Question

$\underset{0}{\overset{2}{\int }}(|2x^2-3x|+[x-\frac{1}{2}])$ dx, Where $\left[t\right]$ is the greatest integer function, is equal to

Answer 1

$\underset{0}{\overset{2}{\int }}|2x^2-3x|dx+\underset{0}{\overset{2}{\int }}[x-\frac{1}{2}]dx$
$=\underset{0}{\overset{3/2}{\int }}(3x-2x^2)dx+\underset{3/2}{\overset{2}{\int }}(2x^2-3x)dx+\underset{0}{\overset{1/2}{\int }}-1dx+\underset{1/2}{\overset{3/2}{\int }}0dx+\underset{3/2}{\overset{2}{\int }}1dx$

${\begin{array}{c}=(\frac{3x^2}{2}-\frac{2x^3}{3})\end{array}|}_0^{3/2}+{\begin{array}{c}(\frac{2x^3}{3}-\frac{3x^2}{2})\end{array}|}_{3/2}^2-\frac{1}{2}+\frac{1}{2}$

$=(\frac{27}{8}-\frac{27}{12})+(\frac{16}{3}-6-\frac{27}{12}+\frac{27}{8})$

$=\frac{19}{12}$