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If f\ a n d\ g are continuious on [0,\ a...

If `f\ a n d\ g` are continuious on `[0,\ a]` and satisfy `f(x)=f(a-x)a n d\ g(x)+g(a-x)=2.` show that `int_0^af(x)g(x)dx=int_0^af(x)dx`

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Let,`int_0^a f(x) dx=int_0^af(a−x) dx`
∴`int_0^af(x)g(x) dx =int_0^af(a−x)g(a−x) dx`
⇒`int_0^af(x)g(x) dx =int_0^af(a−x)(2−g(x)} dx`
⇒`int_0^af(x)g(x) dx =2int_0^af(x) dx−∫a0f(x)g(x) dx`
⇒`2int_0^af(x)g(x) dx =2int_0^af(x) dx`
⇒`int_0^af(x)g(x) dx =int_0^af(x) dx`
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