\(\int_0^1|5 x-3| d x=\)

Let \(\mathrm{I}=\int_0^1|5 \mathrm{x}-3| \mathrm{dx}\)
\( 5 x-3=0 \Rightarrow x=\frac{3}{5} \)
\( \therefore I=\int_0^{\frac{3}{5}}-(5 x-3) d x+\int_{\frac{3}{5}}^1(5 x-3) d x \)
\( =\frac{-5}{2}\left[x^2\right]_0^{\frac{3}{5}}+3[x]_0^{\frac{3}{5}}+\frac{5}{2}\left[x^2\right]_{\frac{3}{5}}^1-3[x]_{\frac{3}{5}}^1 \)
\( =\left(\frac{-5}{2}\right)\left(\frac{9}{25}\right)+3\left(\frac{3}{5}\right)+\frac{5}{2}\left(1-\frac{9}{25}\right)-3\left(1-\frac{3}{5}\right) \)
\( =\frac{-45}{50}+\frac{9}{5}+\left(\frac{5}{2}\right)\left(\frac{16}{25}\right)-3\left(\frac{2}{5}\right)=\frac{-45}{50}+\frac{3}{5}+\frac{8}{5} \)
\( =\frac{-45+110}{50}=\frac{65}{50}=\frac{13}{10}\)