The value of int_0^{npi + v} {|sin x|,dx} is | vedclass

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

Question

(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n

Vedclass · Accepted Answer

$2n + 1 - \cos v$

Vedclass · Answer

(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$. Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n \int_0^{\pi} \sin x \, dx = n [-\cos x]_0^{\pi} = n(1 - (-1)) = 2n$. Now,consider the integral $\int_{n\pi}^{n\pi + v} |\sin x| \, dx$. Let $x = n\pi + t$,then $dx = dt$. When $x = n\pi$,$t = 0$. When $x = n\pi + v$,$t = v$. Since $|\sin(n\pi + t)| = |\sin(n\pi) \cos t + \cos(n\pi) \sin t| = |(-1)^n \sin t| = |\sin t|$. So,$\int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_0^v |\sin t| \, dt$. Assuming $0 \le v \le \pi$,we have $\int_0^v \sin t \, dt = [-\cos t]_0^v = 1 - \cos v$. Therefore,$\int_0^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v$.

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

Similar Questions

The value of the integral $\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$,where $p$ and $q$ are integers,is equal to:

$\int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx = $

$\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x=$

The value of the integral $\int_0^\pi(1-|\sin 8 x|) d x$ is

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module