Get Answers to all your Questions

header-bg qa

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

Answers (1)

\begin{aligned} &y=\sqrt{x}\\ &\text { Squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\

The above equation depicts a parabola which does not define negative values for x and therefore it lies to the right of the Y axis and passes through origin. 

For plotting y = \sqrt x, both the values of x and y have to be positive hence this graph lies in the 1st quadrant only. 

The point cannot be negative because the square root symbol itself depicts the positive root. 

The equation x = 2y + 3 is a straight line. 

For finding the point of interaction, solve the two equations simultaneously. 
\\ \text { Put } x=2 y+3 \text { in } y^{2}=x \\ \Rightarrow y^{2}=2 y+3 \\ \Rightarrow y^{2}-2 y-3=0 \\ \Rightarrow y^{2}-3 y+y-3=0 \\ \Rightarrow y(y-3)+1(y-3)=0 \\ \Rightarrow(y+1)(y-3)=0 \\ \Rightarrow y=-1 \text { and } y=3 \\ \text { Put } y=3 \text { in } y^{2}=x \\ \Rightarrow x=3^{2} \\ \Rightarrow x=9 \\

Y = -1 won’t make a difference since we are looking for the 1st quadrant only. 

Therefore, the point of interaction are (9, 3) 

Below is the figure shows the area to be calculated

Point of interaction of straight line on X axis can be calculated by substituting y = 0 in the equation of the line 
              \item X = 2(0) + 3 => x = 3


The area enclosed = area under y = \sqrt x – area under the straight line   \ldots (1) \\


For the area under y = \sqrt x  integrate the equation from 0 to 9 
\begin{aligned} &\Rightarrow \int_{0}^{9} y d x=\int_{0}^{9} x^{\frac{1}{2}} d x\\ &\Rightarrow \int_{0}^{9} y d x=\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[9^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[\left(3^{2}\right)^{\frac{3}{2}}\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3} \times 3^{3} \end{aligned} \\

For the area under straight line 

Y = (x-3) / 2 


Integrating the equation from 3 to 9

\\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2} \int_{3}^{9}(x-3) d x \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left[\frac{x^{2}}{2}-3 x\right]_{3}^{9} \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{9^{2}}{2}-3(9)\right)-\left(\frac{3^{2}}{2}-3(3)\right)\right) \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{81}{2}-27\right)-\left(\frac{9}{2}-9\right)\right) \\ \Rightarrow \int_{0}^{9} y d x=\frac{1}{2} \times 18

\Rightarrow \int_{0}^{9} \mathrm{ydx}=9

Using (i)

 \Rightarrow  area bounded  =18-9=9  unit ^{2}

Hence area bounded by the curve and a straight line is 9 unit ^{2}

Posted by


View full answer