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Area Under the Curves question

2023 · 6 Apr · Shift 1 · Q37
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  5. /2023 · 6 Apr · Shift 1 · Q37

Area Under the Curves question

2023 · 6 Apr · Shift 1 · Q37

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
If the area of the region S={(x,y):2y−y2≤x2≤2y,x≥y}S=\left\{(x, y): 2 y-y^{2} \leq x^{2} \leq 2 y, x \geq y\right\}S={(x,y):2y−y2≤x2≤2y,x≥y} is equal to n+2n+1−πn−1\frac{n+2}{n+1}-\frac{\pi}{n-1}n+1n+2​−n−1π​, then the natural number nnn is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 5

  1. Interpret the region

We need the area of

S={(x,y):2y−y2≤x2≤2y, x≥y}.S=\{(x,y): 2y-y^2\le x^2\le 2y,\ x\ge y\}.S={(x,y):2y−y2≤x2≤2y, x≥y}.

So the point must satisfy both:

  • x2≤2y  ⟹  y≥x22,x^2\le 2y \implies y\ge \frac{x^2}{2},x2≤2y⟹y≥2x2​,
  • x2≥2y−y2.x^2\ge 2y-y^2.x2≥2y−y2.

Rewrite the second inequality:

x2≥2y−y2=1−(y−1)2.x^2\ge 2y-y^2 = 1-(y-1)^2.x2≥2y−y2=1−(y−1)2.

Equivalently,

x2+(y−1)2≥1.x^2+(y-1)^2\ge 1.x2+(y−1)2≥1.

So the region lies outside the circle centered at (0,1)(0,1)(0,1) with radius 111.

Also, from x2≤2yx^2\le 2yx2≤2y we must have y≥0y\ge 0y≥0.

Finally, we also need x≥y.x\ge y.x≥y.

  1. Express in terms of yyy

For fixed yyy, the inequalities become

Hence 2y−y2≤∣x∣≤2y,\sqrt{2y-y^2}\le |x|\le \sqrt{2y},2y−y2​≤∣x∣≤2y​, whenever 2y−y2≥02y-y^2\ge 02y−y2≥0, i.e. 0≤y≤20\le y\le 20≤y≤2.

But because of the condition x≥yx\ge yx≥y, only the positive side can survive. So we take x≥0,x∈[2y−y2,2y],x\ge 0,\qquad x\in [\sqrt{2y-y^2},\sqrt{2y}],x≥0,x∈[2y−y2​,2y​], with also x≥y.x\ge y.x≥y.

Thus for fixed yyy, the actual lower bound is x≥max⁡(y,2y−y2),x\ge \max\left(y,\sqrt{2y-y^2}\right),x≥max(y,2y−y2​), while upper bound is x≤2y.x\le \sqrt{2y}.x≤2y​.

So the horizontal length is L(y)=2y−max⁡(y,2y−y2).L(y)=\sqrt{2y}-\max\left(y,\sqrt{2y-y^2}\right).L(y)=2y​−max(y,2y−y2​).

We must find where yyy exceeds 2y−y2\sqrt{2y-y^2}2y−y2​.

  1. Compare yyy and 2y−y2\sqrt{2y-y^2}2y−y2​

Solve

Squaring:

  ⟹  2y2−2y=0  ⟹  2y(y−1)=0.\implies 2y^2-2y=0 \implies 2y(y-1)=0.⟹2y2−2y=0⟹2y(y−1)=0.

Hence

Now for 0<y<10<y<10<y<1, 2y−y2>y,\sqrt{2y-y^2}>y,2y−y2​>y, and for 1<y≤21<y\le 21<y≤2, y>2y−y2.y>\sqrt{2y-y^2}.y>2y−y2​.

Therefore,

  • for 0≤y≤1,0\le y\le 1,0≤y≤1, lower bound is 2y−y2,\sqrt{2y-y^2},2y−y2​,
  • for 1≤y≤2,1\le y\le 2,1≤y≤2, lower bound is y.y.y.

So area is

A=∫01(2y−2y−y2)dy+∫12(2y−y)dy.A=\int_0^1\left(\sqrt{2y}-\sqrt{2y-y^2}\right)dy +\int_1^2\left(\sqrt{2y}-y\right)dy.A=∫01​(2y​−2y−y2​)dy+∫12​(2y​−y)dy.
  1. Evaluate the first integral

Let

First term:

∫012y dy=2∫01y1/2dy=2⋅23=223.\int_0^1\sqrt{2y}\,dy=\sqrt{2}\int_0^1 y^{1/2}dy =\sqrt{2}\cdot \frac{2}{3} =\frac{2\sqrt2}{3}.∫01​2y​dy=2​∫01​y1/2dy=2​⋅32​=322​​.

Second term:

2y−y2=1−(y−1)2.2y-y^2=1-(y-1)^2.2y−y2=1−(y−1)2.

Thus

∫012y−y2 dy=∫011−(y−1)2 dy.\int_0^1\sqrt{2y-y^2}\,dy =\int_0^1\sqrt{1-(y-1)^2}\,dy.∫01​2y−y2​dy=∫01​1−(y−1)2​dy.

This is the area of a quarter of the unit circle, so

∫011−(y−1)2 dy=π4.\int_0^1\sqrt{1-(y-1)^2}\,dy=\frac{\pi}{4}.∫01​1−(y−1)2​dy=4π​.

Hence

I1=223−π4.I_1=\frac{2\sqrt2}{3}-\frac{\pi}{4}.I1​=322​​−4π​.
  1. Evaluate the second integral

Let

Now,

∫122y dy=2∫12y1/2dy=2⋅23(23/2−1).\int_1^2\sqrt{2y}\,dy=\sqrt2\int_1^2 y^{1/2}dy =\sqrt2\cdot \frac{2}{3}\left(2^{3/2}-1\right).∫12​2y​dy=2​∫12​y1/2dy=2​⋅32​(23/2−1).

Since 23/2=22,2^{3/2}=2\sqrt2,23/2=22​,

∫122y dy=223(22−1)=83−223.\int_1^2\sqrt{2y}\,dy =\frac{2\sqrt2}{3}(2\sqrt2-1) =\frac{8}{3}-\frac{2\sqrt2}{3}.∫12​2y​dy=322​​(22​−1)=38​−322​​.

Also,

∫12y dy=[y22]12=4−12=32.\int_1^2 y\,dy=\left[\frac{y^2}{2}\right]_1^2=\frac{4-1}{2}=\frac32.∫12​ydy=[2y2​]12​=24−1​=23​.

Therefore,

I2=(83−223)−32=76−223.I_2=\left(\frac{8}{3}-\frac{2\sqrt2}{3}\right)-\frac32 =\frac{7}{6}-\frac{2\sqrt2}{3}.I2​=(38​−322​​)−23​=67​−322​​.
  1. Total area
A=I1+I2=(223−π4)+(76−223)=76−π4.A=I_1+I_2 =\left(\frac{2\sqrt2}{3}-\frac{\pi}{4}\right)+\left(\frac{7}{6}-\frac{2\sqrt2}{3}\right) =\frac{7}{6}-\frac{\pi}{4}.A=I1​+I2​=(322​​−4π​)+(67​−322​​)=67​−4π​.

So the area is

A=76−π4.A=\frac{7}{6}-\frac{\pi}{4}.A=67​−4π​.
  1. Match with the given form

Given

A=n+2n+1−πn−1.A=\frac{n+2}{n+1}-\frac{\pi}{n-1}.A=n+1n+2​−n−1π​.

Comparing,

n+2n+1=76,1n−1=14.\frac{n+2}{n+1}=\frac76, \qquad \frac{1}{n-1}=\frac14.n+1n+2​=67​,n−11​=41​.

From the second equation,

Check first equation:

n+2n+1=76  ⟹  n=5,\frac{n+2}{n+1}=\frac{7}{6} \implies n=5,n+1n+2​=67​⟹n=5,

which matches.

Therefore, 5.\boxed{5}.5​.

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