Question

If $f(x) + f(\pi-x) = \pi^{2}$ then $_{0}\int^{\pi}f(x) sinx dx$ ?

• A. π2
• B. $\frac{\pi_2}{4}$
• C. 2π2
• D. $\frac{\pi_2}{8}$

Answer 1

Answer: (A)

π2

Answer 2

The correct option is (A): π2
$I=\int_0^\pi f(x)sinxdx$
$I=\int_0^\pi f(\pi-x)sinxdx$
$2I=\int_0^\pi sin(x)(f(x)+f(\pi-x))dx$
$2I=\pi^2\int_0^\pi sinxdx$
$2I=2\pi^2\int_0^\pi sinxdx$
$I=\pi^2(-cos x)^\frac{\pi}{2}_0$
$I=\pi^2$