$\int_0^2|2 x-3| d x=$

Question

Aicharya · Accepted Answer

Let When $x=\frac{3}{2}, 2 x-3=0$ $\begin{aligned} & 	herefore \mathrm{I}=\int_0^{3 / 2}(3-2 \mathrm{x}) \mathrm{dx}+\int_{\frac{3}{2}}^2(2 \mathrm{x}-3) \mathrm{dx}=[3 \mathrm{x}]_0^{3 / 2}-\frac{2}{2}\left[\mathrm{x}^2ight]_{3 / 2}^2-[3 \mathrm{x}]_{3 / 2}^2 \ & =\left(\frac{9}{2}ight)-\left(\frac{9}{4}ight)+\left(4-\frac{9}{4}ight)-3\left(2-\frac{3}{2}ight)=\frac{5}{2} \end{aligned}$

$\int_0^2|2 x-3| d x=$

Solution