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  • please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 1 maths textbook solution

please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 1 maths textbook solution

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best_answer

\frac{32}{3} sq.units

Hint:

Find the point of intersection of parabola and the given line and then find the required region.

Given:

               x=4y-y^{2}and   x=2y-3

Solution:

To find the point at intersection of the parabola x=4y-y^{2}and the line x=2y-3

Let us substitute x=2y-3 in equation of the parabola

               \begin{aligned} &2 y-3=4 y-y^{2} \\ &y^{2}-2 y-3=0 \\ &(y+1)(y-3)=0 \\ &y=-1,3 \end{aligned}

Therefore, the point of intersection are D(-1,-5) and A(3,3)

The area of the required region ABCDOA,

                A=\int_{-1}^{3}\left | x_{1}-x_{2} \right |dywhere  x=4y-y^{2}and   x=2y-3

\begin{aligned} &=\int_{-1}^{3}\left(x_{1}-x_{2}\right) d y \quad\left[\because x_{1}>x_{2}\right] \\ &=\int_{-1}^{3}\left[\left(4 y-y^{2}\right)-(2 y-3)\right] d y \\ \end{aligned}

\begin{aligned} &=\int_{-1}^{3}\left[4 y-y^{2}-2 y+3\right] d y \\ &=\int_{-1}^{3}\left[-y^{2}+2 y+3\right] d y \\ \end{aligned}

\begin{aligned} &=\left[\frac{-y^{3}}{3}+y^{2}+3 y\right]_{-1}^{3} \\ &=\left[\frac{-3^{3}}{3}+3^{2}+3(3)-\left(\frac{1}{3}+1-3\right)\right] \\ \end{aligned}

\begin{aligned} &=\left[-3^{2}+3^{2}+9-\frac{1}{3}-1+3\right] \\ &=11-\frac{1}{3} \\ &=\frac{32}{3} s q \cdot \text { units } \end{aligned}

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