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  • please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 4 maths textbook solution

please solve RD sharma class 12 chapter 20 Area of bounded Region exercise 20.4 question 4 maths textbook solution

Answers (1)

18 sq.units

Hint:

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

Given:

                x-y=4 and y^{2}=2x

Solution:

 

The parabola y^{2}=2x opens towards the positive x-axis and its focus is \left ( \frac{1}{2},0 \right )

The straight linex-y=4 passes through (4,0) and (0,-4)

Solving  x-y=4 and y^{2}=2x, we get

               \begin{aligned} &y^{2}=2(y+4) \\ &y^{2}-2 y-8=0 \\ &(y-4)(y+2)=0 \\ &y=4,-2 \end{aligned}

So, the points of intersection of given parabola and the line are(8,4) and (2,-2)

Required area,

\begin{aligned} &=\int_{-2}^{4} x_{\text {line }} d y-\int_{-2}^{4} x_{\text {parabola }} d y \\ \end{aligned}

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

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

\begin{aligned} &=\frac{1}{2}(64-6)-\frac{1}{6}[64-(-8)] \\ &=30-12 \\ &=18 \mathrm{sq} \cdot \text { units } \end{aligned}

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