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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 :

Similar Questions

Evaluate the integral: $\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x$

The value of the integral $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ is:

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

The value of $\int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$ is

The value of $\int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module