Q114 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

$\int_0^3\left|x^2-3 x+2\right| d x=$

  • Correct answer
    A
    $\frac{11}{6}$
  • B
    $\frac{5}{6}$
  • C
    $\frac{3}{2}$
  • D
    $\frac{2}{3}$

Solution

$\int_0^3\left|x^2-3 x+2\right| d x$ $\begin{aligned} & \left|x^2-3 x+2\right|=\left\{\begin{array}{l}x^2-3 x+2, x \leq 1 \text { or } x>2 \\ -\left(x^2-3 x+2\right), 1 < x \leq 2\end{array}\right. \\ & =\int_0^1\left(x^2-3 x+2\right) d x+\int_1^2-\left(x^2-3 x+2\right) d x+\int_2^3\left(x^2-3 x+2\right) d x \\ & =\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_0^1-\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_1^2+\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_2^3 \\ & =\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left(\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\right) \\ & =\left(\frac{2-9+12}{6}\right)-\left(\frac{14+9-24}{6}\right)+\left(\frac{102-81-16}{6}\right) \\ & =\frac{5}{6}+\frac{1}{6}+\frac{5}{6}=\frac{11}{6} .\end{aligned}$

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

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