Question

If $f\left(x\right)=\{\begin{array}{c}2x^2+1,x\le 1\\ 4x^3-1,x>1\end{array}$, then ${\int }_0^{2}f\left(x\right)dx$ is

• A. $\frac{47}{3}$
• B. $\frac{50}{3}$
• C. $\frac{1}{3}$
• D. $\frac{47}{2}$

Answer 1

The correct option is A $\frac{47}{3}$

Given,

$f\left(x\right)=\{\begin{array}{c}2x^2+1,x\le 1\\ 4x^3-1,x>1\end{array}$

$\therefore {\int }_0^{2}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx$

$={\int }_0^1(2x^2+1)dx+{\int }_1^2(4x^3-1)dx$

$={[\frac{2x^3}{3}+x]}_0^1+{[\frac{4x^4}{4}-x]}_1^2$

$=\frac{2}{3}{\left(1\right)}^3+1-(0+0)+[{\left(2\right)}^4-2-\{{\left(1\right)}^4-1\}]$

$=\frac{2}{3}+1+[16-2-0]$

$=\frac{2}{3}+15=\frac{2+45}{3}$

$=\frac{47}{3}$sq units