The value of int_{-1}^{2} frac{|x|}{x} d x is | vedclass

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(B) Let $I = \int_{-1}^{2} \frac{|x|}{x} d x$.We know that $|x| = \begin{cases} x, & 	ext{if } x \geq 0 \ -x, & 	ext{if } x < 0 \end{cases}$.Splitt

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(B) Let $I = \int_{-1}^{2} \frac{|x|}{x} d x$.We know that $|x| = \begin{cases} x, & 	ext{if } x \geq 0 \ -x, & 	ext{if } x Splitting the integral at $x = 0$,we get:$I = \int_{-1}^{0} \frac{-x}{x} d x + \int_{0}^{2} \frac{x}{x} d x$$I = \int_{-1}^{0} (-1) d x + \int_{0}^{2} (1) d x$$I = -[x]_{-1}^{0} + [x]_{0}^{2}$$I = -(0 - (-1)) + (2 - 0)$$I = -(1) + 2$$I = 1$.

The value of $\int_{-1}^{2} \frac{|x|}{x} d x$ is

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