Q. If $f$ is a continuous function, then
A) $\int_{- 2}^{2}{f(x)dx = \int_{0}^{2}{\left\lbrack f(x) - f( - x) \right\rbrack dx}}$
B) $\int_{- 3}^{5}{2f(x)dx = \int_{- 6}^{10}{f(x - 1)dx}}$
C) $\int_{- 3}^{5}{f(x)dx = \int_{- 4}^{4}{f(x - 1)dx}}$
D) $\int_{- 3}^{5}{f(x)dx = \int_{- 2}^{6}{f(x - 1)dx}}$
Explanation
(b)
Given, $f(x)$ is a continuous function.
Let us consider $f(x) = x$
$\therefore\int_{- 3}^{5}{2x}\ dx = \left\lbrack x^{2} \right\rbrack_{- 3}^{5} = 16$
and $\int_{- 6}^{10}{(x - 1)}dx = \left\lbrack \frac{x^{2}}{2} - x \right\rbrack_{- 6}^{10} = 16$
$\therefore\int_{- 3}^{5}{2f(x)dx = \int_{- 6}^{10}{f(x - 1)dx}}\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.