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

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(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,$\

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$4$

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(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

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