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+1 vote
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in Mathematics by (29.3k points)

Let the area of the region enclosed by the curves \(y=3 x, 2 y=27-3 x \,and\, y=3 x-x \sqrt{x}\) be A. Then 10 A is equal to

(1) 184 

(2) 154 

(3) 172 

(4) 162

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2 Answers

0 votes
by (15.5k points)

RohitGanguly, hello

Then 10A is equal to: (4) 162

+1 vote
by (30.2k points)
edited by

Correct option is (4) 162  

\(y=3 x, 2 y=27-3 x\) and \(y=3 x-x \sqrt{x}\)

the region enclosed by the curves

\(A=\int_{0}^{3} 3 x-(3 x-x \sqrt{x}) d x+\int_{3}^{9}\left(\frac{27-3 x}{2}-(3 x-x \sqrt{x})\right) d x\)

\(A=\int_{0}^{3} \mathrm{x}^{3 / 2} \mathrm{dx}+\int_{3}^{9} \frac{27}{2}-\frac{9 \mathrm{x}}{2}+\mathrm{x}^{3 / 2} \mathrm{dx}\)

\(\mathrm{A}=\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_{0}^{3}+\frac{27}{2}[\mathrm{x}]_{3}^{9}-\frac{9}{2}\left[\frac{\mathrm{x}^{2}}{2}\right]_{3}^{9}+\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_{3}^{9}\)

\(\mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+\frac{27}{2}(6)-\frac{9}{4}(72)+\frac{2}{5}\left(9^{5 / 2}-3^{5 / 2}\right)\)

\(\mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+81-162+\frac{2}{5} \times 3^{5}-\frac{2}{5} \times 3^{5 / 2}\)

\(\mathrm{A}=\frac{486}{5}-81=\frac{81}{5}\)

\(10 \mathrm{~A}=162\)

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