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  • Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 2

Need solution for RD sharma maths class 12 chapter 20 area of bounded region exercise 20.2 question 2

Answers (1)

\frac{56}{3}sq.units

Hint:

To find the area under two or more than two curves, the first crucial step is to find the intersection points of the curves.

Given:

               x^{2}=16y, y=1, y=4and y-axis in the first quadrant

Solution:

               y=\frac{x^{2}}{16} (Curve)

               y=4 (Line )

               y=1 (Line)

Between curve A and line C

               16=x^{2}\\ x=4

Between curve A and line B

               4=\frac{x^{2}}{16}\\ x=\sqrt{64}\\ x=8

Required area can be calculated by breaking the problem into two parts

  1. Calculate area under the curve A and Line B
  2. Subtract the area enclosed by curve A and line C from the above area.

1. \int_{0}^{8}\left(4-\frac{x^{2}}{16}\right) d x= Area under B and A

               \begin{aligned} &=[4 x]_{0}^{8}-\left[\frac{x^{3}}{16(3)}\right]_{0}^{8} \\ &=\frac{64}{3} s q \cdot \text { units } \end{aligned}

2. \int_{0}^{4}\left(1-\frac{x^{2}}{16}\right) d x=   Area under C and A

               \begin{aligned} &=[x]_{0}^{4}-\left[\frac{x^{3}}{16(3)}\right]_{0}^{4} \\ &=\frac{8}{3} s q \cdot \text { units } \end{aligned}

The required area under the curve

               \frac{64}{3}-\frac{8}{3} =\frac{56}{3} sq.units

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