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Question

Prove the following:
[f(x)]nf(x)dx=[f(x)]n+1n+1+C, if n1

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Solution

I=[f(x)]nf(x)dx.
Let f(x)=t,f(x)dx=dt
I=tndt
=tn+1n+1+c
=[f(x)]n+1n+1+c

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