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Q.

The area of the region bounded by the curves y = 9x² and y = 5x² + 4 (in square units) is

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a

64

b

$\frac{16}{3}$

c

$\frac{64}{3}$

d

$\frac{32}{3}$

answer is D.

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Detailed Solution

Question Image

$9 x^{2} = 5 x^{2} + 4 \Rightarrow 4 x^{2} = 4 \Rightarrow x = \pm 1$

$A r e a = \int_{- 1}^{1} (5 x^{2} + 4) - (9 x^{2}) d x = \int_{- 1}^{1} (4) - (4 x^{2}) d x = 2 \times 4 \int_{0}^{1} (1 - x^{2}) d x$

$a r e a = 8 {(x - \frac{x^{3}}{3})}_{0}^{1} = 8 (1 - \frac{1}{3}) = \frac{16}{3}$

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