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Question
Options
`15/16`
`3/16`
`-3/16`
`-16/3`
Solution
`-16/3`
`I=int_0^1x/(1-x)^(5/4)dx`
Put, 1 - x = t ⇒ x = 1 - t
⇒ dx = -dt
| x | 0 | 1 |
| t | 1 | 0 |
`I=int_1^0((1-t)(-dt))/t^(5/4)`
`I=int_0^1(1-t)/t^(5/4)dt`
`I=int_0^1(t^(-5/4)-t^(-1/4))dt`
`I=[t^(-1/4)/(-1/4)-t^(3/4)/(3/4)]_0^1`
`I=-4-4/3`
`I=-16/3`
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