Question

Let $f$ and $g$ be two continuous functions. Then ${\int }_{-\pi /2}^{\pi /2}\{f\left(x\right)+f(-x)\}\{g\left(x\right)-g(-x)\}dx$ is equal to

• A. $\pi$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

Answer: (D)

$0$

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f\left(x\right)+f(-x))(g\left(x\right)-g(-x))dx$ ...(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(-x)+f\left(x\right))(g(-x)-g\left(x\right))dx$ ...(2)
Adding (1) and (2), we get
$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(-x)+f\left(x\right))((g\left(x\right)-g(-x))+(g(-x)-g\left(x\right)))dx=0\Rightarrow I=0$