int_{-1}^1 frac{|x|}{x} , dx is equal to | vedclass

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

Question

(C) We are given the integral $I = \int_{-1}^1 \frac{|x|}{x} \, dx$. First,we define the function $f(x) = \frac{|x|}{x}$. Based on the definition of t

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) We are given the integral $I = \int_{-1}^1 \frac{|x|}{x} \, dx$. First,we define the function $f(x) = \frac{|x|}{x}$. Based on the definition of the absolute value function,we have: $f(x) = \begin{cases} -\frac{x}{x} = -1, & x  0 \end{cases}$ Since the function is discontinuous at $x = 0$,we split the integral at $x = 0$: $I = \int_{-1}^0 (-1) \, dx + \int_0^1 (1) \, dx$ Evaluating the integrals: $I = [-x]_{-1}^0 + [x]_0^1$ $I = (-(0) - (-(-1))) + (1 - 0)$ $I = (-1) + 1 = 0$ Thus,the value of the integral is $0$.

$\int_{-1}^1 \frac{|x|}{x} \, dx$ is equal to

Similar Questions

If $a_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin x} dx$,then $a_2 - a_1, a_3 - a_2, a_4 - a_3$ are in

$\int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = $

$ \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $ is equal to

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

$\int_0^1 |5x - 3| dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module