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

If $f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$,then $\int_1^4 f(x) dx=$

$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $

Let $[ t ]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_{1}^{2} |2x - [3x]| dx$ is

The value of $\int_{0}^{\sqrt{2}} [x^2] \, dx$,where $[.]$ denotes the greatest integer function.

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module