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Question

Show that:
aaf(x)dx=2a0f(x)dx, if f(x) is an even function. I=0, if f(x) is an odd function.

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Solution

Let I=aaf(x)dx=0af(x)dx+a0f(x)dx
Now, if f(x) is even, f(x)=f(x)

I=0af(x)dx+a0f(x)dx

I=0(a)f(x)(dx)+a0f(x)dx
(substituting x by x)

I=a0f(x)dx+a0f(x)dx

but f(x)=f(x)
I=a0f(x)dx+a0f(x)dx

I=2a0f(x)dx if f(x) is even.

Now, if f(x) is odd, f(x)=f(x)
I=0af(x)dx+a0f(x)dx

I=0(a)f(x)(dx)+a0f(x)dx (substituting x by x)

I=a0f(x)dx+a0f(x)dx
but f(x)=f(x)

I=a0f(x)dx+a0f(x)dx

I=0 if f(x) is odd.
Hence, proved.

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