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

To solve the problem, we need to evaluate the definite integral \(\int_{0}^{\pi} f(x) \sin x \, dx\), given the condition \(f(x) + f(\pi-x) = \pi^{2}\).

We begin by understanding the given function property. The property \(f(x) + f(\pi - x) = \pi^{2}\) suggests that the function \(f(x)\) is symmetric around \(\pi/2\). This identity helps simplify the integral.

Consider the integral:

\(\int_{0}^{\pi} f(x) \sin x \, dx\)

Using the substitution \(x = \pi - t\), hence \(dx = -dt\). Change limits: when \(x = 0\), \(t = \pi\) and when \(x = \pi\), \(t = 0\). Thus,

\(\int_{0}^{\pi} f(x) \sin x \, dx = \int_{\pi}^{0} f(\pi-t) \sin(\pi-t) (-dt)\)

Since \(\sin(\pi - t) = \sin t\), the above integral becomes:

\(\int_{0}^{\pi} f(\pi-t) \sin t \, dt = \int_{0}^{\pi} (\pi^2 - f(t)) \sin t \, dt\)

Thus, the integral becomes:

\(\int_{0}^{\pi} (\pi^2 - f(t)) \sin t \, dt = \pi^2 \int_{0}^{\pi} \sin t \, dt - \int_{0}^{\pi} f(t) \sin t \, dt\)

Let \(I = \int_{0}^{\pi} f(t) \sin t \, dt\), therefore:

\(I = \pi^2 \int_{0}^{\pi} \sin t \, dt - I\)

This implies:

\(2I = \pi^2 \int_{0}^{\pi} \sin t \, dt\)

Now evaluate \(\int_{0}^{\pi} \sin t \, dt\):

\(\int_{0}^{\pi} \sin t \, dt = [-\cos t]_{0}^{\pi} = [-\cos(\pi) + \cos(0)] = [1 + 1] = 2\)

Plug this back into our equation:

\(2I = \pi^2 \times 2\)

Thus, \(I = \pi^2\).

Therefore, the value of the integral \(\int_{0}^{\pi} f(x) \sin x \, dx\) is:

π²