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  • Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 4

Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 4

Answers (1)

\frac{1}{96}Hint:

Find the latus rectum and its intersection points with the parabola and then integrate the expression to find the area enclosed by the curves.

Given:

               x^{2}+16y=0

Solution:

Curve is x^{2}+16y=0

To find the enclosed area

\begin{aligned} &2 \times \int_{0}^{8}\left(\frac{-x^{2}}{16}-(-4)\right) d x \\ \end{aligned}

\begin{aligned} &=2 \times\left[\frac{-x^{3}}{16(3)}+4 x\right]_{0}^{8} \\ &=2 \times\left[\frac{8^{3}-0^{3}}{48}-4(8-0)\right] \\ \end{aligned}

\begin{aligned} &=2 \times\left[\frac{512}{48}-32\right] \\ &=2 \times\left[\frac{32}{3}-32\right] \\ \end{aligned}

\begin{aligned} &=2 \times \frac{64}{3} \\ &=\frac{128}{3} \end{aligned}

So, the answer is \begin{aligned} &\frac{128}{3} \end{aligned} sq. Units

 

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