# 1. Introduction

It is possible to determine a function ƒ (x) from its derivative ƒ (x) by calculating the anti-derivative or integral of ƒ (x), i.e.:

 if  dF/dx   =  ƒ (x),  then  ƒ (x)  =     ∫   ƒ (x) dx + C

- where C is an integration constant (see the module on Indefinite Integration).

In this module we will see how to use integration to calculate the area under a curve.

As a revision exercise, try this quiz on indefinite integration:

Quiz 1: Select the indefinite integral of (3x21/2x) dx with respect to x:

(a) 6x1/2 + C Incorrect - please try again!
(b) 3/2x3x2 + C Incorrect - please try again!
(c) x2 + 1/4x2 + C Incorrect - please try again!
(d) x31/4x2 + C Correct - well done!

Hint: If n ≠ −1, the integral of xn is xn + 1/(n + 1).

Explanation: To find the indefinite integral (3x21/2x) dx we use the sum rule for integrals, rewriting it as the sum of two integrals:

 ∫ (3x2 − 1/2x) dx = ∫ 3x2 dx − ∫ (− 1/2x) dx = 3 ∫ x2 dx − 1/2 ∫ x dx

Using xndx = 1/n+1 xn+1; n ≠ −1 with n = 2 in the first integral and n = 1 in the second gives:

 3 ∫ x2 dx − 1/2 ∫ x dx = 3 × 1/1 + 2 x1+2 − 1/2 × 1/1 + 1 x1+1 + C = 3/3 x3 − 1/2(1 + 1) x2 + C = x3 − 1/4 x2 + C

It can be checked that differentiation of this result gives 3x2 −  1/2x

# 2. Definite Integration

We define the definite integral of the function ƒ(x) with respect to x from x = a to x = b to be:

 b∫a ƒ (x) dx = ƒ (x)b a = ƒ (b) − ƒ (a)

- where ƒ (x) is the anti-derivative of ƒ(x). We call a and b the lower and upper limits of integration, respectively. The function being integrated, ƒ(x), is called the integrand. Note the minus sign!

Note: integration constants are not written in definite integrals since they always cancel in them:

 b∫a ƒ (x) dx = ƒ (x)b a = (ƒ (b) + C) − (ƒ (a) + C) = ƒ (b) − ƒ (a)

Example 1: Calculate the definite integral 21 x3dx

Solution: From the rule xndx = 1/n+1 xn+1 we have:

 2∫1 = 1/3 + 1 x3+1 2 1 = 1/4 x42 1 = 1/4 × 24 − 1/4 × 14 = 1/4 × 16 − 1/4 = 4 − 1/4 = 15/4

Exercise 1: Calculate the following definite integrals:

(a) 30 xdx

Solution: To calculate 30 xdx use the formula:

xndx = 1/n+1 xn+1

with n = 1. This yields:

 3∫0 x dx = 1/1 + 1 x1+1 3 0 = 1/2x2 3 0 = 1/2 × (3)2 − 1/2 × (0)2 = 1/2 × 9 − 0 = 9/2
(b) 2−1 xdx

Solution: To calculate 2−1 xdx use the formula for the definite integral:

xndx = 1/n+1 xn+1

with n = 1. This yields:

 2∫−1 x dx = 1/1 + 1 x1+1 2 −1 = 1/2 x2 2 −1 = 1/2 × (2)2 − 1/2 × (−1)2 = 1/2 × 4 − 1/2 × (+1) = 2 − 1/2 = 3/2
(c) 21 (x2x) dx

Solution: To evaluate the definite integral 21 (x2x) dx we rewrite it as the sum of two integrals and use xndx = 1/n+1 xn+1, with n = 2 in the first integral and n = 1 in the second one:

 2∫1 x2 dx − 2∫1 x dx = 1/3 x3 2 1 − 1/2 x2 2 1 = 1/3 23 − 1/3 13 − ( 1/2 22 − 1/2 12 ) = 1/3 × 8 − 1/3 × 1 − ( 1/2 × 4 − 1/2 × 1 ) = 7/3 − 3/2 = 14/6 − 9/6 = 5/6
(d) 2−1 (x2x) dx

Solution: To find the integral 2−1 (x2x) dx, we rewrite it as the sum of two integrals and use the result of the previous part to write is as:

 2∫−1x2 dx − 2∫−1x dx = 1/3 x3 2 −1 − 1/2 x2 2 −1 = 1/3 23 − 1/3 (−1)3 − (1/2 22 − 1/2 (−1)2) = 1/3 × 8 + 1/3 × 1 − ( 1/2 × 4 − 1/2 × 1 ) = 9/3 − 3/2 = 3 − 3/2 = 3/2

Click on questions to reveal their solution

# 3. The Area Under a Curve

 The definite integral of a function ƒ(x) which lies above the x-axis can be interpreted as the area under the curve of ƒ(x). Thus the area shaded blue in the figure on the right is given by the definite integral: b∫a ƒ (x) dx = ƒ (x)b a = ƒ (b) − ƒ (a) This is demonstrated as follows:
 Consider the area, A, under the curve y = ƒ (x)in the figure on the right. If we increase the value of x by δx, then the increase in area, δA, is approximately: δA = y δx → δA/δx = y Here we approximate the area of the thin strip by a rectangle of width δx and height y. In the limit as the strips become thin, δx → 0, this means: dA/dx = limδx → 0 δA/δx = y

The function (height of the curve) is the derivative of the area and the area below the curve is an anti-derivative or integral of the function.

Note: So far we have assumed that y = ƒ (x) lies above the x-axis.

 Example 2: Consider the integral 3∫0 x dx. The integrand y = x (a straight line) is shown in the figure on the right. The area A underneath the line is the blue shaded triangle. The area of any triangle is half its base times the height. For the blue shaded triangle, this is: A = 1/2 × 3 × 3 = 9/2 As expected, the integral yields the same result: 3∫0 x dx = x2/2 3 0 = 32/2  − 02/2  = 9/2 − 0 = 9/2

Here is a quiz on this relation between definite integrals and the area under a curve:

Quiz 2:

 Select the value of the definite integral 3∫1 2 dx, shown here: (a)6Incorrect - please try again! (b)2Incorrect - please try again! (c)4Correct - well done! (d)8Incorrect - please try again! Hint: 2 may be written as 2x0, since x0 = 1.

Explanation: Using 2 = 2x0, the integral is:

 3∫12 dx = 2 3∫1 x0 dx = 2x3 1 = 2 × 3 − 2 × 1 = 6 − 2 = 4

This can be confirmed by inspecting the figure. The area under the curve between the integration limits is the area of a square of side 2, which has area 22 = 4.

 Example 3: Consider the two lines: y = 3 and y = −3. Let us integrate these functions in turn from x = 0 to x = 2. For y = +3: 2∫0 (+3) dx = 3x2 0 = 3 × 2 − 3 × 0 = 6 - and 6 is indeed the area of the rectangle of height 3 and length 2. However, for y = −3: 2∫0 (−3) dx = −3x2 0 = −3 × 2 − (−3 × 0) = −6

Although both rectangles have the same area, the sign of this result is negative because the curve y = −3 lies below the x-axis. This indicates the sign convention:

If a function lies below the x-axis, its integral is negative
If a function lies above the x-axis, its integral is positive
 Exercise 2: From the figure on the right, what can you say about the signs of the following definite integrals? (a) B∫A y1 dx Solution: The sign of the definite integral, B∫A y1 dx, must be negative. This is because the function y1(x) is negative for all values of x between A and B. The area is all below the x-axis. (b) D∫B y1 dx Solution: The sign of the definite integral, D∫B y1 dx, must be positive. This is because, between the integration limits B and D, there is more area above the x-axis than below it. (c) O∫A y2 dx Solution: The sign of the definite integral, O∫A y2 dx, must be positive. This is because, between the integration limits A and O, there is more area above the x-axis than below it. (d) D∫C y2 dx Solution: The sign of the definite integral, D∫C y2 dx, must be negative. This is because, between the integration limits C and D, the integrand y2(x) is always negative. Click on questions to reveal their solution

Example 4: To calculate −2−6 6x2dx, use axndx = a/n+1 xn+1. Thus:

 −2∫−6 6x2 dx = 6/2 + 1 x2+1 −2 −4 = 6/3 x3 −2 −4 = 2x3 −2 −4 = 2 × (−2)3 − 2 × (−4)3 = −16 + 128 = 112

Note: Even though the integration range is for negative x (from −4 to −2), the integrand, ƒ(x) = 6x2, is a positive function. The definite integral of a positive function is positive. Similarly, it is negative for a negative function.

Quiz 3: Select the definite integral of y = 5x4 with respect to x if the lower limit of the integral is x = −2 and the upper limit is x = −1:

(b)31Correct - well done!

Explanation: The definite integral of y = 5x4 with respect to x if the lower limit of the integral is x = −2 and the upper limit is x = −1 can be written as:

−1−2 5x4dx

From the basic result B Aaxndx = a/n+1 xn+1 B A we obtain:

 −1∫−2 5x4 dx = 5/5 x5 −1 −2 = (−1)5 − (−2)5 = −1 − (−32) = −1 + 32 = 31

Note: Since the integrand 5x4 is positive for all x, the negative suggested solutions could not be correct.

Exercise 3: Use the integrals listed below to calculate the following definite integrals:

 ƒ (x) ∫ ƒ (x) dx xn for n ≠ −1 sin (ax) cos (ax) eax 1/x 1/n + 1 xn+1 − 1/a cos (ax) 1/a sin (ax) 1/a eax ln (x)

(a)9 4 3 t dt

Solution: To calculate the definite integral 9 4 3 t dt we rewrite it as:

9 4 3 t dt = 3 × 9 4 t½ dt

and use xndx = 1/n+1 xn+1, with n = ½:

 3 × 9 ∫4 t½ dt = 1/½ + 1 t½+1 9 4 = 3 × 1/ 3 ⁄ 2 t3 ⁄ 2 9 4 = 3 × 2/ 3 t3 ⁄ 2 9 4 = 2t3 ⁄ 2 9 4 = 2 × (9)3 ⁄ 2 − 2 × (4)3 ⁄ 2 = 2 × (9½)3 − (4½)3 = 2 × (9)3 ⁄ 2 − 2 × (4)3 ⁄ 2 = 2 × 33 − 2 × (2)3 = 2 × 27 − 2 × 8 = 54 − 16 = 38

Note: Dividing by a fraction is equivalent to multiplying by its inverse (see the module on Fractions).

(b)1 −1 (x2 − 2x + 4) dx

Solution: To calculate the definite integral 1 −1 (x2 − 2x + 4) dx we rewrite it as a sum of integrals:
1 −1 x2dx − 2 × 1 −1 xdx + 4 × 1 −1dx, and use xndx = 1/n+1 xn+1, with n = 2 in the first integral:

1 −1 x2dx = 1/2 + 1 x2+1 1 −1 = 1/3 x3 1 −1 = 1/3 (13 − (−1)3) = 2/3,

with n = 1 in the second integral:

2 × 1 −1 xdx = 2 × 1/1 + 1 x1+1 1 −1 = x2 1 −1 = −(12 − (−1)2) = 0,

and with n = 0 in the second integral:

4 × 1 −1 1 dx = 4 × 1/0 + 1 x0+1 1 −1 = 4x1 −1 = 4(1 − (−1)) = 8

Summing up these numbers we obtain 2 ⁄ 3 + 0 + 8 = 26 ⁄ 3.

(c)π 0 sin (x) dx

Solution: To calculate the definite integral π0 sin (x) dx we note from the table that:

sin (x) dx = − 1/a cos (ax)

This yields (with a = 1):

 π∫0 sin (x) dx = − cos (x)π 0 = − (cos (π) − cos (0)) = − ((−1) − 1) = 2

Note: It is worth emphasizing that the angles in calculus formulae for trigonometric functions are measured in radians.

(d)3 0 4 e2xdx

Solution: To calculate the definite integral 30 4 e2xdx rewrite it as:

30 4 e2xdx = 4 × 30 e2xdx

and use from the table:

eaxdx = 1/a eax

This gives, for a = 2:

 4 × 3∫0 e2x dx = 4 × 1/2  e2x3 0 = 2  e2x3 0 = 2 e(2×3) − 2 e(2×0) = 2 e6 − 2 e0 = 2 e6 − 2 ×1 = 2 e6 − 2
(e)2 1 3/t dt

Solution: To evaluate the definite integral 2 1 3/t dt we write:

2 1 3/t dt = 3 ×  2 1 1/t dt

and use:

1/t dt = ln (t)

This yields:

 3 ×  2 ∫1 1/t dt = 3 × ln (t) 2 1 = 3 × ln (2) − 3 × ln (1) = 3 × ln (2) − 3 × 0 = 3 ln (2)

Note: ln (0) = 1, since e0 = 1.

(f)π ⁄ 2 π ⁄ 4 2 cos (4w)dw

Solution: To find the definite integral π ⁄ 2 π ⁄ 4 2 cos (4w)dw use

π ⁄ 2 π ⁄ 4 2 cos (4w)dw = 2 × π ⁄ 2 π ⁄ 4 cos (4w)dw

and

cos (ax) dx = 1/a sin (ax). This gives for a = 4:
 2 × π ⁄ 2 ∫π ⁄ 4 cos (4w)dw = 2 × 1/4 sin (4w) π ⁄ 2 π ⁄ 4 = 1/2 sin (4w) π ⁄ 2 π ⁄ 4 = 1/2 sin (4 × π/2) − 1/2 sin (4 × π/4) = 1/2 sin (2π) − 1/2 sin (π) = 1/2 × 0 − 1/2 × 0 = 0

Click on questions to reveal their solution

Quiz 4: Which is the correct result for the definite integral 2b ax2dx?

(a) 8/3b31/3a3 Correct - well done!
(b) 4b − 2a Incorrect - please try again!
(c) 8/3b3 + 1/3a3 Incorrect - please try again!
(d) 1/3b31/3a3 Incorrect - please try again!

Explanation: To caclulate the definite integral 2ba x2dx use the basic indefinite integral:

xndx = 1/n+1 xn+1

with n = 2. This gives:

 2b∫a x2 dx = 1/2 + 1 x(2+1) 2b a = 1/3x3 2b a = 1/3 × (2b)3 − 1/3 × 1(a)3 = 1/3 × (2)3 × b3 − 1/3 × a3 = 1/3 × 8 × b3 − 1/3 × a3 = 8/3b3 − 1/3a3

Quiz 5: Which is the correct answer for the definite integral 32 1/x2dx?

(a)−1Correct - well done!

Explanation: To evaluate the definite integral 32 1/x2dx = 32 x−2dx

use BA xndx = 1/n+1 xn+1 B A with n = −2:

 3∫2 x−2 dx = 1/−2 + 1 x(−2+1) 3 2 = 1/(−1)x−1 3 2 = (−1) × 1/x 3 2 = − 1/x 3 2 = − 1/3 − (1/2) = − 1/3 + 1/2 = − 2/6 + 3/6 = −2 + 3/6 =1/6

# 4. Quiz on Definite Integrals

Choose the solutions from the options given

1.What is the area under the curve of the following positive function y = 10x4 + 3x2 between x = −1 and x = 2?
(a)75
(b)53
(c)69
(d)57
2.What is the definite integral of 3 sin (2x) from x = 0 to x = π ⁄ 2?
(a)−3
(b)0
(c)3
(d)5 ⁄ 2
3.What is the (non-zero) value of b for which the definite integral b0 (2s − 3)ds vanishes?
(a)1
(b)5
(c)3
(d)2
4.What is the definite integral of 2−2 e2xdx?
(a) 2(e4 − e−4)
(b) 1/2 (e44e )
(c) 0
(d) 1/2 (e4 − e−4)