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If fa n dg are continuous function on [0...

If `fa n dg` are continuous function on `[0,a]` satisfying `f(x)=f(a-x)a n dg(x)(a-x)=2,` then show that `int_0^af(x)g(x)dx=int_0^af(x)dxdot`

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The correct Answer is:
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`int_(0)^(a)f(x)g(x)dx`
`=int_(0)^(a)f(a-x)g(a-x)dx`
`=int_(0)^(a)f(x).{2-g(x)}dx`
`=2int_(0)^(a)f(x)dx-int_(0)^(a)f(x)g(x)dx`
or `2int_(0)^(a)f(x)f(x)dx=2int_(0)^(a)f(x)dx`
or `int_(0)^(a)f(x)g(x)dx=int_(0)^(a)f(x)dx`
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