Question

$\int\limits_{2-a}^{2+a} f\left(x\right) dx$ is equal to (where $f\left(2 - a\right)=f\left(2 + a\right) \forall a\in R$)

• A. $2 \int\limits_{2}^{2+a} f\left(x\right) dx$
• B. $2 \int\limits_{0}^{a} f\left(x\right) dx$
• C. $2 \int\limits_{0}^{2} f\left(x\right) dx$
• D. None of these

Answer 1

Answer: (A)

$2 \int\limits_{2}^{2+a} f\left(x\right) dx$

Answer 2

$\because f\left(2 - a \right)=f\left(2 + a\right)$ $\therefore$ Function is symmetrical about the line $x = 2$ Then $\int\limits_{2-a}^{2+a} f\left(x\right) dx = 2 \int\limits_{2}^{2+a} f\left(x\right) dx$