Show that$\int\limits_0^a {f\left( x \right)g\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} }$ ,

if f and g are defined as $f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 4$

Let $I = \int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)\left[ {4 - g\left( {a - x} \right)} \right]dx} }$

$= 4\int\limits_0^a {f\left( {a - x} \right)dx} - \int\limits_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx}$

Let $a - x = t$ $\Rightarrow$ $- dx = dt$

When $x = 0,t = a$and when $x = a,t = 0$

$\Rightarrow$ $I = - 4\int\limits_a^0 {f\left( t \right)dt} + \int\limits_a^0 {f\left( t \right)g\left( t \right)dt} = 4\int\limits_0^a {f\left( t \right)dt} - \int\limits_0^a {f\left( t \right)g\left( t \right)dt}$

$= 4\int\limits_0^a {f\left( x \right)dx} - \int\limits_0^a {f\left( x \right)g\left( x \right)dx} = 4\int\limits_0^a {f\left( x \right)dx - I}$

$\Rightarrow$ $2I = 4\int\limits_0^a {f\left( x \right)dx}$

Hence, $I = 2\int\limits_0^a {f\left( x \right)dx}$

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