Question

IF $f\left(x\right)=x^2$ for $0\le x\le 1,\sqrt{x}$ for $1\le x\le 2$ then ${\int }_0^{2}f\left(x\right)dx=$

• A. $\frac{4\sqrt{2}}{3}$
• B. $\frac{4\sqrt{2}-1}{3}$
• C. $\frac{\sqrt{2}}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{4\sqrt{2}-1}{3}$

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

$={\int }_0^1{x}^{2}dx+{\int }_1^2\sqrt{x}dx$

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

$=\frac{1}{3}\left(1\right)+\frac{2}{3}\left(2\sqrt{2}-1\right)$

$=\frac{4\sqrt{2}-2+1}{3}$

$=\frac{4\sqrt{2}-1}{3}$