Question

Calculate the area under the curve $y=2\sqrt{x}$ and included between the lines $x=0,x=4$.

• (A) $\frac{32}{5}$
• (B) $\frac{32}{3}$
• (C) $\frac{31}{2}$
• (D) $\frac{3}{2}$

Answer 1

The Correct Option is B

Explanation:
Given here, $y=2\sqrt{x}$$\therefore {\mathrm{y}}^2=4\mathrm{x}(\mathrm{x}\ge 0)$We need to determine area under curve ${\mathrm{y}}^2=4\mathrm{x}(\mathrm{x}\ge 0)$ included between the lines $\mathrm{x}=0,\mathrm{x}=4$ as:

So, the area under curve will be given as:$={\int }_0^{4}2\sqrt{\mathrm{x}}\mathrm{dx}$$=2{\int }_0^4{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx}$We know that:$\int x^{n}dx=\frac{x^{n+1}}{n+1}$$x^{m+n}=\left(x^m\right)\left(x^n\right)$Therefore,$2{\int }_0^4{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx}$$=2{\left[\frac{x^{1+\frac{1}{2}}}{1+\left(\frac{1}{2}\right)}\right]}_0^4$$=2{\left[\frac{x^{1+\left(\frac{1}{2}\right)}}{\frac{3}{2}}\right]}_0^4$$=2\left[\frac{2}{3}((4{)}^{1+\left(\frac{1}{2}\right)}-0)\right]$$=2\left[\frac{2}{3}((4{)}^1(4{)}^{\frac{1}{2}})\right]$$=\frac{(2)(2)(4)(2)}{3}$$=\frac{32}{3}$ sq unitsHence, the correct option is (B).