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# Find the area bounded by the curve $y=\sqrt x,x=2y+3$ in the first quadrant and $x$-axis.

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

• 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.