Mathematics - Definite Integration Question with Solution | TestHub

${\int }_0^2\left(\left|2x^2-3x\right|+\left[x-\frac{1}{2}\right]\right)dx$

$={\int }_0^2\left|2x^2-3x\right|dx+{\int }_0^2\left[x-\frac{1}{2}\right]dx$

Let $I_1={\int }_0^2\left|2x^2-3x\right|dx={\int }_0^{\frac{3}{2}}\left(3x-2x^2\right)dx+{\int }_{\frac{3}{2}}^2\left(2x^2-3x\right)dx$

$={\left(\frac{3x^2}{2}-\frac{2x^3}{3}\right)}_0^{\frac{3}{2}}+{\left(\frac{2x^3}{3}-\frac{3x^2}{2}\right)}_{\frac{3}{2}}^2$

$=\left(\frac{27}{8}-\frac{9}{4}\right)+\left(\frac{8}{3}-6-\frac{9}{4}+\frac{27}{8}\right)=\frac{9}{4}-\frac{10}{3}=\frac{19}{12}$

Let $I_2={\int }_0^2\left[x-\frac{1}{2}\right]dx$

Assume $x-\frac{1}{2}=t$

i.e. $I_2={\int }_{\frac{-1}{2}}^{\frac{3}{2}}\left[t\right]dt$

$={\int }_{-\frac{1}{2}}^0\left(-1\right)dt+{\int }_0^{1}0·dt+{\int }_1^{\frac{3}{2}}1·dt$

$={\left(-t\right)}_{-\frac{1}{2}}^0+{\left(t\right)}_1^{\frac{3}{2}}=-\frac{1}{2}+\left(\frac{3}{2}-1\right)=0$

Hence, ${\int }_0^2\left(\left|2x^2-3x\right|+\left[x-\frac{1}{2}\right]\right)dx=\frac{19}{12}$