If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ∫0a f(x)g(x)dx is equal to

A
$\int_{0}^{a} f (x) dx$
B
$2 \int_{0}^{a} f (x) dx$
C
$2 \int_{0}^{a} f (x) g (x) dx$
D
None of these

Solution:

Let,

$\begin{array}{l} I = \int_{0}^{a} f (x) g (x) dx - - - - (i) \\ I = \int_{0}^{a} f (a - x) g (a - x) dx \\ [∵ \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \end{array}$

$\begin{array}{l} \Rightarrow I = \int_{0}^{a} f (x) {4 - g (x)} dx - - - (ii) \\ [∵ f (x) = f (a - x) and g (x) + g (a - x) = 4 (given)] \end{array}$

On adding Eqs. (i) and (ir), we get

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

If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then $\int_{0}^{a} f (x) g (x) dx$ is equal to

Solution:

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