# 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 $x=5$.

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: Given: y=\sqrt{x-1} in the interval [1,5]. Squaring on both sides we get, y^2=x-1 Clearly this is a parabola with its centre (0,1). It is open rightward and the focus is (a,0).Where 'a' can be obtained as follows : Let y^2=X. When y=y. and X=x-1. The vertex of the parabola is (1,0). Here 4a=1 \Rightarrow a=\large\frac{1}{4}. Now this equating to (x-1) Hence if x-1=\large\frac{1}{4},then x=\large\frac{1}{4}$$+1=\large\frac{5}{4}$
Therefore $a=\large\frac{5}{4}$
Step 2:
Hence the focus of the parabola is $(\large\frac{5}{4}$,0).
Now let us sketch this graph.
This required area is the shaded portion bounded between the lines $x=1$ and $x=5$
To find the area of the required portion,
$A=\int_1^5 ydx.$
$\;\;=\int_1^5\sqrt{x-1}dx=\int(x-1)^{\large\frac{1}{2}}$$dx. Step 3: On integrating we get, \;\;=\begin{bmatrix}\large\frac{(x-1)^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_1^5 \;\;=\large\frac{2}{3}\begin{bmatrix}\normalsize (x-1)^{\Large\frac{3}{2}}\end{bmatrix}_5^1 Now applying limits we get, \;\;=\large\frac{2}{3}\begin{bmatrix}\normalsize(5-1)^{\large\frac{3}{2}}\normalsize-(1-1)^{\large\frac{3}{2}}\end{bmatrix} \;\;=\large\frac{2}{3}\normalsize(4)^{\large\frac{3}{2}}=\large\frac{2}{3}$$\times 2^3$
$\;\;=\large\frac{16}{3}$sq.units.
Hence the required area is $\large\frac{16}{3}$sq.units.
edited Apr 29, 2013