Class 12 Maths, Worked in Full Past papersGet updates
CBSE Sample Paper 2021-22 (Term 2) · Section C · 4 marks

Q12: Area enclosed by x+y=2x + y = 2⁠, y2=xy^2 = x and the xx-axis, or of {0≤y≤3x, x2+y2≤4}\{0 \le y \le \sqrt3x,\ x^2 + y^2 \le 4\}

Sample 2021-22 T2Ch 8 · Application of IntegralsCh 7 · Integrals
The question

(a) Using integration, find the area of the region in the first quadrant enclosed by the line x+y=2x + y = 2⁠, the parabola y2=xy^2 = x and the xx-axis.

OR

(b) Using integration, find the area of the region {(x,y):0≤y≤3x, x2+y2≤4}\{(x, y): 0 \le y \le \sqrt3x,\ x^2 + y^2 \le 4\}

Exam answer

(a)

Solving x+y=2x + y = 2 and y2=xy^2 = x⁠: y2+y−2=0y^2 + y - 2 = 0⁠, so y=1y = 1 or y=−2y = -2⁠. The points of intersection are (1,1)(1, 1) and (4,−2)(4, -2)⁠.

Why?

Put x=y2x = y^2 into x+y=2x + y = 2⁠: y2+y−2=0y^2 + y - 2 = 0⁠, which factorises as (y−1)(y+2)=0(y - 1)(y + 2) = 0⁠. Then x=y2x = y^2 gives x=1x = 1 and x=4x = 4⁠.

y2+y−2=0y2+2y−y−2=0y(y+2)−1(y+2)=0(y−1)(y+2)=0y=1: x=(1)2=1y=−2: x=(−2)2=4\begin{aligned} y^2 + y - 2 &= 0 \\ y^2 + 2y - y - 2 &= 0 \\ y(y + 2) - 1(y + 2) &= 0 \\ (y - 1)(y + 2) &= 0 \\[4pt] y = 1&:\ x = (1)^2 = 1 \\ y = -2&:\ x = (-2)^2 = 4 \end{aligned}
X Y O 1 2 (1, 1) (4, −2) y² = x x + y = 2
The parabola y2=xy^2 = x and the line x+y=2x + y = 2⁠; the region is shaded.

In the shaded region the upper boundary is y=xy = \sqrt x for 0≤x≤10 \le x \le 1 and y=2−xy = 2 - x for 1≤x≤21 \le x \le 2⁠.

Why?

In the first quadrant, y2=xy^2 = x gives the upper half of the parabola, y=xy = \sqrt x⁠. The line meets the xx-axis where y=0y = 0⁠, at x=2x = 2⁠.

From x=0x = 0 to 11 the region lies under the parabola; from 11 to 22 it lies under the line. So the area is split at x=1x = 1⁠.

Watch out: the point (4,−2)(4, -2) is below the xx-axis, so it does not bound this region.

Area=∫01x dx+∫12(2−x) dx=23[x32]01+[2x−x22]12=23(1−0)+(4−2)−(2−12)=23+12\begin{aligned} \text{Area} &= \int_0^1 \sqrt x\,dx \\ &\quad + \int_1^2 (2 - x)\,dx \\ &= \frac23\left[x^{\frac32}\right]_0^1 \\ &\quad + \left[2x - \frac{x^2}{2}\right]_1^2 \\ &= \frac23(1 - 0) \\ &\quad + (4 - 2) - \left(2 - \frac12\right) \\ &= \frac23 + \frac12 \end{aligned}
Why?

Standard integrals:

∫x dx=x3/23/2=23x32∫(2−x) dx=2x−x22\begin{aligned} \int \sqrt x\,dx &= \frac{x^{3/2}}{3/2} = \frac23x^{\frac32} \\ \int (2 - x)\,dx &= 2x - \frac{x^2}{2} \end{aligned}

At the limits, 13/2=11^{3/2} = 1⁠, and the second bracket is 2−32=122 - \frac32 = \frac12⁠. Then 23+12=4+36=76\frac23 + \frac12 = \frac{4 + 3}{6} = \frac76⁠.

23[13/2−03/2]=23(1−0)=23x=2: 2(2)−222=4−2=2x=1: 2(1)−122=2−12=322−32=42−32=1223+12=46+36=76\begin{aligned} \frac23\left[1^{3/2} - 0^{3/2}\right] &= \frac23(1 - 0) = \frac23 \\[4pt] x = 2&:\ 2(2) - \frac{2^2}{2} = 4 - 2 = 2 \\ x = 1&:\ 2(1) - \frac{1^2}{2} = 2 - \frac12 = \frac32 \\ &2 - \frac32 = \frac42 - \frac32 = \frac12 \\[4pt] &\frac23 + \frac12 = \frac46 + \frac36 = \frac76 \end{aligned}

∴\therefore Required area =76= \frac76 sq. units.

(b)

Solving y=3xy = \sqrt3x and x2+y2=4x^2 + y^2 = 4⁠: 4x2=44x^2 = 4⁠, so x=±1x = \pm 1⁠. The points of intersection are (1,3)(1, \sqrt3) and (−1,−3)(-1, -\sqrt3)⁠.

Why?

Put y=3xy = \sqrt3x into the circle: x2+3x2=4x^2 + 3x^2 = 4⁠, so x2=1x^2 = 1 and x=±1x = \pm 1⁠. Then y=3xy = \sqrt3x gives y=±3y = \pm\sqrt3⁠.

x2+(3x)2=4x2+3x2=44x2=4x2=1x=±1\begin{aligned} x^2 + (\sqrt3x)^2 &= 4 \\ x^2 + 3x^2 &= 4 \\ 4x^2 &= 4 \\ x^2 &= 1 \\ x &= \pm 1 \end{aligned}
X Y O 1 2 −2 y = √3x x² + y² = 4
The circle x2+y2=4x^2 + y^2 = 4 and the line y=3xy = \sqrt3x⁠; the region is shaded.

The region lies in the first quadrant, below the line y=3xy = \sqrt3x for 0≤x≤10 \le x \le 1 and below the circle for 1≤x≤21 \le x \le 2⁠.

Why?

y≥0y \ge 0 keeps the region above the xx-axis, and y≤3xy \le \sqrt3x then needs x≥0x \ge 0⁠. So the region is in the first quadrant, inside the circle and below the line.

Up to x=1x = 1⁠, where they meet, the line is below the circle. After that the circle is the upper boundary, until it meets the xx-axis at x=2x = 2⁠.

Area=∫013x dx+∫124−x2 dx=32[x2]01+12[x4−x2+4sin⁡−1x2]12=32+12[2π−3−2π3]=32+2π3−32\begin{aligned} \text{Area} &= \int_0^1 \sqrt3x\,dx \\ &\quad + \int_1^2 \sqrt{4 - x^2}\,dx \\ &= \frac{\sqrt3}{2}\left[x^2\right]_0^1 \\ &\quad + \frac12\bigg[x\sqrt{4 - x^2} \\ &\qquad + 4\sin^{-1}\frac{x}{2}\bigg]_1^2 \\ &= \frac{\sqrt3}{2} + \frac12\left[2\pi - \sqrt3 - \frac{2\pi}{3}\right] \\ &= \frac{\sqrt3}{2} + \frac{2\pi}{3} - \frac{\sqrt3}{2} \end{aligned}
Why?

Standard integral:

∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+C\begin{aligned} &\int \sqrt{a^2 - x^2}\,dx \\ &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \end{aligned}

With a=2a = 2⁠, a22=42=2\frac{a^2}{2} = \frac42 = 2⁠, so the result is 12[x4−x2+4sin⁡−1x2]\frac12\big[x\sqrt{4 - x^2} + 4\sin^{-1}\frac{x}{2}\big]⁠.

The line part: ∫3x dx=3⋅x22\int \sqrt3x\,dx = \sqrt3 \cdot \frac{x^2}{2}⁠, and 32(12−02)=32\frac{\sqrt3}{2}(1^2 - 0^2) = \frac{\sqrt3}{2}⁠.

With a=2a = 2⁠: at x=2x = 2 the bracket is 0+4⋅π2=2π0 + 4 \cdot \frac{\pi}{2} = 2\pi⁠; at x=1x = 1 it is 3+4⋅π6\sqrt3 + 4 \cdot \frac{\pi}{6}⁠, that is, 3+2π3\sqrt3 + \frac{2\pi}{3}⁠.

x=2: 24−4+4sin⁡−11=0+4⋅π2=2πx=1: 14−1+4sin⁡−112=3+4⋅π6=3+2π3\begin{aligned} x = 2&:\ 2\sqrt{4 - 4} + 4\sin^{-1}1 \\ &= 0 + 4 \cdot \frac{\pi}{2} = 2\pi \\[4pt] x = 1&:\ 1\sqrt{4 - 1} + 4\sin^{-1}\frac12 \\ &= \sqrt3 + 4 \cdot \frac{\pi}{6} = \sqrt3 + \frac{2\pi}{3} \end{aligned}

Upper minus lower, then halve:

2π−(3+2π3)=2π−3−2π312[2π−3−2π3]=π−32−π3=3π3−π3−32=2π3−32\begin{aligned} &2\pi - \left(\sqrt3 + \frac{2\pi}{3}\right) \\ &= 2\pi - \sqrt3 - \frac{2\pi}{3} \\[4pt] &\frac12\left[2\pi - \sqrt3 - \frac{2\pi}{3}\right] \\ &= \pi - \frac{\sqrt3}{2} - \frac{\pi}{3} \\ &= \frac{3\pi}{3} - \frac{\pi}{3} - \frac{\sqrt3}{2} \\ &= \frac{2\pi}{3} - \frac{\sqrt3}{2} \end{aligned}

Add the line part: 32+2π3−32=2π3\frac{\sqrt3}{2} + \frac{2\pi}{3} - \frac{\sqrt3}{2} = \frac{2\pi}{3}⁠, as the 32\frac{\sqrt3}{2} terms cancel.

Check: the region is a sector of angle π3\frac{\pi}{3} (as tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠) in a circle of radius 2. Its area is 12r2θ=2π3\frac12 r^2\theta = \frac{2\pi}{3}⁠.

∴\therefore Required area =2π3= \dfrac{2\pi}{3} sq. units.

Where this comes from

Chapter 8, Application of Integrals: Area: parabola and a line; circle and a line.

Question type: (a) A curve and a line; (b) A curve and a line + Circles (combined)

How to spot it: click to see

(a) Type 7. A curve and a line: top minus bottom

Spot it: the region is shut in by a curve and a line, like a parabola and y=xy = x⁠, sometimes written as a set with ≤\le signs. Your book has no worked question of this kind, but CBSE sample papers ask it.

Here: The line x+y=2x + y = 2 and the parabola y2=xy^2 = x meet at (1,1)(1, 1) in the first quadrant. With the xx-axis below, the top edge is x\sqrt x from 0 to 1, then 2−x2 - x from 1 to 2.

(b) Type 7. A curve and a line: top minus bottom

Spot it: the region is shut in by a curve and a line, like a parabola and y=xy = x⁠, sometimes written as a set with ≤\le signs. Your book has no worked question of this kind, but CBSE sample papers ask it.

Here: Read the set: under the line y=3xy = \sqrt3x⁠, inside the circle. They meet at (1,3)(1, \sqrt3)⁠, so the top edge is the line from 0 to 1, then the circle from 1 to 2.

(b) Type 3. Circles: the square-root card, one quarter at a time

Spot it: x2+y2=a2x^2 + y^2 = a^2⁠, a circle with its centre at the origin and radius aa⁠.

Here: The circle piece is ∫124−x2 dx\int_1^2\sqrt{4 - x^2}\,dx⁠, with radius 2 and the square-root card; it starts at x=1x = 1⁠, so it is less than a quarter.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Application of Integrals in past papers: question 14 of 17 · see them all