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  • Find the area of the region enclosed by the parabola x^2 = y and the line y = x + 2

Find the area of the region enclosed by the parabola x^2 = y and the line y = x + 2

Answers (1)

The equation x\textsuperscript{2} = y depicts a parabola and has no negative values for y, therefore, it lies above X axis and passing through origin (0, 0). 

 

And y = x + 2 is a straight line. 

 

Below figure shows the area to be found out, 

Let’s find the point of interaction of the two equations. 

\\\text{Put } y=x+2 in x^{2}=y\\ \Rightarrow x^{2}=x+2\\ \Rightarrow x^{2}-x-2=0\\ \Rightarrow x^{2}-2 x+x-2=0\\ \Rightarrow x(x-2)+1(x-2)=0\\ \Rightarrow(x+1)(x-2)=0\\ \Rightarrow x=-1 \text{ and } x=2\\ \text{Put } x=-1 \text{ and } x=2 in x^{2}=y \text{ we get } y=1 \text{ and } y=4 respectively \\

The point of interaction was found to be (-1, 1) and (2, 4) 

 

Area between the line and parabola = area under line – area under parabola 

 

Let us find area under line  y=x+2

\\ \Rightarrow y=x+2\\ \text{Integrate from -1 to 2} \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}(x+2) d x\\ \Rightarrow \int_{-1}^{2} y \mathrm{~d} x=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}\\ \Rightarrow \int_{-1}^{2} y d x=\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{(-1)^{2}}{2}+2(-1)\right)\right]\\ \Rightarrow \int_{-1}^{2} y d x=6-\left(\frac{1}{2}-2\right) \\ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=6-\left(\frac{-3}{2}\right)\\ \Rightarrow \int_{-1}^{2} y \mathrm{dx}=6+\frac{3}{2}\\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\frac{15}{2}\\

Now let us find area under the parabola

\\ x^{2}=y \Rightarrow y=x^{2}\\ \text{Integrate from -1 to 2 }\\ \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2} x^{2} d x \\

\\\begin{aligned} &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=3\\ &\text { Using (i) }\\ &\text { area enclosed by line and parabola }=\frac{15}{2}-3=\frac{9}{2} \text { unit }^{2} \end{aligned} \\

Therefore, the area was found to be 

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