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Let $f$ and $g$ be continuous functions on $\left [ 0,a \right ]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$ , then $\int_{0}^{a}f(x)g(x)dx$ is equal to :
Option 1)
$-3\int_{0}^{a}f(x)dx$

Option 2)
$2\int_{0}^{a}f(x)dx$
Option 3)
$\int_{0}^{a}f(x)dx$
Option 4)
$4\int_{0}^{a}f(x)dx$

Answers (1)

best_answer

lower and upper limit -

$\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}$

$= F\left ( b \right )-F\left ( a \right )$

- wherein

Where a is lower and b is upper limit.

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

Since $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

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

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

Option 1)

$-3\int_{0}^{a}f(x)dx$

Option 2)
$2\int_{0}^{a}f(x)dx$
Option 3)

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

Option 4)

$4\int_{0}^{a}f(x)dx$

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