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

The Correct Option is A

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

Answer 2

We start with the given equation: \[ I = \int_0^{\pi} f(x)\sin x \, dx \quad \text{(1)} \] By applying the property of the function, we get: \[ I = \int_0^{\pi} f(\pi - x)\sin (\pi - x) \, dx = \int_0^{\pi} f(\pi - x) \sin x \, dx \quad \text{(2)} \] Adding (1) and (2): \[ 2I = \int_0^{\pi} \left( f(x) + f(\pi - x) \right) \sin x \, dx \] Substituting \(f(x) + f(\pi - x) = \pi^2\): \[ 2I = \int_0^{\pi} \pi^2 \sin x \, dx \] Thus, \[ 2I = \pi^2 \int_0^{\pi} \sin x \, dx = \pi^2 \times 2 \] So, \[ I = \pi^2 \]