If int_0^{2 pi} |x sin x| , dx = k pi,then k | vedclass

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

Question

(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$. In the inte

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$. In the interval $[0, \pi]$,$\sin x \ge 0$,and in the interval $[\pi, 2 \pi]$,$\sin x \le 0$. Thus,$I = \int_0^{\pi} x \sin x \, dx - \int_{\pi}^{2 \pi} x \sin x \, dx$. Using integration by parts,$\int x \sin x \, dx = -x \cos x + \sin x$. Evaluating the first integral: $\int_0^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi} = (-\pi(-1) + 0) - (0 + 0) = \pi$. Evaluating the second integral: $\int_{\pi}^{2 \pi} x \sin x \, dx = [-x \cos x + \sin x]_{\pi}^{2 \pi} = (-2 \pi(1) + 0) - (-\pi(-1) + 0) = -2 \pi - \pi = -3 \pi$. Therefore,$I = \pi - (-3 \pi) = 4 \pi$. Given $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,we have $4 \pi = k \pi$,which implies $k = 4$.

If $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,then $k =$

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x=$ . . . . . . .

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

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

If $\int_{-\infty}^{\infty} f(x) dx = 1$,then $\int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$ is equal to

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module