int_0^pi |cos x| , dx = | vedclass

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

Question

(C) We need to evaluate the integral $I = \int_0^\pi |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_0^\pi |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral: $I = \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^\pi (-\cos x) \, dx$. Evaluating the first part: $\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$. Evaluating the second part: $\int_{\pi/2}^\pi -\cos x \, dx = -[\sin x]_{\pi/2}^\pi = -(\sin(\pi) - \sin(\pi/2)) = -(0 - 1) = 1$. Adding both parts: $I = 1 + 1 = 2$.

$\int_0^\pi |\cos x| \, dx = $

Similar Questions

$\int_0^1 {{e^{2\ln x}}dx} = $

$\int_a^b \frac{\log x}{x} \, dx = $

If $p(x)$ is a cubic polynomial with $p(1)=3, p(0)=2$ and $p(-1)=4$,then $\int_{-1}^1 p(x) dx$ is

$\int_0^{\pi / 4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ is equal to

$\int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module