Question

Let $f:R\to Randg:R\to R$ be continuous functions. Then, the value of the integral
${\int }_{-\pi }^{\pi }\left(f\left(x\right)+f(-x)\right]\left(g\left(x\right)-g(-x)\right]\mathrm{dx}$ is

• A. \N
• B. 1
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A \N

$\mathrm{Let}h\left(x\right)=\left(f\left(x\right)+f(-x)\right]\left(g\left(x\right)-g(-x)\right]\therefore h(-x)=\left(f(-x)+f\left(x\right)\right]\left(g(-x)-g\left(x\right)\right]\Rightarrow h(-x)=-h\left(x\right)\mathrm{Hence},h\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}.\therefore {\int }_{-\pi }^{\pi }\left(f\left(x\right)+f(-x)\right]\left(g\left(x\right)-g(-x)\right]\mathrm{dx}=0$