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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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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 $x=5$.

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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.
answered Apr 29, 2013 by sreemathi.v
edited Apr 29, 2013 by sreemathi.v
 

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