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  • please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 3 sub question 2 maths textbook solution

please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 3 sub question 2 maths textbook solution

Answers (1)

\frac{23}{3}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 vertical strips:

The required area is the shaded region

A=\int_{0}^{4}y_{2}dx-\int_{1}^{4}y_{1}dxwhere  y_{2}=2\sqrt{2} and y_{1}=2x-4

     \begin{aligned} &=\int_{0}^{4}(2 \sqrt{x}) d x-\int_{1}^{4}(2 x-4) d x \\ &=\left[\frac{4}{3} x^{\frac{3}{2}}\right]_{0}^{4}-\left[x^{2}-4 x\right]_{1}^{4} \\ \end{aligned}

\begin{aligned} &=\left[\frac{4}{3}(4)^{\frac{3}{2}}\right]-\left[\frac{4}{3}(0)^{\frac{3}{2}}\right]-\left[4^{2}-4(4)\right]-[1-4(1)] \\ \end{aligned}

\begin{aligned} &=\left[\frac{32}{3}-0\right]-[0-(1-4)] \\ &=\frac{32}{3}-3 \\ &=\frac{23}{3} s q \cdot \text { units } \end{aligned}

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