Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^2& \text{for}0\le x\le 1\\ \sqrt{x}& \text{for}1\le x\le 2\end{array}$ then ${\int }_0^{2}f\left(x\right)dx=$

• A. $\left(1/3\right)(4\sqrt{2}+1)$
• B. $\left(1/3\right)(4\sqrt{2}-1)$
• C. $\left(1/3\right)(2\sqrt{2}-1)$
• D. $\left(1/2\right)(3\sqrt{2}-1)$

Answer 1

Answer: (B)

$\left(1/3\right)(4\sqrt{2}-1)$

given, $f\left(x\right)=\{\begin{array}{ccc}{x}^2& 0⩽x& ⩽1\\ \sqrt{x}& 1⩽x& ⩽2\end{array}$

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

now

$\Rightarrow$ ${\int }_0^{1}f\left(x\right)dx={\int }_0^1{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{3}$

and

$\Rightarrow$ ${\int }_1^{2}f\left(x\right)dx={\int }_1^2\sqrt{x}dx={\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_1^2=\frac{2}{3}(2\sqrt{2}-1)$

then

$\Rightarrow$ ${\int }_0^{2}f\left(x\right)dx=\frac{1}{3}+\frac{2}{3}\left(2\sqrt{2}\right)-\frac{2}{3}$

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

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

$\Rightarrow$ ${\int }_0^{2}f\left(x\right)dx=\frac{1}{3}(4\sqrt{2}-1)$