2. nite Integration
R Horan & M Lavelle
The aim of this package is to provide a short self
assessment programme for students who want to
be able to calculate basic de
3. nite integrals.
Copyright 2004 rhoran@plymouth.ac.uk , mlavelle@plymouth.ac.uk
c
Last Revision Date: September 9, 2005 Version 1.0
5. nite Integration
3. The Area Under a Curve
4. Final Quiz
Solutions to Exercises
Solutions to Quizzes
The full range of these packages and some instructions,
should they be required, can be obtained from our web
page Mathematics Support Materials.
6. Section 1: Introduction 3
1. Introduction
It is possible to determine a function F (x) from its derivative f (x) by
calculating the anti-derivative or integral of f (x), i.e.,
dF
if = f (x) ; then F (x) = f (x)dx + C
dx
where C is an integration constant (see the package on inde
7. nite
integration). In this package we will see how to use integration to
calculate the area under a curve.
As a revision exercise, try this quiz on inde
9. nite integral of (3x2 1 x)dx with respect to x
2
1 3 3
(a) 6x + C ; (b) x x2 + C ;
2 2
1 2 1
(c) x + x + C ;
2
(d) x3 x2 + C :
4 4
Hint: If n T= 1 ; the integral of x is x
n n+1 =(n + 1).
14. nite integral of the function f (x) with respect to
x from a to b to be
b b
f (x)dx = F (x) = F (b) F (a) ;
a a
where F (x) is the anti-derivative of f (x). We call a and b the lower
and upper limits of integration respectively. The function being inte-
grated, f (x), is called the integrand. Note the minus sign!
Note integration constants are not written in de
15. nite integrals since
they always cancel in them:
b b
f (x)dx = F (x)
a a
= (F (b) + C ) (F (a) + C )
= F (b) + C F (a) C
= F (b) F (a) :
18. nite integral 1
x3 dx.
‚ a
From the rule axn dx = xn+1 we have
n+1
2
1 3+1 2
3
x dx = x
1 3+1 1
2
1 1 1
= x4 = ¢ 24 ¢ 11
4 1 4 4
1
=
4
¢ 16 1 = 4 1 = 15 :
4 4 4
Exercise 1. Calculate the following de
19. nite integrals: (click on the
green letters for the solutions)
‚3 ‚2
(a) 0
xdx ; (b) 1 xdx ;
‚2 ‚2
(c) 1
(x2 x)dx ; (d) 1 (x
2
x)dx :
20. Section 3: The Area Under a Curve 6
3. The Area Under a Curve
The de
21. nite integral of a function f (x) which lies above the x axis
can be interpreted as the area under the curve of f (x).
Thus the area shaded blue below
y
y = f (x)
A
x
0 a b
is given by the de
22. nite integral
b b
f (x)dx = F (x)
= F (b) F (a) :
a a
This is demonstrated on the next page.
23. Section 3: The Area Under a Curve 7
Consider the area, A, under the curve, y = f (x). If we increase
the value of x by x, then the increase in area, A, is approximately
y
A = y x A A = y :
x
Here we approximate the y
area of the thin strip by
a rectangle of width x
and height y . In the limit A A
as the strips become thin, x x+x
x
x 3 0, this means:
0
dA A
= lim = y:
dx x 30 x
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.
N.B. so far we have assumed that y = f (x) lies above the x axis.
24. Section 3: The Area Under a Curve 8
‚3
Example 2 Consider the integral 0
xdx. The integrand y = x (a
straight line) is sketched below. The area 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
1 9
A= ¢3¢3= :
2 2
y
y =x
3
A
x
0 3
As expected, the integral yields the same result:
3 3
= 3 0 = 9 0= 9:
x2 2 2
xdx =
0 2 0 2 2 2 2
25. Section 3: The Area Under a Curve 9
Here is a quiz on this relation between de
26. nite integrals and the area
under a curve.
Quiz Select the value of the de
27. nite integral
3
2dx ;
1
which is sketched in the following diagram:
y
y =2
2
A
x
0 1 3
(a) 6 ; (b) 2 ; (c) 4 ; (d) 8 :
Hint: 2 may be written as 2x , since x = 1.
0 0
28. Section 3: The Area Under a Curve 10
Example 3 Consider the two lines: y = 3 and y = 3.
Let us integrate these functions in y 6
turn from x = 0 to x = 2. y = +3
3
a) For y = 3:
-
2
2
(+3)dx = 3x = 3¢2 3¢0 = 6 : 0 j j
0
0 1 2 x
and 6 is indeed the area of the rect- 3
angle of height 3 and length 2. y = 3
b) However, for y = 3:
2
2
( 3)dx = 3x
= 3¢2 ( 3¢0) = 6 :
0 0
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.
29. Section 3: The Area Under a Curve 11
Exercise 2.
y 6
y1 (x)
A B 0 C D
-
x
y2 (x)
From the diagram above, what can you say about the signs of the
following de
30. nite integrals? (Click on the green letters for the solu-
tions)
‚B ‚D
(a) A y1 (x)dx ; (b) B y1 (x)dx ;
‚0 ‚D
(c) A y2 (x)dx ; (d) C y2 (x)dx :
31. Section 3: The Area Under a Curve 12
‚ ‚ a
Example 4 To calculate 42 6x2 dx, use axn dx = xn+1 . Thus
n+1
2 2
6
6x2 dx = x2+1
4 2+1 4
6 2 2
= x3 = 2x3
3 4
4
= 2 ¢ ( 2)3 2 ¢ ( 4)3 = 16 + 128 = 112 :
Note that even though the integration range is for negative x (from
4 to 2), the integrand, f (x) = 6x2 , is a positive function. The
de
32. nite integral of a positive function is positive. (Similarly it is
negative for a negative function.)
Quiz Select the de
33. nite 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
(a) 31 ; (b) 31 ; (c) 29 ; (d) 27 :
34. Section 3: The Area Under a Curve 13
Exercise 3. Use the integrals listed below to calculate the following
de
35. nite integrals. (Click on the green letters for the solutions)
1
f (x) xn for n T= 1 sin(ax) cos(ax) eax
x
‚ 1 1 1 1
f (x)dx
n+1
xn+1 a cos(ax) a
sin(ax)
a
eax ln(x)
‚9 p ‚1
(a) 4
3 tdt ; (b) 1 (x
2
2x + 4)dx ;
‚ ‚3
(c) 0
sin(x)dx ; (d) 0
4e2x dx ;
‚2 3 ‚
(e) 1
dt ; (f) 2 2 cos(4w)dw :
t 4
36. Section 3: The Area Under a Curve 14
Quiz Find the correct result for the de
37. nite integral
2b
x2 dx :
a
8 3 1 3
(a) b a ; (b) 4b 2a ;
3 3
8 3 1 3 1 3 1 3
(c) b + a ; (d) b a :
3 3 3 3
Quiz Select the correct result for the de
38. nite integral
3
1
dx ;
2 x2
from the answers oered below
1 1 1
(a) 1 ; (b) ; (c) ; (d) :
5 36 6
39. Section 4: Final Quiz 15
4. Final Quiz
Begin Quiz 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 de
40. nite integral of 3 sin(2x) from x = 0 to x = =2 ?
5
(a) 3 ; (b) 0 ; (c) 3 ; (d) :
2
3. Find the (non-zero) value of b for which the de
41. nite integral
‚b
0
(2s 3)ds vanishes
(a) 1 ; (b) 5 ; (c) 3 ; (d) 2 :
‚
4. Select below the de
42. nite 2
integral ¡ 2 e2x dx with respect to x. ¡
¡
(a) 2 e4 e 4 ; (b) 1 e4 4
pe ; (c) 0 ; (d) 1 e4 e 4 :
2 2
End Quiz
43. Solutions to Exercises 16
Solutions to Exercises
‚
Exercise 1(a) To calculate 3
0
xdx,
use the formula
1 n+1
xn dx = x
n+1
with n = 1. This yields
3
1 1+1 3 1 2 3
= x
xdx = x
0 1+1 0
2 0
1
=
2
¢ (3)2 1 ¢ (0)2
2
1
=
2
¢9 0= 9: 2
Click on the green square to return
£
44. Solutions to Exercises 17
‚
Exercise 1(b) To calculate 1 xdx, use the formula for the inde
45. nite
2
integral
1
xn dx = xn+1
n+1
with n = 1. This yields
2
1 1+1 2 2
= 1 x2
xdx = x
1 1+1 1 2 1
1
=
2
¢ (2)2 1 ¢ ( 1)2
2
1 1
=
2
¢ 4 2 ¢ (+1)
1 3
= 2 = :
2 2
Click on the green square to return
£
50. nd the integral 1 (x2 x)dx we rewrite it as the
2
sum of two integrals and use the result of the previous part to write
it as
2 2 2
1 3 2
1 x2
x dx
2
x dx = x
1 1 3 1 2 1
1
=
3
¢ 23 1 ¢ ( 1)3 1 ¢ 22 1 ¢ ( 1)2
3 2 2
1
=
3
¢8+ 1 ¢1 1 ¢4 2 ¢1 = 9 2
3 2
1
3
3
3 3
= 3 = :
2 2
Click on the green square to return £
52. nite integral, A y1 (x)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.
Click on the green square to return
£
54. nite integral, B y1 (x)dx, must be positive. This
is because, between the integration limits B and D, there is more area
above the x axis than below the x axis.
Click on the green square to return
£
56. nite integral, A y2 (x)dx, must be positive. This
is because, between the integration limits A and 0, there is more area
above the x axis than below it.
Click on the green square to return £
58. nite integral, C y2 (x)dx, must be negative. This
is because, between the integration limits C and D, the integrand
y2 (x) is always negative.
Click on the green square to return £
62. nite integral 1 (x2 2x +4)dx we
1
‚1 2 ‚1 ‚1
rewrite it as a sum of integrals 1 x dx 2 ¢ 1 xdx + 4 ¢ 1 1dx
‚ 1 n+1
and use xn dx x= with n = 2 in the
63. rst integral,
n+1
1 1
1 2+1 1
= 1 x3 = 1 13 ( 1)3 = 2 ;
¡
x2 dx = x
1 2+1 1 3 1 3 3
with n = 1 in the second integral
1
1 1+1 1 1 ¡
2¢ xdx = 2 ¢ x = x2 = 12 ( 1)2 = 0 ;
1 1+1 1 1
and with n = 0 in the last integral
1
1 0+1 1 1
4¢ 1dx = 4 ¢ x = 4x = 4 (1 ( 1)) = 8 :
1 0+1 1 1
Summing up these numbers we obtain 2=3 + 0 + 8 = 26=3.
Click on the green square to return £
65. nite integral 0
sin(x)dx we note
from the table that
1
sin(ax)dx = cos(x):
a
This yields (with a = 1)
sin(x)dx = cos(x)
0 0
= (cos() cos(0)) = (( 1) 1) = 2 :
N.B. It is worth emphasizing that the angles in calculus formulae for
trigonometric functions are measured in radians.
Click on the green square to return
£
75. nite integral (3x2 1 x)dx we
2
use the sum rule for integrals, rewriting it as the sum of two integrals
1 1
(3x x) dx =
2
3x dx + ( x) dx
2
2
2
1
= 3 x dx 2
x dx :
2
‚
Using xn dx = n+1 xn+1 ; n T= 1 with n = 2 in the
76. rst integral and
1
with n = 1 in the second one gives
1 1 1+2 1 1
3 x dx
2
x dx = 3 ¢ x 2 ¢ 1 + 1 x1+1 + C
2 1+2
3 1 1
= x3 x2 + C = x3 x2 + C :
3 2(1 + 1) 4
Check that dierentiation of this result gives 3x 2 x.
2 1
End Quiz
77. Solutions to Quizzes 31
Solution to Quiz: Using 2 = 2x0 , the integral
3
3 3
2dx = 2 x0 dx = 2x
1 1 1
= 2¢3 2¢1=6 2
= 4:
Indeed from the diagram
y
y =2
2
A
x
0 1 3
the area under the curve between the integration limits is the area of
a square of side 2. This has area 2 ¢ 2 = 4. End Quiz
79. nite integral of y = 5x4 with respect to x
if the lower limit of the integral is x = 2 and the upper limit x = 1
can be written as 1
5x4 dx :
2
B
‚B a
From the basic result n
A ax dx = xn+1
we obtain
n+1 A
5 5 1
1 1
5x4 dx = x = x5
2 5 2
2
= ( 1)5 ( 2)5
= 1 ( 32)
= 1 + 32 = 31 :
Note that since the integrand 5x4 is positive for all x, the negative
suggested solutions could not be correct. End Quiz
