Question

lf $f:R\to R$ and $g:R\to R$ are continuous then ${\int }_{-\pi /2}^{\pi /2}\left(f\right(x)+f(-x\left)\right)\left(g\right(x)-g(-x\left)\right)dx=$

• A. $\pi$
• B. 1
• C. -1
• D. 0

Answer 1

The correct option is 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
$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(f(-x)+f\left(x\right))((g\left(x\right)-g(-x))+(g(-x)-g\left(x\right)))dx=0\Rightarrow I=0$