int_{-1}^{2} {|x|,dx} = (in /2) | vedclass

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(A) We need to evaluate the definite integral $I = \int_{-1}^{2} {|x|\,dx}$.Since the integrand $|x|$ changes its definition at $x = 0$,we split the i

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$5$

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(A) We need to evaluate the definite integral $I = \int_{-1}^{2} {|x|\,dx}$.Since the integrand $|x|$ changes its definition at $x = 0$,we split the integral into two parts:$I = \int_{-1}^{0} {|x|\,dx} + \int_{0}^{2} {|x|\,dx}$.For $x \in [-1, 0]$,$|x| = -x$,and for $x \in [0, 2]$,$|x| = x$.So,$I = \int_{-1}^{0} {-x\,dx} + \int_{0}^{2} {x\,dx}$.Evaluating the first part: $\int_{-1}^{0} {-x\,dx} = -\left[ \frac{x^2}{2} \right]_{-1}^{0} = -\left( 0 - \frac{(-1)^2}{2} \right) = -\left( 0 - \frac{1}{2} \right) = \frac{1}{2}$.Evaluating the second part: $\int_{0}^{2} {x\,dx} = \left[ \frac{x^2}{2} \right]_{0}^{2} = \left( \frac{2^2}{2} - 0 \right) = \frac{4}{2} = 2$.Adding both parts: $I = \frac{1}{2} + 2 = \frac{5}{2}$.

$\int_{-1}^{2} {|x|\,dx} =$ (in $/2$)

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