Let f and g be continuous functions on [0, a] such that f(x)=f(a−x) and g(x)+g(a−x)=4 then ∫0a f(x)g(x)dx is equal to

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f (x) = f (a - x)$ and $g (x) + g (a - x) = 4$ then $\int_{0}^{a} f (x) g (x) d x$ is equal to

A
$2 \int_{0}^{a} f (x) d x$
B
$4 \int_{0}^{a} f (x) d x$
C
$- 3 \int_{0}^{a} f (x) d x$
D
$\int_{0}^{a} f (x) d x$

Solution:

Here, $f (x) = f (a - x)$ and $g (x) + g (a - x) = 4$

Let $I = \int_{0}^{a} f (x) g (x) d x = \int_{0}^{a} f (a - x) g (a - x) d x$

$= \int_{0}^{a} f (x) (4 - g (x)) d x = 4 \int_{0}^{a} f (x) d x - \int_{0}^{a} f (x) g (x) d x$

$= 4 \int_{0}^{a} f (x) d x - I \Rightarrow I = 2 \int_{0}^{a} f (x) d x$

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