The area bounded by curves $3 x^{2} + 5 y = 32$ and $y = | x - 2 |$ is

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17/2

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33/2

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Solution

The correct option is C 33/2
$\begin{matrix} 3 x^{2} + 5 y = 32 D r a w a p a r a b o l a . s y m m e t r i c a l a b o u t n e g a t i v e x - a x i s A l s o, y = | x - 2 | \Rightarrow y = (\begin{matrix} x - 2, x \geq 2 2 - x, x < 2 \end{matrix}) y = x - 2, x \geq 2 r e p r e s e n t s a l i n e, c u t t i n g t h e x - a x i s a t A (2, 0) a n d t h e p a r a b o l a C (3, 1) A n d, y = 2 - x, x < 2 r e p r e s e n t s a l i n e, c u t t i n g t h e p a r a b o l a a t (- 2, 4) S h a d e d a r e a A E Y C A = \int_{- 2}^{2} [(\frac{32 - 32 x^{2}}{5}) - (2 - x)] d x + \int_{2}^{3} [(\frac{32 - 32 x^{2}}{5}) - (x - 2)] d x = \int_{- 2}^{3} (\frac{32 - 32 x^{2}}{5}) d x - \int_{- 2}^{2} (2 - x) d x + \int_{2}^{3} (x - 2) d x = \frac{1}{5} {[32 - x]}_{- 2}^{3} - {[2 x - \frac{x^{2}}{2}]}_{- 2}^{2} - {[\frac{x^{2}}{2} - 3 x]}_{2}^{3} = \frac{1}{5} [32 \times 3 - 27 + 64 - 8] - (4 - 2 + 4 + 2) - (\frac{9}{2} - 6 - 2 + 4) = \frac{1}{5} [96 - 35 + 64] - (8) - (\frac{9}{2} - 4) = \frac{125}{5} - 4 - \frac{9}{2} = 25 - 4 - \frac{9}{2} = \frac{42 - 9}{2} = \frac{33}{2} s q . u n i t s . H e n c e, o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

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