The value of int_0^pi left| sin x - frac{2x}{ | vedclass

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

Question

(B) Let $I = \int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$. Consider the function $f(x) = \sin x - \frac{2x}{\pi}$. At $x = \frac{\pi}{2}$,$f(

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) Let $I = \int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$. Consider the function $f(x) = \sin x - \frac{2x}{\pi}$. At $x = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) - \frac{2(\pi/2)}{\pi} = 1 - 1 = 0$. For $0 \le x \le \frac{\pi}{2}$,$\sin x \ge \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \sin x - \frac{2x}{\pi}$. For $\frac{\pi}{2} \le x \le \pi$,$\sin x \le \frac{2x}{\pi}$,so $|\sin x - \frac{2x}{\pi}| = \frac{2x}{\pi} - \sin x$. Thus,$I = \int_0^{\pi/2} (\sin x - \frac{2x}{\pi}) dx + \int_{\pi/2}^{\pi} (\frac{2x}{\pi} - \sin x) dx$. Evaluating the integrals: $I = [-\cos x - \frac{x^2}{\pi}]_0^{\pi/2} + [\frac{x^2}{\pi} + \cos x]_{\pi/2}^{\pi}$. $I = [(-\cos(\frac{\pi}{2}) - \frac{(\pi/2)^2}{\pi}) - (-\cos(0) - 0)] + [(\frac{\pi^2}{\pi} + \cos(\pi)) - (\frac{(\pi/2)^2}{\pi} + \cos(\frac{\pi}{2}))]$. $I = [(0 - \frac{\pi}{4}) - (-1)] + [(\pi - 1) - (\frac{\pi}{4} + 0)]$. $I = 1 - \frac{\pi}{4} + \pi - 1 - \frac{\pi}{4} = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.

The value of $\int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$ is

Similar Questions

If $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$,then the value of $k$ is

If $[x]$ is the greatest integer function not greater than $x$,then $\int_{0}^{11} [x] dx$ is equal to

If $[\cdot]$ denotes the greatest integer function,then $\int_1^2 [x^2] dx =$

If $f(x) = \begin{cases} \frac{6 x^2 + 1}{4 x^3 + 2 x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \le x \le 2 \end{cases}$ then $\int_0^2 f(x) dx =$

$\int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module