Q63 / 150 · MHT CET 2021 (22 Sep Shift 2) · Mathematics

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

  • A
    $\frac{3}{10}$
  • Correct answer
    B
    $\frac{5}{2}$
  • C
    $\frac{10}{3}$
  • D
    $\frac{2}{5}$

Solution

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

Hence, the correct answer is option: 'B'.

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