Question

The value of definite integral ${\int }_{-1}^2\left(x^3-x\right|\mathrm{dx}$ is

• A. $2$
• B. $3$
• C. $\frac{13}{4}$
• D. $\frac{11}{4}$

Answer 1

The correct option is D $\frac{11}{4}$

We have, ${\int }_{-1}^2\left(x^3-x\right|dxClearly,f\left(x\right)=x^3-x=x(x-1)(x+1)Thesignsoff\left(x\right)fordifferentvaluesofxareshownas$

$\mathrm{We}\mathrm{observe}\mathrm{that}:f\left(x\right)>0\mathrm{for}\mathrm{all}x\in \left(-1,0\right)\cup \left(1,2\right)\mathrm{and},f\left(x\right)<0\mathrm{for}\mathrm{all}x\in (0,1)\left(f\left(x\right)\right|=(\begin{array}{c}{x}^3-x,\mathrm{if}x\in \left(-1,0\right)\cup \left(1,2\right)\\ -\left(x^3-x\right),\mathrm{if}x\in \left(0,1\right)\end{array}\therefore I={\int }_{-1}^0|x^3-x|\mathrm{dx}+{\int }_0^1|x^3-x|\mathrm{dx}+{\int }_1^2|x^3-x|\mathrm{dx}\Rightarrow I={\int }_{-1}^0(x^3-x)\mathrm{dx}-{\int }_0^1(x^3-x)\mathrm{dx}+{\int }_1^2(x^3-x)\mathrm{dx}\Rightarrow I={\left(\frac{x^4}{4}-\frac{x^2}{2}\right]}_{-1}^0-{\left(\frac{x^4}{4}-\frac{x^2}{2}\right]}_0^1+{\left(\frac{x^4}{4}-\frac{x^2}{2}\right]}_1^2\Rightarrow I=-(\frac{1}{4}-\frac{1}{2})-(\frac{1}{4}-\frac{1}{2})+(\frac{16}{4}-\frac{4}{4})-(\frac{1}{4}-\frac{1}{2})\Rightarrow I=\frac{3}{4}+2=\frac{11}{4}$