Mathematics - Definite Integration Question with Solution | TestHub

$I={\int }_0^3\left|x^2-3x+2\right|dx$

$={\int }_0^3\left|\left(x-1\right)\left(x-2\right)\right|dx$

$={\int }_0^1\left(x^2-3x+2\right)dx-{\int }_1^2\left(x^2-3x+2\right)dx+{\int }_2^3\left(x^2-3x+2\right)dx$

$={\left[\frac{x^3}{3}-3\left(\frac{x^2}{2}\right)+2x\right]}_0^1-{\left[\frac{x^3}{3}-3\left(\frac{x^2}{2}\right)+2x\right]}_1^2+{\left[\frac{x^3}{3}-3\left(\frac{x^2}{2}\right)+2x\right]}_2^3$

$=\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left[\left(\frac{8}{3}-6+4\right)-\left(\frac{1}{3}-\frac{3}{2}+2\right)\right]+\left[\left(9-\frac{27}{2}+6\right)-\left(\frac{8}{3}-6+4\right)\right]$

$=\left(\frac{2}{3}-3+4\right)-\left(\frac{16}{3}-12+8\right)+\left(9-\frac{27}{2}+6\right)$

$=20-\frac{109}{6}=\frac{11}{6}$

So,

$12I=12\times \frac{11}{6}=22$