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$\int_0^2|2 x-3| d x=$
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1724 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{2}$
Let
When $x=\frac{3}{2}, 2 x-3=0$
$$
\begin{aligned}
& \therefore \mathrm{I}=\int_0^{3 / 2}(3-2 \mathrm{x}) \mathrm{dx}+\int_{\frac{3}{2}}^2(2 \mathrm{x}-3) \mathrm{dx}=[3 \mathrm{x}]_0^{3 / 2}-\frac{2}{2}\left[\mathrm{x}^2\right]_{3 / 2}^2-[3 \mathrm{x}]_{3 / 2}^2 \\
& =\left(\frac{9}{2}\right)-\left(\frac{9}{4}\right)+\left(4-\frac{9}{4}\right)-3\left(2-\frac{3}{2}\right)=\frac{5}{2}
\end{aligned}
$$
When $x=\frac{3}{2}, 2 x-3=0$
$$
\begin{aligned}
& \therefore \mathrm{I}=\int_0^{3 / 2}(3-2 \mathrm{x}) \mathrm{dx}+\int_{\frac{3}{2}}^2(2 \mathrm{x}-3) \mathrm{dx}=[3 \mathrm{x}]_0^{3 / 2}-\frac{2}{2}\left[\mathrm{x}^2\right]_{3 / 2}^2-[3 \mathrm{x}]_{3 / 2}^2 \\
& =\left(\frac{9}{2}\right)-\left(\frac{9}{4}\right)+\left(4-\frac{9}{4}\right)-3\left(2-\frac{3}{2}\right)=\frac{5}{2}
\end{aligned}
$$
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