Question

$\int \left[3.3\right]|x{|}^{3}dx=$
where [.] and |.| represent Greatest integer function and Modulus function

• A. $\frac{-3x^4}{4}+c$
• B. $\frac{x^3}{3}+c$
• C. $\frac{3x|x{|}^3}{4}+c$
• D. $\frac{3x^4}{4}+c$

Answer 1

The correct option is C $\frac{3x|x{|}^3}{4}+c$

$\left[3.3\right]=3$

$I=\int 3{\left|x\right|}^{3}dxI=3\int \left|x\right|{\left|x\right|}^{2}dx{\left|x\right|}^2=x^{2}I=3\int \left|x\right|x^{2}dx$

Using integration by parts;

$u=\left|x\right|u^{\prime }=\frac{\left|x\right|}{x}v=x^2\int vdx=\frac{x^3}{3}I=3\left|x\right|\frac{x^3}{3}-3\int \frac{\left|x\right|}{x}\frac{x^3}{3}dx+CI=\left|x\right|x^3-\int \left|x\right|x^{2}dx+C\int \left|x\right|x^{2}dx=\frac{I}{3}I=\left|x\right|x^3-\frac{I}{3}+C\frac{4}{3}I=\left|x\right|x^3+CI=\frac{3x\cdot x^2\left|x\right|}{4}+CI=\frac{3x\cdot |{x|}^2\left|x\right|}{4}+C\{|x{|}^2=x^2\}\therefore I=\frac{3x{|x|}^3}{4}+C$