3}\sqrt{p^3}\end{aligned}}

\boldsymbol{\begin{aligned}&\frac{2}{3}\sqrt{p^3}=2p\\&\sqrt{p^3}=3p\\&p^3=9p^2\\&p^3-9p^2=0\\&p^2(p-9)=0\\&p-9=0\\&p=9\text{ (discarding }p=0\text{)}\end{aligned}}


Have you definitely finished all there is here on definite integration? Yes? Then head over to our revision post on indefinite integration or check out the rest of our blogs here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free KS5 Maths resources before you subscribe too.

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