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Area Under the Curves question

2022 · 29 Jul · Shift 1 · Q36
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  5. /2022 · 29 Jul · Shift 1 · Q36

Area Under the Curves question

2022 · 29 Jul · Shift 1 · Q36

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the region {(x,y):∣x−1∣≤y≤5−x2}\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}{(x,y):∣x−1∣≤y≤5−x2​} is equal to :
  1. A
    52sin⁡−1(35)−12\frac{5}{2} \sin ^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}25​sin−1(53​)−21​
  2. B
    5π4−32\frac{5 \pi}{4}-\frac{3}{2}45π​−23​
  3. C
    3π4+32\frac{3 \pi}{4}+\frac{3}{2}43π​+23​
  4. D
    5π4−12\frac{5 \pi}{4}-\frac{1}{2}45π​−21​
View written solutionFree

Correct answer: D

  1. Interpret the region

We need the area of

{(x,y):∣x−1∣≤y≤5−x2}.\{(x,y): |x-1| \le y \le \sqrt{5-x^2}\}.{(x,y):∣x−1∣≤y≤5−x2​}.

So the region lies:

  • above the curve y=∣x−1∣y=|x-1|y=∣x−1∣
  • below the upper semicircle y=5−x2y=\sqrt{5-x^2}y=5−x2​, i.e. x2+y2=5,y≥0.x^2+y^2=5,\quad y\ge 0.x2+y2=5,y≥0.

Thus, area exists only where

∣x−1∣≤5−x2.|x-1| \le \sqrt{5-x^2}.∣x−1∣≤5−x2​.
  1. Find intersection points

Solve

∣x−1∣=5−x2.|x-1| = \sqrt{5-x^2}.∣x−1∣=5−x2​.

Squaring both sides:

(x−1)2=5−x2.(x-1)^2 = 5-x^2.(x−1)2=5−x2.

Expand:

x2−2x+1=5−x2x^2-2x+1 = 5-x^2x2−2x+1=5−x2 2x2−2x−4=02x^2-2x-4=02x2−2x−4=0 x2−x−2=0x^2-x-2=0x2−x−2=0 (x−2)(x+1)=0.(x-2)(x+1)=0.(x−2)(x+1)=0.

So,

x=−1,  2.x=-1,\;2.x=−1,2.

Hence the required area is

∫−12(5−x2−∣x−1∣)dx.\int_{-1}^{2}\left(\sqrt{5-x^2}-|x-1|\right)dx.∫−12​(5−x2​−∣x−1∣)dx.
  1. Break the absolute value

Since

∣x−1∣={1−x,x≤1,x−1,x≥1,|x-1|= \begin{cases} 1-x,& x\le 1,\\ x-1,& x\ge 1, \end{cases}∣x−1∣={1−x,x−1,​x≤1,x≥1,​

we get

A=∫−11(5−x2−(1−x))dx+∫12(5−x2−(x−1))dx.A=\int_{-1}^{1}\left(\sqrt{5-x^2}-(1-x)\right)dx +\int_{1}^{2}\left(\sqrt{5-x^2}-(x-1)\right)dx.A=∫−11​(5−x2​−(1−x))dx+∫12​(5−x2​−(x−1))dx.

Combine the linear parts:

A=∫−125−x2 dx−[∫−11(1−x)dx+∫12(x−1)dx].A=\int_{-1}^{2}\sqrt{5-x^2}\,dx -\left[\int_{-1}^{1}(1-x)dx+\int_{1}^{2}(x-1)dx\right].A=∫−12​5−x2​dx−[∫−11​(1−x)dx+∫12​(x−1)dx].

Now,

∫−11(1−x)dx=[x−x22]−11=2,\int_{-1}^{1}(1-x)dx=\left[x-\frac{x^2}{2}\right]_{-1}^{1}=2,∫−11​(1−x)dx=[x−2x2​]−11​=2,

and

∫12(x−1)dx=[x22−x]12=12.\int_{1}^{2}(x-1)dx=\left[\frac{x^2}{2}-x\right]_{1}^{2}=\frac12.∫12​(x−1)dx=[2x2​−x]12​=21​.

So total subtraction is

2+12=52.2+\frac12=\frac52.2+21​=25​.

Thus,

A=∫−125−x2 dx−52.A=\int_{-1}^{2}\sqrt{5-x^2}\,dx-\frac52.A=∫−12​5−x2​dx−25​.
  1. Evaluate the semicircle integral

Use the standard formula

∫a2−x2 dx=x2a2−x2+a22sin⁡−1(xa).\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).∫a2−x2​dx=2x​a2−x2​+2a2​sin−1(ax​).

Here a=5a=\sqrt5a=5​, so

∫5−x2 dx=x25−x2+52sin⁡−1(x5).\int \sqrt{5-x^2}\,dx=\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt5}\right).∫5−x2​dx=2x​5−x2​+25​sin−1(5​x​).

Therefore,

I=∫−125−x2 dxI=\int_{-1}^{2}\sqrt{5-x^2}\,dxI=∫−12​5−x2​dx

is

I=[x25−x2+52sin⁡−1(x5)]−12.I=\left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt5}\right)\right]_{-1}^{2}.I=[2x​5−x2​+25​sin−1(5​x​)]−12​.

At x=2x=2x=2:

221=1,\frac{2}{2}\sqrt{1}=1,22​1​=1,

so contribution is

1+52sin⁡−1(25).1+\frac{5}{2}\sin^{-1}\left(\frac{2}{\sqrt5}\right).1+25​sin−1(5​2​).

At x=−1x=-1x=−1:

−124=−1,\frac{-1}{2}\sqrt{4}=-1,2−1​4​=−1,

and contribution is

−1+52sin⁡−1(−15)=−1−52sin⁡−1(15).-1+\frac{5}{2}\sin^{-1}\left(\frac{-1}{\sqrt5}\right) =-1-\frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt5}\right).−1+25​sin−1(5​−1​)=−1−25​sin−1(5​1​).

Hence

I=2+52[sin⁡−1(25)+sin⁡−1(15)].I=2+\frac{5}{2}\left[\sin^{-1}\left(\frac{2}{\sqrt5}\right)+\sin^{-1}\left(\frac{1}{\sqrt5}\right)\right].I=2+25​[sin−1(5​2​)+sin−1(5​1​)].

Now let

α=sin⁡−1(25),β=sin⁡−1(15).\alpha=\sin^{-1}\left(\frac{2}{\sqrt5}\right), \qquad \beta=\sin^{-1}\left(\frac{1}{\sqrt5}\right).α=sin−1(5​2​),β=sin−1(5​1​).

Then

sin⁡α=25,cos⁡α=15,\sin\alpha=\frac{2}{\sqrt5},\quad \cos\alpha=\frac{1}{\sqrt5},sinα=5​2​,cosα=5​1​, sin⁡β=15,cos⁡β=25.\sin\beta=\frac{1}{\sqrt5},\quad \cos\beta=\frac{2}{\sqrt5}.sinβ=5​1​,cosβ=5​2​.

So

α+β=π2.\alpha+\beta=\frac{\pi}{2}.α+β=2π​.

Thus

I=2+52⋅π2=2+5π4.I=2+\frac{5}{2}\cdot\frac{\pi}{2}=2+\frac{5\pi}{4}.I=2+25​⋅2π​=2+45π​.

Therefore,

A=(2+5π4)−52=5π4−12.A=\left(2+\frac{5\pi}{4}\right)-\frac52 =\frac{5\pi}{4}-\frac12.A=(2+45π​)−25​=45π​−21​.
  1. Match with the options
A=5π4−12\boxed{A=\frac{5\pi}{4}-\frac12}A=45π​−21​​

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer is also D, so they agree.

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