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$\int_0^3\left|x^2-3 x+2\right| d x=$
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Verified Answer
The correct answer is:
$\frac{11}{6}$
Let $I=\int_0^3\left|x^2-3 x+2\right| d x$
$\because \quad x^2-3 x+2=(x-2)(x-1)$
So, $\quad x^2-3 x+2 < 0$ for $x \in(1,2)$
and $x^2-3 x+2 \geq 0$ for $x \in R-(1,2)$
So,
$$
\begin{aligned}
& I= \int_0^1\left(x^2-3 x+2\right) d x-\int_1^2\left(x^2-3 x+2\right) d x \\
&+\int_2^3\left(x^2-3 x+2\right) d x \\
&=\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_0^1-\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_1^2+ \\
& \qquad\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_2^3 \\
&=\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left[\left(\frac{8}{3}-\frac{12}{2}+4\right)-\left(\frac{1}{3}-\frac{3}{2}+2\right)\right] \\
& \quad+\left[\left(\frac{27}{3}-\frac{27}{2}+6\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right)\right]=\frac{11}{6}
\end{aligned}
$$
$\because \quad x^2-3 x+2=(x-2)(x-1)$
So, $\quad x^2-3 x+2 < 0$ for $x \in(1,2)$
and $x^2-3 x+2 \geq 0$ for $x \in R-(1,2)$
So,
$$
\begin{aligned}
& I= \int_0^1\left(x^2-3 x+2\right) d x-\int_1^2\left(x^2-3 x+2\right) d x \\
&+\int_2^3\left(x^2-3 x+2\right) d x \\
&=\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_0^1-\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_1^2+ \\
& \qquad\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_2^3 \\
&=\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left[\left(\frac{8}{3}-\frac{12}{2}+4\right)-\left(\frac{1}{3}-\frac{3}{2}+2\right)\right] \\
& \quad+\left[\left(\frac{27}{3}-\frac{27}{2}+6\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right)\right]=\frac{11}{6}
\end{aligned}
$$
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