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  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 27.

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 27.

Answers (1)

Answer:

\frac{64}{3}\; \text{sq. units}

Hint:

use concept.

Given:

The given equations are y=x-1 and (y-1)^2=4(x+1)

Solution:

We have.y=x-1 and (y-1)^2=4(x+1)

\therefore (x-1-1)^2=4(x+1)

\begin{aligned} &(x-2)^{2}=4(x+1) \\ &x^{2}+4-4 x=4 x+4 \\ &x^{2}+4-4 x-4 x-4=0 \\ &x^{2}-8 x=0 \\ &x=0 \text { or } x=8 \\ &\therefore y=-1 \text { or } 7 \end{aligned}

Consider a horizontal strip of length \left |x_2-x_1 \right |  and width dy where P(x_2,y)  lies on straightline and Q(x_1,y) lies on the parabola.

Area of approximating rectangle=\left |x_2-x_1 \right |  and it moves from y =1 to y =- 7

\text { Required area }=\operatorname{area}(O A D O)=\int_{1}^{7}\left|x_{2}-x_{1}\right| d y

\begin{aligned} &=\int_{-1}^{7}\left|x_{2}-x_{1}\right| d y \ldots \ldots \ldots\left\{\therefore\left|x_{2}-x_{1}\right|=x_{2}-x_{1} \text { as } x_{2}>x_{1}\right\} \\ &=\int_{-1}^{7}\left[(1+y)-\frac{1}{4}\left\{(y-1)^{2}-4\right\}\right] d y \\ &=\int_{-1}^{7}\left\{1+y-\frac{1}{4}(y-1)^{2}+1\right\} d y \\ &=\int_{-1}^{7}\left\{2+y-\frac{1}{4}(y-1)^{2}\right\} d y \end{aligned}

\begin{aligned} &=\left[2 y+\frac{y^{2}}{2}-\frac{1}{12}(y-1)^{3}\right]_{-1}^{7} \\ &=\left[14+\frac{49}{2}-\frac{1}{2} \times 6 \times 6 \times 6\right]-\left[-2+\frac{1}{2}+\frac{1}{12} \times 2 \times 2 \times 2\right] \\ &=\left[14+\frac{49}{2}-18\right]-\left[-2+\frac{1}{2}+\frac{2}{3}\right] \\ &=\left[\frac{41}{2}+\frac{5}{6}\right] \\ &=\frac{64}{3} \text { sq.units } \end{aligned}

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