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  • please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 3 sub question 1 maths textbook solution

please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 3 sub question 1 maths textbook solution

Answers (1)

9 sq.units

Hint:

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

Given:

              y=2x-4 and y^{2}=4x

Solution:

Substitute y=2x-4 in y^{2}=4x

\begin{aligned} &(2 x-4)^{2}=4 x \\ &4 x^{2}+16-16 x=4 x \\ &4 x^{2}-20 x+16=0 \\ \end{aligned}

\begin{aligned} &x^{2}-5 x+4=0 \\ &(x-1)(x-4)=0 \\ &x=1,4 \\ &y=-2,4 \end{aligned}

Therefore point of intersection are (1,-2) and (4,4)

Using horizontal strips:

The required area is the shaded region

\begin{aligned} A &=\int_{-2}^{4}\left(x_{1}-x_{2}\right) d y \end{aligned}  where  \begin{aligned} x_{1}=\frac{y+4}{2} \end{aligned} and \begin{aligned} x_{2}=\frac{y^{2}}{4} \\ \end{aligned}

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

\begin{aligned} &=\left[\frac{y^{2}}{4}+2 y-\frac{y^{3}}{12}\right]_{-2}^{4} \\ \end{aligned}

\begin{aligned} &=\left[\frac{4^{2}}{4}+2(4)-\frac{4^{3}}{12}\right]-\left[\frac{(-2)^{2}}{4}+2(-2)-\frac{(-2)^{3}}{12}\right] \\ \end{aligned}

\begin{aligned} &=12-\frac{16}{3}+3-\frac{2}{3} \\ &=9 \mathrm{sq} \cdot \text { units } \end{aligned}

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