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If int0^ooe^(-ax)dx=1/a, then int0^oo(x^...

If `int_0^ooe^(-ax)dx=1/a`, then `int_0^oo(x^n)e^(-ax)dx` is

A

`((-1)^(n)n!)/(a^(n+1))`

B

`((-1)^(n)(n-1)!)/(a^(n))`

C

`(n!)/(a^(n+1))`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I_(n)=int_(0)^(oo)x^(n)e^(-ax)dx`
`=[x^(n).(e^(-ax))/(-a)]_(0)^(oo)-int_(0)^(oo)nx^(n-1).(e^(-ax))/(-a)dx`
`=-(1)/(a)underset(xrarroo)(lim)(x^(n))/(e^(ax))+(n)/(a)I_(n-1)`
`I_(n)=(n)/(a)I_(n-1)" "[because underset(xrarroo)(lim)(x^(n))/(e^(ax))=0]`
`=(n)/(a).(n-1)/(a)I_(n-2)`
`=(n(n-1)(n-2))/(a^(3))I_(n-3)`
`=(n!)/(a^(n))int_(0)^(oo)e^(-ax)dx=(n!)/(a^(n))I_(0)=(n!)/(a^(n+1))`
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