$\int_0^1|5 x-3| \mathrm{d} x=$
$\int_0^1|5 x-3| \mathrm{d} x=$
- $\frac{23}{10}$
- $\frac{13}{10}$
- $\frac{31}{10}$
- $\frac{1}{2}$
Solution
$\begin{aligned} & \int_0^1|5 x-3| \mathrm{d} x=\int_0^{3 / 5}(3-5 x) \mathrm{d} x+\int_{3 / 5}^1(5 x-3) \mathrm{d} x \\ & =\left[3 x-\frac{5 x^2}{2}\right]_0^{3 / 5}+\left[\frac{5 x^2}{2}-3 x\right]_{3 / 5}^1 \\ & =\left(\frac{9}{5}-\frac{9}{10}\right)-0+\left(\frac{5}{2}-3\right)-\left(\frac{9}{10}-\frac{9}{5}\right) \\ & =\frac{13}{10}\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 2)
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