int_{pi}^{16pi} |sin x| dx = | vedclass

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

Question

(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$. We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$. We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for a periodic function with period $T$. Here, the interval length is $16\pi - \pi = 15\pi$. Thus, $\int_{\pi}^{16\pi} |\sin x| dx = 15 \int_{0}^{\pi} |\sin x| dx$. Since $\sin x \ge 0$ for $x \in [0, \pi]$, we have $|\sin x| = \sin x$. Therefore, $15 \int_{0}^{\pi} \sin x dx = 15 [-\cos x]_{0}^{\pi}$. $= 15 [-\cos(\pi) - (-\cos(0))] = 15 [-(-1) - (-1)] = 15 [1 + 1] = 15 	imes 2 = 30$.

$\int_{\pi}^{16\pi} |\sin x| dx = $

Similar Questions

$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} \,d x=$

If $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$,then $I = $

The value of $\int_{-2}^{2}(a x^{3}+b x+c) d x$ depends on the

The value of the integral $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ is

The value of $\int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module