Page 53 - Integration Problems
P. 53
Example 14:
Evaluate the integral,
0
∫
√1 − 4
−3
Make the substitution
= 1 − 4
= −4
= −
4
Also,
1 −
=
4
The integral becomes,
0
0 1 − 1
∫ = ∫ ( ) (− )
√1 − 4 −3 4 √ 4
−3
This becomes,
1 0 1 −
− ∫
16 √
−3
Expanding
0
1 1
− ∫ [ − ]
16 √ √
−3
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