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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 4 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 4 Maths textbook solution.

Answers (1)

Answer:\frac{28}{3}   sq.units

Given:y=4-x^2  and the lined y =0, y =3

Hint:

Bounded area ,A = 2 times [area between the equation 1 and y axis from y =0 to y = 3  ]

Solution:

The given equation are ,

y=4-x^2    …(1)

y=0    ..(2)

And y=3    …(3)

Equation 1 represents a parabola with vertex (0,4) and passes through (0,2),(0,2)

Equation 2 is x - axis   and cutting the parabola at C(2,0) and D(-2,0)

Equatin 3 is a line parellel to x - axis  cutting the parabola at A(3,1)and B(-3,1)

On solving these equations ,we get pont of intersection of a parabola with the other two lines are A(3,1) B(-3,1),c(2,0)and D(-2,0) these are shown in the graph below

https://www.sarthaks.com/?qa=blob&qa_blobid=6555794839998903474

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

Hence ,bounded area ,A = 2 times [area between the equation 1 and y axis from y = 0 to y =3 ]

A=-2 \int_{0}^{3} \sqrt{4-y} d y

On intregrating the above definate integration,

\begin{aligned} &=-2\left[\frac{(4-y)^{3 / 2}}{3 / 2}\right]_{0}^{3} \\ &=-2 \frac{2}{3}\left[4^{3 / 2}-1^{3 / 2}\right] \end{aligned}                      ,  as area in not negative.

=\frac{28}{3} sq units

The area bounded by the curved  y=4-x^2  and the lined y=0, y=3 is \frac{28}{3} sq units

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