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JEE Mains · Maths · STD 12 - 7.2 definite integral

અહી વિધેય \(f(x)\) એ  \(f(x)+f(\pi-x)=\) \(\pi^2, \forall x \in R\) નું સમાધાન કરે છે . તો  \(\int \limits_0^\pi f(x) \sin x d x\) ની કિમંત મેળવો.

  1. A \(\frac{\pi^2}{4}\)
  2. B \(\frac{\pi^2}{2}\)
  3. C \(2 \pi^2\)
  4. D \(\pi^2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\pi^2\)

Step-by-step Solution

Detailed explanation

\(f(x)+f(\pi-x)=\pi^2\) \(I=\int \limits_0^\pi f(x) \sin x d x\) Applying King's Rule \(I=\int \limits_0^\pi f(\pi-x) \cdot \sin (\pi-x) d x\) \(2 I=\int \limits_0^\pi[f(x)+f(\pi-x)] \sin x d x\) \(2 I=\int \limits_0^\pi \pi^2 \sin x d x\)…
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