Area Between Two Curves PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2014 Board
00:00
Find the area enclosed by the parabolas $y^2 = 4x$ and $x^2 = 4y$. 5 Marks
Step 1: Find Intersection Points
The curves are $y^2 = 4x$ and $x^2 = 4y \Rightarrow y = x^2/4$.
Substitute $y$ in first eq: $(x^2/4)^2 = 4x \Rightarrow x^4/16 = 4x \Rightarrow x^4 = 64x$.
$x(x^3 - 64) = 0 \Rightarrow x = 0$ or $x = 4$.
Intersection points are $(0,0)$ and $(4,4)$.
Step 2: Set up the Integral
Area $= \int_0^4 (y_{upper} - y_{lower}) \, dx$.
Upper curve is $y = 2\sqrt{x}$ and lower curve is $y = x^2/4$.
Area $= \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) \, dx$.
Step 3: Integrate and Evaluate
$= [2 \cdot \frac{2}{3}x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3}]_0^4$.
$= [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_0^4 = (\frac{4}{3} \cdot 8 - \frac{64}{12}) - (0)$.
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ sq. units.
16/3 sq. units
Q2
2008 Board
00:00
Find the area enclosed between the curves $y = x^2$ and $y = 2x$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$.
Points of intersection are $x=0$ and $x=2$.
Step 2: Set up the Integral
Area $= \int_0^2 (2x - x^2) \, dx$.
Step 3: Integrate and Evaluate
$= [x^2 - \frac{x^3}{3}]_0^2 = (4 - 8/3) - (0) = 4/3$ sq. units.
4/3 sq. units
Q3
2010 Board
00:00
Find the area enclosed by the curve $y = |x|$ and $y = 1$. 3 Marks
Step 1: Analyze the Region
$y = |x|$ is symmetric about the y-axis. The region is bounded by $y=x$ and $y=-x$ from below, and $y=1$ from above.
Step 2: Find Intersection Points
$|x| = 1 \Rightarrow x = \pm 1$.
Step 3: Set up the Integral
Using symmetry, Area $= 2 \times \int_0^1 (y_{upper} - y_{lower}) \, dx$.
Area $= 2 \int_0^1 (1 - x) \, dx$.
Step 4: Integrate and Evaluate
$= 2 [x - \frac{x^2}{2}]_0^1 = 2(1 - 1/2) = 1$ sq. unit.
1 sq. unit
Q4
2026 Board
00:00
Find the area enclosed between the curves $x = y^2$ and $x = 4 - y^2$. 4 Marks
Step 1: Find Intersection Points
$y^2 = 4 - y^2 \Rightarrow 2y^2 = 4 \Rightarrow y^2 = 2 \Rightarrow y = \pm \sqrt{2}$.
Step 2: Set up the Integral
The curves are symmetric about the x-axis. Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (x_{right} - x_{left}) \, dy$.
Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (4 - y^2 - y^2) \, dy = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2y^2) \, dy$.
Step 3: Integrate and Evaluate
$= [4y - \frac{2y^3}{3}]_{-\sqrt{2}}^{\sqrt{2}}$.
$= (4\sqrt{2} - \frac{4\sqrt{2}}{3}) - (-4\sqrt{2} + \frac{4\sqrt{2}}{3})$.
$= \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3}$ sq. units.
16√2/3 sq. units
Q5
2025 Board
00:00
Find the area enclosed by the curves $y = x^2$ and $y = 4x - x^2$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 4x - x^2 \Rightarrow 2x^2 - 4x = 0 \Rightarrow 2x(x-2) = 0$.
Intersection points are $x=0$ and $x=2$.
Step 2: Set up the Integral
Area $= \int_0^2 (y_{upper} - y_{lower}) \, dx = \int_0^2 (4x - x^2 - x^2) \, dx$.
Area $= \int_0^2 (4x - 2x^2) \, dx$.
Step 3: Integrate and Evaluate
$= [2x^2 - \frac{2x^3}{3}]_0^2 = (8 - 16/3) - (0) = 8/3$ sq. units.
8/3 sq. units
Q6
2023 Board
00:00
Find the area enclosed between the parabolas $y = x^2$ and $y = 4 - x^2$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 4 - x^2 \Rightarrow 2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$.
Step 2: Set up the Integral
Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (4 - x^2 - x^2) \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx$.
Step 3: Integrate and Evaluate
$= [4x - \frac{2x^3}{3}]_{-\sqrt{2}}^{\sqrt{2}} = (4\sqrt{2} - \frac{4\sqrt{2}}{3}) - (-4\sqrt{2} + \frac{4\sqrt{2}}{3}) = \frac{16\sqrt{2}}{3}$.
16√2/3 sq. units
Q7
2018 Board
00:00
Find the area enclosed between the curve $y = x^2$ and $y = |x|$. 3 Marks
Step 1: Analyze Symmetry
Both curves are symmetric about the y-axis. Area $= 2 \times$ area in the first quadrant.
Step 2: Find Intersection in 1st Quadrant
$x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x=0, 1$.
Step 3: Set up and Evaluate Integral
Area $= 2 \int_0^1 (x - x^2) \, dx = 2 [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = 2(1/2 - 1/3) = 1/3$.
1/3 sq. units
Q8
2013 Board
00:00
Find the area enclosed between the parabolas $y = x^2$ and $y = 2 - x^2$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 2 - x^2 \Rightarrow 2x^2 = 2 \Rightarrow x = \pm 1$.
Step 2: Set up and Evaluate Integral
Area $= \int_{-1}^1 (2 - x^2 - x^2) \, dx = \int_{-1}^1 (2 - 2x^2) \, dx$.
$= [2x - \frac{2x^3}{3}]_{-1}^1 = (2 - 2/3) - (-2 + 2/3) = 4/3 + 4/3 = 8/3$.
8/3 sq. units
Q9
2010 Board
00:00
Find the area between the curve $y = x^2$ and the line $y = 2x + 3$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$.
Points: $x = -1, 3$.
Step 2: Set up and Evaluate Integral
Area $= \int_{-1}^3 (2x+3 - x^2) \, dx = [x^2 + 3x - \frac{x^3}{3}]_{-1}^3$.
$= (9+9-9) - (1-3+1/3) = 9 - (-5/3) = 32/3$.
32/3 sq. units
Q10
2009 Board
00:00
Find the area enclosed by the parabolas $x^2 = y$ and $y^2 = x$. 4 Marks
Step 1: Find Intersection Points
$(x^2)^2 = x \Rightarrow x^4 - x = 0 \Rightarrow x(x^3-1) = 0 \Rightarrow x=0, 1$.
Step 2: Set up and Evaluate Integral
Area $= \int_0^1 (\sqrt{x} - x^2) \, dx = [\frac{2x^{3/2}}{3} - \frac{x^3}{3}]_0^1 = 2/3 - 1/3 = 1/3$.
1/3 sq. units
Q11
2025 Board
00:00
Find the area enclosed by the curves $y = \sqrt{x}$ and $y = x$. 3 Marks
Step 1: Find Intersection Points
$\sqrt{x} = x \Rightarrow x = x^2 \Rightarrow x(x-1) = 0 \Rightarrow x=0, 1$.
Step 2: Set up and Evaluate Integral
Area $= \int_0^1 (\sqrt{x} - x) \, dx = [\frac{2x^{3/2}}{3} - \frac{x^2}{2}]_0^1 = 2/3 - 1/2 = 1/6$.
1/6 sq. unit
Q12
2023 Board
00:00
Find the area enclosed between $y^2 = 4x$ and $y = 2x$. 3 Marks
Step 1: Find Intersection Points
$(2x)^2 = 4x \Rightarrow 4x^2 = 4x \Rightarrow x(x-1) = 0 \Rightarrow x=0, 1$.
Step 2: Set up and Evaluate Integral
Area $= \int_0^1 (2\sqrt{x} - 2x) \, dx = [\frac{4x^{3/2}}{3} - x^2]_0^1 = 4/3 - 1 = 1/3$.
1/3 sq. units
Q13
2022 Board
00:00
Find the area enclosed between $y = x^2$ and $y = 2x + 8$. 4 Marks
Step 1: Find Intersection Points
$x^2 = 2x + 8 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0$.
Points: $x = -2, 4$.
Step 2: Set up and Evaluate Integral
Area $= \int_{-2}^4 (2x+8 - x^2) \, dx = [x^2 + 8x - \frac{x^3}{3}]_{-2}^4$.
$= (16+32-64/3) - (4-16+8/3) = (48-64/3) - (-12+8/3) = 60 - 72/3 = 36$.
36 sq. units
Q14
2021 Board
00:00
Find the area enclosed by $y = |x|$ and $y = 2$. 2 Marks
Step 1: Analyze Symmetry
Area $= 2 \times$ area in the first quadrant.
Step 2: Set up Integral
In first quadrant, boundary is $y=x$. Upper line is $y=2$.
Area $= 2 \int_0^2 (2 - x) \, dx = 2 [2x - \frac{x^2}{2}]_0^2 = 2(4 - 2) = 4$.
4 sq. units
Q15
2019 Board
00:00
Find the area enclosed between $x = y^2$ and $x = 2y$. 3 Marks
Step 1: Find Intersection Points
$y^2 = 2y \Rightarrow y(y-2) = 0 \Rightarrow y=0, 2$.
Step 2: Set up and Evaluate Integral
Area $= \int_0^2 (2y - y^2) \, dy = [y^2 - \frac{y^3}{3}]_0^2 = 4 - 8/3 = 4/3$.
4/3 sq. units
Q16
2018 Board
00:00
Find the area enclosed by $y = x^2 - 2x$ and the x-axis. 3 Marks
Step 1: Find Intersection Points
$x^2 - 2x = 0 \Rightarrow x(x-2) = 0 \Rightarrow x=0, 2$.
Step 2: Set up Integral
The curve is below the x-axis in $[0, 2]$. Area $= |\int_0^2 (x^2 - 2x) \, dx|$.
Step 3: Evaluate
$= |[\frac{x^3}{3} - x^2]_0^2| = |8/3 - 4| = |-4/3| = 4/3$.
4/3 sq. units
Q17
2017 Board
00:00
Find the area enclosed by $y = \sin x$ and $y = \cos x$ in $0 \le x \le \pi/2$. 4 Marks
Step 1: Find Intersection
$\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \pi/4$.
Step 2: Analyze Boundaries
From $0$ to $\pi/4$, $\cos x \ge \sin x$. From $\pi/4$ to $\pi/2$, $\sin x \ge \cos x$.
Step 3: Set up Integral
Area $= \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$.
Step 4: Evaluate
$= [\sin x + \cos x]_0^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$.
$= (\sqrt{2}-1) + (-1 - (-\sqrt{2})) = 2\sqrt{2}-2$.
2√2 - 2 sq. units
Q18
2016 Board
00:00
Find the area enclosed by $x^2 + y^2 = 4$ and $x + y = 2$. 4 Marks
Step 1: Find Intersection
$x^2 + (2-x)^2 = 4 \Rightarrow x^2 + 4 + x^2 - 4x = 4 \Rightarrow 2x^2 - 4x = 0 \Rightarrow x=0, 2$.
Step 2: Set up Integral
Area $= \int_0^2 (\sqrt{4-x^2} - (2-x)) \, dx$.
Step 3: Evaluate
$= [\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}(x/2) - (2x - \frac{x^2}{2})]_0^2$.
$= (0 + \pi - 4 + 2) - 0 = \pi - 2$.
π - 2 sq. units
Q19
2015 Board
00:00
Find the area enclosed by $x = y^2$ and $x = 2 - y^2$. 4 Marks
Step 1: Find Intersection
$y^2 = 2 - y^2 \Rightarrow 2y^2 = 2 \Rightarrow y = \pm 1$.
Step 2: Set up and Evaluate Integral
Area $= \int_{-1}^1 (2 - 2y^2) \, dy = [2y - \frac{2y^3}{3}]_{-1}^1 = 4 - 4/3 = 8/3$.
8/3 sq. units
Q20
2013 Board
00:00
Find the area enclosed by $y = |x-1|$ and coordinate axes. 2 Marks
Step 1: Identify the Region
The line $y = |x-1|$ hits y-axis at $(0,1)$ and x-axis at $(1,0)$.
The region is a right-angled triangle with vertices $(0,0), (0,1), (1,0)$.
Step 2: Calculate Area
Area $= 1/2 \times base \times height = 1/2 \times 1 \times 1 = 0.5$.
0.5 sq. unit
Q21
2023 Board
00:00
Find the area enclosed between $y = x^2$ and $y = 4 - x^2$. 4 Marks
Step 1: Find Intersection
$x^2 = 4 - x^2 \Rightarrow 2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$.
Step 2: Set up Integral
Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx$.
Step 3: Evaluate
$= [4x - \frac{2x^3}{3}]_{-\sqrt{2}}^{\sqrt{2}} = 2(4\sqrt{2} - \frac{4\sqrt{2}}{3}) = \frac{16\sqrt{2}}{3}$.
16√2/3 sq. units
Q22
2022 Board
00:00
Find the area enclosed by $x^2 = y$ and $y = 2x$. 3 Marks
Step 1: Find Intersection
$x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x=0, 2$.
Step 2: Set up and Evaluate
Area $= \int_0^2 (2x - x^2) \, dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - 8/3 = 4/3$.
4/3 sq. units
Q23
2019 Board
00:00
Find the area enclosed between $y = \sqrt{x}$ and $y = x^2$. 3 Marks
Step 1: Find Intersection
$\sqrt{x} = x^2 \Rightarrow x = x^4 \Rightarrow x(x^3-1) = 0 \Rightarrow x=0, 1$.
Step 2: Set up and Evaluate
Area $= \int_0^1 (\sqrt{x} - x^2) \, dx = [\frac{2x^{3/2}}{3} - \frac{x^3}{3}]_0^1 = 2/3 - 1/3 = 1/3$.
1/3 sq. units
Q24
2019 Board
00:00
Find the area enclosed by $y^2 = 4x$ and $y = x$. 3 Marks
Step 1: Find Intersection
$x^2 = 4x \Rightarrow x(x-4) = 0 \Rightarrow x=0, 4$.
Step 2: Set up and Evaluate
Area $= \int_0^4 (2\sqrt{x} - x) \, dx = [\frac{4x^{3/2}}{3} - \frac{x^2}{2}]_0^4 = 32/3 - 8 = 8/3$.
8/3 sq. units
Q25
2012 Board
00:00
Find the area enclosed between $x = y^2$ and $x = 2y$. 3 Marks
Step 1: Find Intersection
$y^2 = 2y \Rightarrow y(y-2) = 0 \Rightarrow y=0, 2$.
Step 2: Set up and Evaluate
Area $= \int_0^2 (2y - y^2) \, dy = [y^2 - \frac{y^3}{3}]_0^2 = 4 - 8/3 = 4/3$.
4/3 sq. units
Q26
2011 Board
00:00
Find the area enclosed by $y = \sqrt{x}$ and $y = x$. 2 Marks
Step 1: Find Intersection
$\sqrt{x} = x \Rightarrow x = x^2 \Rightarrow x=0, 1$.
Step 2: Set up and Evaluate
Area $= \int_0^1 (\sqrt{x} - x) \, dx = [\frac{2x^{3/2}}{3} - \frac{x^2}{2}]_0^1 = 2/3 - 1/2 = 1/6$.
1/6 sq. unit