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Question

If f(a+bx)=f(x) , simplify baxf(x)dx

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Solution

I=bax.f(x)dx ......(1)
Replace xa+bx
I=ba(a+bx).f(a+bx)dx
I=ba(a+bx).f(a+bx)dx
I=ba(a+bx).f(x)dx .........(2)
Adding (1) and (2) we get
2I=bax.f(x)dx+ba(a+bx).f(x)dx
2I=(a+b)baf(x)dx
I=a+b2baf(x)dx


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