JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Definite Integrals

We split the integral into two parts.

Part 1: ${\int }_0^2|2x^2-3x|dx$

$2x^2-3x=x(2x-3)=0$ at $x=0$ and $x=\frac{3}{2}$. So $2x^2-3x<0$ for $0<x<\frac{3}{2}$ and $2x^2-3x>0$ for $\frac{3}{2}<x<2$.

${\int }_0^2|2x^2-3x|dx={\int }_0^{3/2}(3x-2x^2)dx+{\int }_{3/2}^2(2x^2-3x)dx$

$={\left[\frac{3x^2}{2}-\frac{2x^3}{3}\right]}_0^{3/2}+{\left[\frac{2x^3}{3}-\frac{3x^2}{2}\right]}_{3/2}^2=\left(\frac{27}{8}-\frac{9}{4}\right)+\left(-\frac{2}{3}+\frac{9}{8}\right)=\frac{9}{8}+\frac{11}{24}=\frac{19}{12}$

Part 2: ${\int }_0^2\left[x-\frac{1}{2}\right]dx$

Split the range according to the value of $x-\frac{1}{2}$:
for $0\le x<\frac{1}{2}$, $-\frac{1}{2}\le x-\frac{1}{2}<0$, so $\left[x-\frac{1}{2}\right]=-1$;
for $\frac{1}{2}\le x<\frac{3}{2}$, $0\le x-\frac{1}{2}<1$, so $\left[x-\frac{1}{2}\right]=0$;
for $\frac{3}{2}\le x\le 2$, $1\le x-\frac{1}{2}\le \frac{3}{2}$, so $\left[x-\frac{1}{2}\right]=1$.

$\therefore {\int }_0^2\left[x-\frac{1}{2}\right]dx=(-1)\left(\frac{1}{2}\right)+(0)(1)+(1)\left(\frac{1}{2}\right)=-\frac{1}{2}+0+\frac{1}{2}=0$

Adding: $\frac{19}{12}+0=\frac{19}{12}$.

Hence option (B).