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

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(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

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$0$

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(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

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