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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 3 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 3 Maths textbook solution.

Answers (1)

Answer:

\frac{1}{6}  sq.units

Given:

y=\sqrt{x} and

y=x

Hint:

Bounded Area ,

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

Solution:

The given eqation are,

y=\sqrt{x}     (i)and  y=x    (ii)

Solving eqation (1) and( 2)

y^2=x=y

y^2=y

y(y-1)=0'

So, y=0

Or y = 1 and x = 0 or x = 1

On solving these two equations we get the points of intersection .

The point are O(0,0)and A(1,1) these are shown in the graph below

https://www.sarthaks.com/?qa=blob&qa_blobid=14342760982036744916

Now the bounded area is the required area to be calculated,

Hence,Bounded Area ,

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

\begin{aligned} &A=\int_{0}^{1} \sqrt{x} d x-\int_{0}^{1} x d x \\ &A=\int_{0}^{1}(\sqrt{x}-x) d x \end{aligned}

On integrating the above definate integration,

\begin{aligned} &=\int_{0}^{1}\left(y_{1}-y_{2}\right) d x \\ &=\int_{0}^{1}(\sqrt{x}-x) d x \\ &=\left[\frac{2}{3} x \sqrt{x}-\frac{x^{2}}{2}\right]_{0}^{1} \\ &=\left[\frac{2}{3} 1 \sqrt{1}-\frac{(1)^{2}}{2}\right]-[0] \\ &=\left[\frac{2}{3}-\frac{1}{2}\right] \end{aligned}

=\frac{1}{6}  sq.units

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