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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 2 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 2 Maths textbook solution.

Answers (1)

Answer: 4 sq.units

Given:

4y^2=9x

3x^2=16y

Hint:

the bounded area ,A=[Area between the curve (1) and x -axis  from 0 to 4]-[Area between the curve (2)and x -axis  from 0 to 4]

Solution:

The given eqation are,

4y^2=9x

y=\frac{3}{2}\sqrt{x}                        .........(1)

And,

3x^2=16y

y=\frac{3x^2}{16}

Equating (1)and (2)

\begin{aligned} \frac{3}{2} \sqrt{x} &=\frac{3 x^{2}}{16} \\ x^{\frac{2}{2}} &=4^{\frac{3}{2}} \\ x &=4 \end{aligned}

When we put x =4  in eqation (I)

Then y =3 ,

When we put x =0 in eqation (I)

Then y = 3 ,

On solving these two equations,we get the point of intersections

The points are shown in the graph below

https://www.sarthaks.com/?qa=blob&qa_blobid=11099588031339998552

Now the bounded area ,

A= [Area between the curve (1) and x- axis from 0 to 4] - [ Area between the curve (2) and x - axis from 0 to 4]

\begin{aligned} &A=\int_{0}^{4} \frac{3}{2} \sqrt{x} d x-\int_{0}^{4} \frac{3 x^{2}}{16} d x \\ &A=\int_{0}^{4}\left(\frac{3}{2} \sqrt{x}-\frac{3 x^{2}}{16}\right) d x \end{aligned}

On integrating the above definate integration ,

The required area

\begin{aligned} A &=\int_{0}^{4}\left[\frac{3 \sqrt{x}}{2}-\frac{3 x^{2}}{16}\right] d x \\ &=\left[x^{3 / 2}-\frac{x^{3}}{16}\right]_{0}^{4} \\ &=\left[(4)^{3 / 2}-\frac{(4)^{3}}{16}\right] \\ &=\left[8-\frac{64}{16}\right] \\ &=[8-4] \\ &=4 \end{aligned}

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