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\(\int_{-1}^2|x| d x=\)
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Verified Answer
The correct answer is:
\(\frac{5}{2}\)
\(\begin{aligned}
I & =\int_{-1}^2|x| d x=\int_{-1}^0(-x) d x+\int_0^2(x) d x \\
& =\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}\right]_0^2=\left(0+\frac{1}{2}\right)+\left(\frac{4}{2}-0\right)=\frac{5}{2}
\end{aligned}\)
I & =\int_{-1}^2|x| d x=\int_{-1}^0(-x) d x+\int_0^2(x) d x \\
& =\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}\right]_0^2=\left(0+\frac{1}{2}\right)+\left(\frac{4}{2}-0\right)=\frac{5}{2}
\end{aligned}\)
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