int_{-1}^2 |x| , dx = | vedclass

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(C) We know that the absolute value function $|x|$ is defined as:$|x| = \begin{cases} -x, & x < 0 \ x, & x \ge 0 \end{cases}$Therefore,the integral c

Vedclass · Accepted Answer

$\frac{5}{2}$

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(C) We know that the absolute value function $|x|$ is defined as:$|x| = \begin{cases} -x, & x Therefore,the integral can be split at $x = 0$:$I = \int_{-1}^2 |x| \, dx = \int_{-1}^0 (-x) \, dx + \int_0^2 (x) \, dx$Evaluating the first part:$\int_{-1}^0 (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 = 0 - \left( -\frac{(-1)^2}{2} \right) = 0 - (-\frac{1}{2}) = \frac{1}{2}$Evaluating the second part:$\int_0^2 (x) \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - 0 = \frac{4}{2} = 2$Adding both parts:$I = \frac{1}{2} + 2 = \frac{1+4}{2} = \frac{5}{2}$

$\int_{-1}^2 |x| \, dx =$

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