Introduction to integration 329
For example, the increase in the value of the integral
x 2 as x increases from 1 to 3 is written as
3
1 x 2 dx.
Applying the limits gives
3
1
x
2 dx =
x 3
3
+ c
3
1
=
3 3
3
+ c
−
1 3
3
+ c
= (9 + c) −
1
3
+ c
= 8
2
3
Note that the c term always cancels out when limits are
applied and it need not be shown with definite integrals.
Problem 17. Evaluate
2
1
3xdx
2
1
3xdx =
3x 2
2
2
1
=
3
2
(2)
2
−
3
2
(1)
2
= 6 − 1
1
2
= 4
1
2
Problem 18. Evaluate
3
−2
(4 − x
2
) dx
3
−2
(4 − x
2
)dx =
4x −
x 3
3
3
−2
=
4(3) −
(3) 3
3
−
4(−2) −
(−2) 3
3
= {12 − 9} −
−8 −
−8
3
= {3} −
−5
1
3
= 8
1
3
Problem 19. Evaluate
2
0
x(3 + 2x)dx
2
0
x(3 + 2x)dx =
2
0
(3x + 2x
2
)dx =
3x 2
2
+
2x 3
3
2
0
=
3(2) 2
2
+
2(2) 3
3
− {0 + 0}
= 6 +
16
3
= 11
1
3
or 11.33
Problem 20. Evaluate
1
−1
x 4 − 5x 2 + x
x
dx
1
−1
x 4 − 5x 2 + x
x
dx
=
1
−1
x 4
x
−
5x 2
x
+
x
x
dx
=
1
−1
x
3
− 5x + 1
dx =
x 4
4
−
5x 2
2
+ x
1
−1
=
1
4
−
5
2
+ 1
−
(−1)
4
4
−
5(−1)
2
2
+ (−1)
=
1
4
−
5
2
+ 1
−
1
4
−
5
2
− 1
= 2
Problem 21. Evaluate
2
1
1
x 2 +
2
x
dx correct
to 3 decimal places
2
1
1
x 2 +
2
x
dx
=
2
1
x
−2
+ 2
1
x
dx =
x
−2+1
−2 + 1
+ 2 ln x
2
1
=
x −1
−1
+ 2 ln x
2
1
=
−
1
x
+ 2 ln x
2
1
=
−
1
2
+ 2 ln2
−
−
1
1
+ 2 ln 1
= 1.886
Problem 22. Evaluate
π/2
0
3 sin2x dx
π/2
0
3 sin2x dx
=
(3)
−
1
2
cos 2x
π/2
0
=
−
3
2
cos 2x
π/2
0
=
−
3
2
cos 2
π
2
−
−
3
2
cos 2 (0)
=
−
3
2
cos π
−
−
3
2
cos 0