Question:
\(\int_{-\pi}^{\pi} \left| \pi - |x| \right| dx\) is equal to :
\(\int_{-\pi}^{\pi} \left| \pi - |x| \right| dx\) is equal to :
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The integrand is even and \(\pi - |x| \ge 0\) on \([-\pi, \pi]\). Double the integral over \([0, \pi]\).
Updated On: Nov 5, 2026
- \(\pi^2\)
- \(2\pi^2\)
- \(\sqrt{2}\,\pi^2\)
- \(\frac{\pi^2}{2}\)
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The Correct Option is A
Solution and Explanation
For \(|x| \le \pi\), \(\pi - |x| \ge 0\), so \(|\pi - |x|| = \pi - |x|\).
The integrand is even, so \(I = 2\int_0^{\pi} (\pi - x)\,dx\).
\(= 2\left[\pi x - \frac{x^2}{2}\right]_0^{\pi} = 2\left(\pi^2 - \frac{\pi^2}{2}\right) = \pi^2\).
Answer: (1) \(\pi^2\)
The integrand is even, so \(I = 2\int_0^{\pi} (\pi - x)\,dx\).
\(= 2\left[\pi x - \frac{x^2}{2}\right]_0^{\pi} = 2\left(\pi^2 - \frac{\pi^2}{2}\right) = \pi^2\).
Answer: (1) \(\pi^2\)
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