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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Area Under Curves

Let the area of the region enclosed by the curves y=3 x, 2 y=27-3 x...

Let the area of the region enclosed by the curves y=3x,2y=27−3xy=3 x, 2 y=27-3 x and y=3x−xxy=3 x-x \sqrt{x} be AA. Then 10A10 A is equal to

Options

  1. A172
  2. B162
  3. C154
  4. D184

Correct answer

B. 162

Step-by-step solution

y=3x,2y=27−3x&y=3x−xxy=3 x, 2 y=27-3 x \& y=3 x-x \sqrt{x}A=∫033x−(3x−xx)dx+∫39(27−3x2−(3x−xx))dxA=∫03x3/2dx+∫39272−9x2+x3/2dxA=[2x5/25]03+272[x]39−92[x22]39+[2x5/25]39A=25(35/2)+272(6)−94(72)+25(95/2−35/2)A=25(35/2)+81−162+25×35−25×35/2A=4865−81=81510A=162Ans. =(2)\begin{aligned} & \mathrm{A}=\int_0^3 3 \mathrm{x}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}}) \mathrm{dx}+\int_3^9\left(\frac{27-3 \mathrm{x}}{2}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}})\right) \mathrm{dx} \\ & \mathrm{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 \\ & \text{Ans. }=(2)\end{aligned}$$
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