The value of int_{-1}^{1} frac{dx}{sqrt{|x|}} | vedclass

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

Question

(C) The integrand $f(x) = \frac{1}{\sqrt{|x|}}$ is an even function because $f(-x) = \frac{1}{\sqrt{|-x|}} = \frac{1}{\sqrt{|x|}} = f(x)$.Therefore,$\

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The integrand $f(x) = \frac{1}{\sqrt{|x|}}$ is an even function because $f(-x) = \frac{1}{\sqrt{|-x|}} = \frac{1}{\sqrt{|x|}} = f(x)$.Therefore,$\int_{-1}^{1} \frac{dx}{\sqrt{|x|}} = 2 \int_{0}^{1} \frac{dx}{\sqrt{x}}$.Now,evaluate the integral: $2 \int_{0}^{1} x^{-1/2} dx = 2 \left[ \frac{x^{1/2}}{1/2} ight]_{0}^{1}$.$= 2 	imes 2 \left[ \sqrt{x} ight]_{0}^{1} = 4 [1 - 0] = 4$.

The value of $\int_{-1}^{1} \frac{dx}{\sqrt{|x|}}$ is

Similar Questions

Let $[x]$ and $\{x\}$ be the integer part and fractional part of a real number $x$ respectively. The value of the integral $\int_0^5 [x]\{x\} dx$ is

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

$\int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx = $

$ \int_{-\pi / 4}^{\pi / 4} \frac{dx}{1+\cos 2x} $ is equal to

Evaluate the definite integral $\int_{0}^{1} \left(x e^{x} + \sin \frac{\pi x}{4}\right) dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module