#### Draw a rough sketch of the curve $y=\sqrt{x-1}$  in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.

$\\ \mathrm{y}=\sqrt{\mathrm{x}-1} \\$

Squaring both sides

$\Rightarrow y^{2}=x-1 \\ \mathrm{y}^{2}=\mathrm{x}-1\text{ is equation of a parabola} \\$

The above equation has no values for x less than 1, therefore parabola will be right of x = 1

Now observe that in  $\mathrm{y}=\sqrt{\mathrm{x}-1} \mathrm{x} \geq 1 \: \: and\: \: \mathrm{y}$  has to positive because of square root hence  $\mathrm{x} \ and\ \mathrm{y}$ both positive hence the parabola will be drawn only in  $1^{\text {st }}$  quadrant

We have to plot the curve in [1,5] so just draw the parabolic curve from  x=1  to  x=5  in  $1^{\text {st }}$ quadrant

x=1  and  x=5  are lines parallel to Y-axis

So we have to integrate  $\mathrm{y}=\sqrt{\mathrm{x}-1}$  from 1 to 5

let us find area under parabolic curve

$\Rightarrow y=\sqrt{x-1}$

Integrate from 1 to 5

$\Rightarrow \int_{1}^{5} y \mathrm{dx}=\int_{1}^{5}(\mathrm{x}-1)^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{1}^{5} y d x=\left[\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{5}\\ \Rightarrow \int_{1}^{5} \mathrm{ydx}=\left[\frac{(\mathrm{x}-1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{5} \\$

$\\\begin{array}{l} \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left[(x-1)^{\frac{3}{2}}\right]_{1}^{5} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left((5-1)^{\frac{3}{2}}-0\right) \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left(2^{2}\right)^{\frac{3}{2}} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3} \times 2^{3} \\ \Rightarrow \quad \int_{1}^{5} y d x=\frac{16}{3} \\ \text { Hence area bounded }=\frac{16}{3} \text { unit }^{2} \end{array} \\$