Evaluate the following integrals: ^2_ 0 |x^2-3x+2 |dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$

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Answer

$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
We know that,
$\big|\text{x}^2-3\text{x}+2\big|\text{dx}=\begin{cases}-(\text{x}^2-3\text{x}+2),&(\text{x}-1)(\text{x}-2)\leq0\text{ or },&1\leq\text{x}\leq2\$\text{x}^2-3\text{x}+2),&\text{x}^2-3\text{x}+2\leq0\text{ or },&\text{x}\in(-\infty,1)(2,\infty)\end{cases}$
$\therefore\ \text{I}=\ \int^\limits2_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\int^\limits1_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}-\int^\limits2_1\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^1_0-\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^2_1$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\Big[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\Big]$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2}$
$\Rightarrow\text{I}=1$

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