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Area Under the Curves question

2021 · 26 Aug · Shift 1 · Q38
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  5. /2021 · 26 Aug · Shift 1 · Q38

Area Under the Curves question

2021 · 26 Aug · Shift 1 · Q38

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
The area of the region S={(x,y):3x2≤4y≤6x+24}S = \{ (x,y):3{x^2} \le 4y \le 6x + 24\}S={(x,y):3x2≤4y≤6x+24} is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 27

  1. We are given the region S={(x,y):3x2≤4y≤6x+24}.S=\{(x,y): 3x^2\le 4y\le 6x+24\}.S={(x,y):3x2≤4y≤6x+24}. This means yyy lies between the curves y=3x24y=\frac{3x^2}{4}y=43x2​ and y=6x+244=3x2+6.y=\frac{6x+24}{4}=\frac{3x}{2}+6.y=46x+24​=23x​+6. So the required area is the area enclosed between the parabola and the line.

  2. First, find the points of intersection: 3x24=3x2+6.\frac{3x^2}{4}=\frac{3x}{2}+6.43x2​=23x​+6. Multiply by 444: 3x2=6x+243x^2=6x+243x2=6x+24 x2=2x+8x^2=2x+8x2=2x+8 x2−2x−8=0x^2-2x-8=0x2−2x−8=0 (x−4)(x+2)=0.(x-4)(x+2)=0.(x−4)(x+2)=0. Hence, the curves intersect at x=−2,  4.x=-2,\;4.x=−2,4.

  3. Determine which curve is above the other between x=−2x=-2x=−2 and x=4x=4x=4. At x=0x=0x=0: 3x24=0,3x2+6=6.\frac{3x^2}{4}=0, \qquad \frac{3x}{2}+6=6.43x2​=0,23x​+6=6. So the line is above the parabola on this interval.

  4. Therefore, the area is A=∫−24(3x2+6−3x24)dx.A=\int_{-2}^{4}\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx.A=∫−24​(23x​+6−43x2​)dx.

  5. Simplify and integrate: A=∫−24(−34x2+32x+6)dx.A=\int_{-2}^{4}\left(-\frac{3}{4}x^2+\frac{3}{2}x+6\right)dx.A=∫−24​(−43​x2+23​x+6)dx. Antiderivative:

=-\frac{x^3}{4}+\frac{3x^2}{4}+6x.$$ So, $$A=\left[-\frac{x^3}{4}+\frac{3x^2}{4}+6x\right]_{-2}^{4}.$$ 6. Evaluate at the limits: For $x=4$, $$-\frac{4^3}{4}+\frac{3\cdot 4^2}{4}+6\cdot 4 =-16+12+24=20.$$ For $x=-2$, $$-\frac{(-2)^3}{4}+\frac{3\cdot (-2)^2}{4}+6(-2) =2+3-12=-7.$$ Thus, $$A=20-(-7)=27.$$ 7. Therefore, the area of the region is $$\boxed{27}.$$ Comparison with stored correct answer: the derived answer is $27$, which matches the stored correct answer.
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