If f(x)=f(a−x) and g(x)+g(a−x)=2 then ∫0af(x).g(x)dx=

I=∫0af(x)g(x)dx=∫0af(a-x)g(a−x)dx. ⇒ I=∫0af(x).2-g(x)dx ⇒i=2∫0af(x)dx.-∫0af(x)g(x)dx ⇒2I=2∫0af(x)dx

If f(x)=f(a−x) and g(x)+g(a−x)=2 then ∫0af(x).g(x)dx=

I=∫0af(x)g(x)dx=∫0af(a-x)g(a−x)dx. ⇒ I=∫0af(x).2-g(x)dx ⇒i=2∫0af(x)dx.-∫0af(x)g(x)dx ⇒2I=2∫0af(x)dx

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If $f (x) = f (a - x) a n d g (x) + g (a - x) = 2$ then $\int_{0}^{a} f (x) . g (x) d x =$

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$\int_{0}^{a} f (x) d x$

a²

answer is D.

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Detailed Solution

${I = \int}_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (a - x) g (a - x) d x . \Rightarrow I = \int_{0}^{a} f (x) . (2 - g (x)) d x \Rightarrow i = {2 \int}_{0}^{a} f (x) d x . - \int_{0}^{a} f (x) g (x) d x \Rightarrow 2 I = {2 \int}_{0}^{a} f (x) d x$