JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\) \(\pi^2, \forall x \in R\). Then \(\int \limits_0^\pi f(x) \sin x d x\) is equal to \(...........\).
- A \(\frac{\pi^2}{4}\)
- B \(\frac{\pi^2}{2}\)
- C \(2 \pi^2\)
- D \(\pi^2\)
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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