$\int_0^1|5 x-3| d x=$

Question

Aicharya · Accepted Answer

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

$\int_0^1|5 x-3| d x=$

Solution