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Area Under the Curves question

2025 · 8 Apr · Shift 2 · Q47
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Area Under the Curves question

2025 · 8 Apr · Shift 2 · Q47

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
Let the area of the bounded region {(x,y):0≤9x≤y2,y≥3x−6}\left\{(x, y): 0 \leq 9 x \leq y^2, y \geq 3 x-6\right\}{(x,y):0≤9x≤y2,y≥3x−6} be AAA. Then 6A6 A6A is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 54

  1. Interpret the region

We need the area of

{(x,y):0≤9x≤y2,y≥3x−6}.\{(x,y): 0\le 9x\le y^2, \quad y\ge 3x-6\}.{(x,y):0≤9x≤y2,y≥3x−6}.

The condition 0≤9x≤y20\le 9x\le y^20≤9x≤y2 means:

  • x≥0x\ge 0x≥0
  • 9x≤y2  ⟹  x≤y299x\le y^2 \implies x\le \dfrac{y^2}{9}9x≤y2⟹x≤9y2​

So the region lies to the right of x=0x=0x=0 and to the left of the parabola

x=y29.x=\frac{y^2}{9}.x=9y2​.

The other condition is

y≥3x−6  ⟹  x≤y+63.y\ge 3x-6 \implies x\le \frac{y+6}{3}.y≥3x−6⟹x≤3y+6​.

So the region is also to the left of the line

x=y+63.x=\frac{y+6}{3}.x=3y+6​.

Hence for a fixed yyy, the allowed xxx values are

0≤x≤min⁡(y29,y+63).0\le x\le \min\left(\frac{y^2}{9},\frac{y+6}{3}\right).0≤x≤min(9y2​,3y+6​).

To get a bounded region, we need the two right-boundary curves to intersect.

  1. Find intersection of the parabola and line

Set

y29=y+63.\frac{y^2}{9}=\frac{y+6}{3}.9y2​=3y+6​.

Multiply by 999:

y2=3y+18.y^2=3y+18.y2=3y+18.

So

y2−3y−18=0y^2-3y-18=0y2−3y−18=0 (y−6)(y+3)=0.(y-6)(y+3)=0.(y−6)(y+3)=0.

Thus,

y=−3,  6.y=-3,\;6.y=−3,6.

The corresponding xxx values are

x=y29.x=\frac{y^2}{9}.x=9y2​.

So:

  • at y=−3y=-3y=−3, x=1x=1x=1
  • at y=6y=6y=6, x=4x=4x=4

Thus the bounded region occurs for

−3≤y≤6.-3\le y\le 6.−3≤y≤6.
  1. Decide which curve is the left/right bound in horizontal slicing

We compare

y29andy+63.\frac{y^2}{9} \quad \text{and} \quad \frac{y+6}{3}.9y2​and3y+6​.

Their difference is

y29−y+63=y2−3y−189=(y−6)(y+3)9.\frac{y^2}{9}-\frac{y+6}{3}=\frac{y^2-3y-18}{9}=\frac{(y-6)(y+3)}{9}.9y2​−3y+6​=9y2−3y−18​=9(y−6)(y+3)​.

For −3<y<6-3<y<6−3<y<6, this is negative, so

y29<y+63.\frac{y^2}{9}<\frac{y+6}{3}.9y2​<3y+6​.

Therefore inside the bounded region,

0≤x≤y29.0\le x\le \frac{y^2}{9}.0≤x≤9y2​.

The line condition is automatically satisfied there.

So the bounded area is simply

A=∫−36y29 dy.A=\int_{-3}^{6} \frac{y^2}{9}\,dy.A=∫−36​9y2​dy.
  1. Compute the area
A=19∫−36y2 dy=19[y33]−36=127(63−(−3)3).A=\frac{1}{9}\int_{-3}^{6} y^2\,dy =\frac{1}{9}\left[\frac{y^3}{3}\right]_{-3}^{6} =\frac{1}{27}\left(6^3-(-3)^3\right).A=91​∫−36​y2dy=91​[3y3​]−36​=271​(63−(−3)3).

Now,

63=216,(−3)3=−27.6^3=216, \qquad (-3)^3=-27.63=216,(−3)3=−27.

Hence

A=127(216−(−27))=24327=9.A=\frac{1}{27}(216-(-27))=\frac{243}{27}=9.A=271​(216−(−27))=27243​=9.

Therefore,

6A=6⋅9=54.6A=6\cdot 9=54.6A=6⋅9=54.
  1. Compare with stored answer

Our derived value is

6A=54.6A=54.6A=54.

The stored correct answer is 151515, which does not match.

So the stored answer appears to be incorrect.

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