int_{-pi}^{pi} |pi - |x|| , dx is equal to : | vedclass

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(A) Let $I = \int_{-\pi}^{\pi} |\pi - |x|| \, dx$. Since the integrand $f(x) = |\pi - |x||$ is an even function,we can write: $I = 2 \int_{0}^{\pi} |\

Vedclass · Accepted Answer

$\pi^{2}$

Vedclass · Answer

(A) Let $I = \int_{-\pi}^{\pi} |\pi - |x|| \, dx$. Since the integrand $f(x) = |\pi - |x||$ is an even function,we can write: $I = 2 \int_{0}^{\pi} |\pi - |x|| \, dx$. For $x \in [0, \pi]$,$|x| = x$,so the expression becomes: $I = 2 \int_{0}^{\pi} |\pi - x| \, dx$. Since $x \leq \pi$ in the interval $[0, \pi]$,$\pi - x \geq 0$,thus $|\pi - x| = \pi - x$. $I = 2 \int_{0}^{\pi} (\pi - x) \, dx$. $I = 2 \left[ \pi x - \frac{x^{2}}{2} \right]_{0}^{\pi}$. $I = 2 \left( (\pi(\pi) - \frac{\pi^{2}}{2}) - (0 - 0) \right)$. $I = 2 \left( \pi^{2} - \frac{\pi^{2}}{2} \right) = 2 \left( \frac{\pi^{2}}{2} \right) = \pi^{2}$.

$\int_{-\pi}^{\pi} |\pi - |x|| \, dx$ is equal to :

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