Question

Calculate the area under the curve $y=2\sqrt{x}$ included between the lines $x=0$ & $x=1$

Answer 1

$\because y=2\sqrt{x}$

Squaring both sides
$\Rightarrow y^2=4x,x\ge 0,y\ge 0$

Graph of above function is a part of parabola lying above $x$ axis

We have ,$y=2\sqrt{x}\cdots \left(1\right)$
$x=0\cdots \left(2\right)$
& $x=1\cdots \left(3\right)$

when $x=0\Rightarrow y=0$ and
when $x=1\Rightarrow y=2$

Hence $(0,0)$and $(1,2)$are the intersection points.

So the area bounded by curves is shaded in the diagram below:


$\Rightarrow \text{Area}={\int }_{x_1}^{x_2}(y_2-y_1)dx$

$\Rightarrow \text{Area}={\int }_0^1(2\sqrt{x}-0)dx$
$[\because x$ varies from $0\text{to}1$ ]

$\Rightarrow \text{Area}=2{\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1$
$[\because {\int }_a^b{x}^{n}dx={\left[\frac{x^{n+1}}{n+1}\right]}_a^b]$

$\Rightarrow \text{Area}=\frac{4}{3}sq.units$

Hence, the required area is $\frac{4}{3}sq.units$.