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

Question

Aicharya · Accepted Answer

Let $I=\int_0^3\left|x^2-3 x+2ight| d x$ $\because \quad x^2-3 x+2=(x-2)(x-1)$ 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ight) d x-\int_1^2\left(x^2-3 x+2ight) d x \ &+\int_2^3\left(x^2-3 x+2ight) d x \ &=\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 xight]_0^1-\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 xight]_1^2+ \ & \qquad\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 xight]_2^3 \ &=\left(\frac{1}{3}-\frac{3}{2}+2ight)-\left[\left(\frac{8}{3}-\frac{12}{2}+4ight)-\left(\frac{1}{3}-\frac{3}{2}+2ight)ight] \ & \quad+\left[\left(\frac{27}{3}-\frac{27}{2}+6ight)-\left(\frac{8}{3}-\frac{12}{2}+4ight)ight]=\frac{11}{6} \end{aligned} $

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

Solution