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int_(-2)^(1)(2-|x|)dx

$\int_{-2}^{1}(2-|x|) d x$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{-2}^{1}(2-|x|) d x

Split Integral: Step $1$ : Break the integral into two parts based on the absolute value function. $\newline$ Since the absolute value function changes at $x = 0$ , split the integral at $x = 0$ . $\newline$ $\int_{-2}^{1}(2-|x|)\,dx = \int_{-2}^{0}(2-|x|)\,dx + \int_{0}^{1}(2-|x|)\,dx$
Evaluate First Part: Step $2$ : Evaluate the first part of the integral from $-2$ to $0$ . $\newline$ For $x$ in $[-2, 0]$ , $|x| = -x$ , so the integral becomes: $\newline$ $\int_{-2}^{0}(2+x)dx$ $\newline$ $= [2x + x^2/2]$ from $-2$ to $0$ $\newline$ $= (2\cdot0 + 0^2/2) - (2\cdot(-2) + (-2)^2/2)$ $\newline$ $= 0 - (-4 + 2)$ $\newline$ $0$ $0$

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