Quiz 1 Quiz 1                              Friday, March 9, 2001



NAME:

Student Number:------            No.----           Marks ----






(25 Marks) Evaluate the following indefinite integral:



x2
(5x3+7)3
 dx.


Solution: (Method one)



x2
(5x3+7)3
 dx
=
1
15

d(5x3+7)
(5x3+7)3
= 1
15
1
(1-3)
(5x3+7)1-3+C = - 1
30
(5x3+7)-2+C.


(Method two)

Let u = 5x3+7. Then du = 15x2dx and x2dx = (1/15)du. Hence, we have



x2
(5x3+7)3
 dx
=
1
15

du
u3
= 1
15
1
1-3
u1-3+C = - 1
30
u-2+C = - 1
30
(5x3+7)-2+C.


(Method three)

Let u = (5x3+7)3. Then u1/3 = 5x3+7. This implies (1/3)u[1/3]-1du = 15x2dx   and (1/45)u-2/3du = x2dx. Hence,



x2
(5x3+7)3
 dx
=
1
45

u-2/3du
u
= 1
45
1
(1-5/3)
u1-5/3+C = - 1
30
u-2/3+C
=
- 1
30
(5x3+7)-2+C.



File translated from TEX by TTH, version 2.79.
On 3 Apr 2001, 10:39.