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Area Under the Curves question

2024 · 30 Jan · Shift 2 · Q57
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Area Under the Curves question

2024 · 30 Jan · Shift 2 · Q57

JEE MainMathematicsArea Under the CurvesNumerical+4 / −1
The area of the region enclosed by the parabola (y−2)2=x−1(y-2)^2=x-1(y−2)2=x−1, the line x−2y+4=0x-2 y+4=0x−2y+4=0 and the positive coordinate axes is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Write the given curves in convenient form

The parabola is

(y−2)2=x−1⇒x=(y−2)2+1=y2−4y+5.(y-2)^2=x-1 \quad \Rightarrow \quad x=(y-2)^2+1=y^2-4y+5.(y−2)2=x−1⇒x=(y−2)2+1=y2−4y+5.

The line is

x−2y+4=0⇒x=2y−4.x-2y+4=0 \quad \Rightarrow \quad x=2y-4.x−2y+4=0⇒x=2y−4.

We are also given the positive coordinate axes, i.e. x=0x=0x=0 and y=0y=0y=0.

  1. Find relevant intersection points

(a) Parabola with xxx-axis

On the xxx-axis, y=0y=0y=0:

x=(0−2)2+1=5.x=(0-2)^2+1=5.x=(0−2)2+1=5.

So point is (5,0)(5,0)(5,0).

(b) Line with xxx-axis

On the xxx-axis, y=0y=0y=0:

x−0+4=0⇒x=−4.x-0+4=0 \Rightarrow x=-4.x−0+4=0⇒x=−4.

This is not on the positive xxx-axis, so not relevant.

(c) Line with yyy-axis

On the yyy-axis, x=0x=0x=0:

−2y+4=0⇒y=2.-2y+4=0 \Rightarrow y=2.−2y+4=0⇒y=2.

So point is (0,2)(0,2)(0,2).

(d) Parabola with yyy-axis

On the yyy-axis, x=0x=0x=0:

(y−2)2=−1,(y-2)^2=-1,(y−2)2=−1,

which has no real solution. So parabola does not meet the yyy-axis.

(e) Parabola with line

Solve

(y−2)2+1=2y−4.(y-2)^2+1=2y-4.(y−2)2+1=2y−4.

Expanding:

y2−4y+5=2y−4y^2-4y+5=2y-4y2−4y+5=2y−4 y2−6y+9=0y^2-6y+9=0y2−6y+9=0 (y−3)2=0.(y-3)^2=0.(y−3)2=0.

Thus y=3y=3y=3, and then

x=2(3)−4=2.x=2(3)-4=2.x=2(3)−4=2.

So they touch at (2,3)(2,3)(2,3).

  1. Understand the enclosed region

The boundary of the enclosed region in the first quadrant is:

  • along the xxx-axis from (0,0)(0,0)(0,0) to (5,0)(5,0)(5,0),
  • along the parabola from (5,0)(5,0)(5,0) to (2,3)(2,3)(2,3),
  • along the line from (2,3)(2,3)(2,3) to (0,2)(0,2)(0,2),
  • along the yyy-axis from (0,2)(0,2)(0,2) to (0,0)(0,0)(0,0).

For integration with respect to yyy:

  • from y=0y=0y=0 to y=2y=2y=2, left boundary is x=0x=0x=0 and right boundary is parabola x=(y−2)2+1x=(y-2)^2+1x=(y−2)2+1.
  • from y=2y=2y=2 to y=3y=3y=3, left boundary is line x=2y−4x=2y-4x=2y−4 and right boundary is parabola x=(y−2)2+1x=(y-2)^2+1x=(y−2)2+1.

So area is

A=∫02[(y−2)2+1−0]dy+∫23[(y−2)2+1−(2y−4)]dy.A=\int_0^2 \big[(y-2)^2+1-0\big]dy+\int_2^3 \big[(y-2)^2+1-(2y-4)\big]dy.A=∫02​[(y−2)2+1−0]dy+∫23​[(y−2)2+1−(2y−4)]dy.
  1. Evaluate the first integral
(y−2)2+1=y2−4y+5.(y-2)^2+1=y^2-4y+5.(y−2)2+1=y2−4y+5.

Hence

I1=∫02(y2−4y+5) dy.I_1=\int_0^2 (y^2-4y+5)\,dy.I1​=∫02​(y2−4y+5)dy.

Antiderivative:

∫(y2−4y+5)dy=y33−2y2+5y.\int (y^2-4y+5)dy=\frac{y^3}{3}-2y^2+5y.∫(y2−4y+5)dy=3y3​−2y2+5y.

Thus

I1=[y33−2y2+5y]02=(83−8+10)−0=143.I_1=\left[\frac{y^3}{3}-2y^2+5y\right]_0^2 =\left(\frac{8}{3}-8+10\right)-0 =\frac{14}{3}.I1​=[3y3​−2y2+5y]02​=(38​−8+10)−0=314​.
  1. Evaluate the second integral

Simplify the integrand:

(y−2)2+1−(2y−4)=y2−4y+5−2y+4=y2−6y+9=(y−3)2.(y-2)^2+1-(2y-4)=y^2-4y+5-2y+4=y^2-6y+9=(y-3)^2.(y−2)2+1−(2y−4)=y2−4y+5−2y+4=y2−6y+9=(y−3)2.

So

I2=∫23(y−3)2 dy.I_2=\int_2^3 (y-3)^2\,dy.I2​=∫23​(y−3)2dy.

Let u=y−3u=y-3u=y−3. Then this is standard:

∫(y−3)2dy=(y−3)33.\int (y-3)^2dy=\frac{(y-3)^3}{3}.∫(y−3)2dy=3(y−3)3​.

Hence

I2=[(y−3)33]23=0−(−13)=13.I_2=\left[\frac{(y-3)^3}{3}\right]_2^3 =0-\left(-\frac{1}{3}\right)=\frac{1}{3}.I2​=[3(y−3)3​]23​=0−(−31​)=31​.
  1. Total area
A=I1+I2=143+13=153=5.A=I_1+I_2=\frac{14}{3}+\frac{1}{3}=\frac{15}{3}=5.A=I1​+I2​=314​+31​=315​=5.

So the required enclosed area is

5.\boxed{5}.5​.
  1. Comparison with stored correct answer

Stored correct answer: 555

Our derived answer is also 555, so they agree.

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