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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20 3 question 21

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 21.

Answers (1)

Answer:

12\sqrt{3}\; \text{sq. unnits} .

Hint:

Given:

Given curve y^2=5x^2 and y=2x^2+9 .

Solution:

 

y^2=5x^2    Represents a parabola with vetex(0,0) and opening upward , symmetrical about +ve  axis

y=2x^2+9  Represents the wider parabola, with vertex at C (0,9)

To find point of intersection, solve the two equations

\begin{aligned} 5 x^{2} &=2 x^{2}+9 \\ 3 x^{2} &=\pm 9 \\ x &=\pm \sqrt{3} \\ y &=15 \end{aligned}

Thus A(\sqrt{3},15) and ’A'(-\sqrt{3},15)  are point of intersection of two parabolas.

Shaded area A’OA=2 x area (OCAO)

Consider a vertical stip of length \left |y_2-y_1 \right | and width dx

Area of approximating  rectangle \left |y_2-y_1 \right |dx

The approximating rectangle moves from x=0  to x=\sqrt{3}

\begin{aligned} &\therefore \text { Area }(O C A O)=\int_{0}^{\sqrt{3}}\left|y_{2}-y_{1}\right| d x \\ &=\left(y_{2}-y_{1}\right) d x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \cdot\left\{\cdot\left|y_{2}-y_{1}\right|=y_{2}-y_{1} a s y_{2}>y_{1}\right\} \\ &=\int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x \\ &=\int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x \\ &=\left[\left(9 x-3 \frac{x^{3}}{3}\right)\right]_{0}^{\sqrt{3}} \\ &=9 \sqrt{3}-3 \sqrt{3} \\ &=6 \sqrt{3} \end{aligned}

\therefore  Shaded area B’A’B=2 area OCAO= 2\times 6\sqrt{3}=12\sqrt{3}\; \text{sq.units}

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