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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area bounded by the curve $y=\sqrt x,x=2y+3$ in the first quadrant and $x$-axis.

$\begin{array}{1 1} \frac{29}{3}\;sq.units. \\ \frac{28}{3}\;sq.units. \\ \frac{27}{3}\;sq.units. \\ \frac{19}{3}\;sq.units\end{array} $

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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}$$+c$.
Step 1:
Given $y=\sqrt x$ and $x=2y+3$
Consider the curve $y=\sqrt x$.On squaring we get $y^2=x$.
Clearly this curve is a parabola,which is open rightwards,with vertex $(0,0)$.
Let us now obtain the points of intersection by solving the two equations.
$y^2=2y+3\Rightarrow y^2-2y-3=0$
On factorising we get,
$(y-3)(y+1)=0$
(i.e) $y=3$ or $-1$ and $x=9$ or $1$
The points of intersection are $(9,3)$ and $(1,-1)$.
Step 2:
The required area is the shaded portion shown in the fig.
The area of the shaded portion $A=\int_1^9(y_2-y_1)dx.$
Where $y_2=\sqrt x$ and $y_1=\big(\large\frac{x-3}{2}\big)$
$A=\int_1^9\sqrt xdx-\int_1^9\big(\large\frac{x-3}{2}\big)$
On integrating we get,
$A=\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}_1^9-\begin{bmatrix}\large\frac{x^2}{2}-\normalsize 3x\end{bmatrix}_1^9$
$\;\;=\large\frac{2}{3}\begin{bmatrix}x^{\large\frac{3}{2}}\end{bmatrix}_1^9-\frac{1}{4}\begin{bmatrix}x^2\end{bmatrix}_1^9+\large\frac{3}{2}\begin{bmatrix}x\end{bmatrix}_1^9$
Step 3:
On applying limits we get,
$\;\;=\large\frac{2}{3}\begin{bmatrix}\normalsize 9^{\Large\frac{3}{2}}-1^{\Large\frac{3}{2}}\end{bmatrix}$$-\large\frac{1}{4}$$[9^2-1^2]+\large\frac{3}{2}$$[9-1]$
$\;\;=\large\frac{2}{3}$$[27-1]-\large\frac{1}{4}$$ [80]+\large\frac{3}{2}$$(8)$
$\;\;=\large\frac{52}{3}$$-20+12$
$\;\;=\large\frac{28}{3}$sq.units.
Hence the required area is $\large\frac{28}{3}$sq.units.
answered May 8, 2013 by sreemathi.v
 

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