Application of Integrals — Class 12 Maths Solution

We have $y = \sqrt x - 1$
$\Rightarrow {y^2} = x - 1$

The graph of above function in parabola with vertex at (1,0) and lying above $x$ -axis.

For $x \in [1,5]$, graph is as shown in the following figure.

From the figure, area of shaded region,
$A = \int_1^5 {{{(x - 1)}^{1/2}}} dx$

$= \left[ {\frac{2}{3}{{(x - 1)}^{3/2}}} \right]_1^5$

$= \left[ {\frac{2}{3}{{(5 - 1)}^{3/2}} - 0} \right] = \frac{{16}}{3}{\rm{sq,}}$ units