JEE MAIN - Mathematics (2003 - No. 20)
If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to
$${{a + b} \over 2}\int\limits_a^b {f\left( {a + b + x} \right)dx} $$
$${{a + b} \over 2}\int\limits_a^b {f\left( {b - x} \right)dx} $$
$${{a + b} \over 2}\int\limits_a^b {f\left( x \right)dx} $$
$$\,{{b - a} \over 2}\int\limits_a^b {f\left( x \right)dx} $$
Explanation
$$I = \int\limits_a^b {xf\left( x \right)} dx$$
$$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$$
$$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$$
$$ = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx} $$
$$\left[ {} \right.$$ As given that $$f\left( {a + b - x} \right) = f\left( x \right)$$ $$\left. {} \right]$$
$$2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx$$
$$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$$
$$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$$
$$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$$
$$ = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx} $$
$$\left[ {} \right.$$ As given that $$f\left( {a + b - x} \right) = f\left( x \right)$$ $$\left. {} \right]$$
$$2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx$$
$$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$$
Comments (0)
