Area Under Simple Curves PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2008 Board
00:00
Find the area bounded by the curve $y^2 = 4x$, lines $x=1, x=4$ and the x-axis in the first quadrant. 4 Marks
Step 1: Identify the Curve
The curve $y^2 = 4x$ is a parabola opening to the right. In the first quadrant, $y = 2\sqrt{x}$.
Step 2: Set up the Integral
The area is bounded between $x=1$ and $x=4$.
Required Area $= \int_1^4 y \, dx = \int_1^4 2\sqrt{x} \, dx$.
Step 3: Integrate and Evaluate
$= 2 [\frac{x^{3/2}}{3/2}]_1^4 = \frac{4}{3} [x^{3/2}]_1^4$.
$= \frac{4}{3} [4^{3/2} - 1^{3/2}] = \frac{4}{3} (8 - 1) = \frac{28}{3}$ sq. units.
28/3 sq. units
Q2
2014 Board
00:00
Find the area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. 5 Marks
Step 1: Standard Form Analysis
$a^2 = 16 \Rightarrow a = 4, b^2 = 9 \Rightarrow b = 3$. The ellipse is symmetric about both axes.
Step 2: Area Calculation Strategy
Total Area $= 4 \times$ Area in first quadrant.
In first quadrant, $y = 3\sqrt{1 - x^2/16} = \frac{3}{4}\sqrt{16-x^2}$.
Step 3: Set up Integral
Area $= 4 \int_0^4 \frac{3}{4}\sqrt{16-x^2} \, dx = 3 \int_0^4 \sqrt{4^2-x^2} \, dx$.
Step 4: Integrate and Evaluate
Using $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$.
$= 3 [\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}(\frac{x}{4})]_0^4$.
$= 3 [(0 + 8\sin^{-1}(1)) - (0)] = 3 [8 \cdot \frac{\pi}{2}] = 12\pi$ sq. units.
12π sq. units
Q3
2009 Board
00:00
Find the area bounded by the curve $y = \sin x$, $x = 0$ and $x = \pi$. 2 Marks
Step 1: Set up Integral
Area $= \int_0^\pi \sin x \, dx$.
Step 2: Integrate and Evaluate
$= [-\cos x]_0^\pi = -(\cos \pi - \cos 0)$.
$= -(-1 - 1) = 2$ sq. units.
2 sq. units
Q4
2018 Board
00:00
Find the area bounded by the circle $x^2 + y^2 = 16$ in the first quadrant. 2 Marks
Step 1: Identify Curve
Radius $a = 4$. In first quadrant, $y = \sqrt{16-x^2}$ and $x$ goes from $0$ to $4$.
Step 2: Set up and Evaluate Integral
Area $= \int_0^4 \sqrt{16-x^2} \, dx = [\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}(\frac{x}{4})]_0^4$.
$= 8(\pi/2) = 4\pi$ sq. units.
4π sq. units
Q5
2024 Board
00:00
Find the area bounded by the parabola $x = y^2$ and the line $x = 4$. 2 Marks
Step 1: Analyze Symmetry
The parabola $x = y^2$ is symmetric about the x-axis. Area $= 2 \times$ area in the first quadrant.
Step 2: Set up Integral
In first quadrant, $y = \sqrt{x}$. Limits are $x=0$ to $x=4$.
Area $= 2 \int_0^4 \sqrt{x} \, dx$.
Step 3: Evaluate
$= 2 [\frac{2}{3}x^{3/2}]_0^4 = \frac{4}{3}(8) = 32/3$ sq. units.
32/3 sq. units
Q6
2023 Board
00:00
Find the area bounded by the circle $x^2 + y^2 = 9$ and coordinate axes in the first quadrant. 2 Marks
Step 1: Set up Integral
Radius is $3$. Area $= \int_0^3 \sqrt{9-x^2} \, dx$.
Step 2: Evaluate
$= [\frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3})]_0^3 = \frac{9}{2}(\frac{\pi}{2}) = \frac{9\pi}{4}$ sq. units.
9π/4 sq. units
Q7
2020 Board
00:00
Find the area under the curve $y = e^x$, from $x = 0$ to $x = 1$. 1 Mark
Step 1: Set up Integral
Area $= \int_0^1 e^x \, dx$.
Step 2: Evaluate
$= [e^x]_0^1 = e^1 - e^0 = e - 1$ sq. units.
e - 1 sq. units
Q8
2017 Board
00:00
Find the area bounded by the parabola $y^2 = 4x$ and the line $x = 4$. 2 Marks
Step 1: Set up Integral
Area $= 2 \int_0^4 y \, dx = 2 \int_0^4 2\sqrt{x} \, dx = 4 \int_0^4 \sqrt{x} \, dx$.
Step 2: Evaluate
$= 4 [\frac{2}{3}x^{3/2}]_0^4 = \frac{8}{3}(8) = 64/3$ sq. units.
64/3 sq. units
Q9
2016 Board
00:00
Find the area bounded by the curve $y = x^2$ and the line $y = 4$. 3 Marks
Step 1: Identify Integration Axis
Integrating along y-axis: $x = \sqrt{y}$. Area $= 2 \int_0^4 \sqrt{y} \, dy$.
Step 2: Evaluate
$= 2 [\frac{2}{3}y^{3/2}]_0^4 = \frac{4}{3}(8) = 32/3$ sq. units.
32/3 sq. units
Q10
2014 Board
00:00
Find the area enclosed by the circle $x^2 + y^2 = a^2$. 5 Marks
Step 1: Area Strategy
Area $= 4 \times$ Area in first quadrant $= 4 \int_0^a \sqrt{a^2-x^2} \, dx$.
Step 2: Evaluate
$= 4 [\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})]_0^a = 4 (0 + \frac{a^2}{2} \cdot \frac{\pi}{2}) = \pi a^2$ sq. units.
πa² sq. units
Q11
2011 Board
00:00
Find the area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum. 4 Marks
Step 1: Identify Latus Rectum
Latus rectum is the line $x = a$.
Step 2: Set up Integral
Area $= 2 \int_0^a y \, dx = 2 \int_0^a 2\sqrt{ax} \, dx = 4\sqrt{a} \int_0^a \sqrt{x} \, dx$.
Step 3: Evaluate
$= 4\sqrt{a} [\frac{2}{3}x^{3/2}]_0^a = \frac{8\sqrt{a}}{3} a^{3/2} = \frac{8}{3}a^2$ sq. units.
8/3 a² sq. units
Q12
2025 Board
00:00
Find the area bounded by the circle $x^2 + y^2 = 25$ and the x-axis. 3 Marks
Step 1: Area Strategy
Bounded by circle and x-axis means the semi-circle (upper or lower). Area $= 2 \int_0^5 \sqrt{25-x^2} \, dx$.
Step 2: Evaluate
$= 2 [\frac{x}{2}\sqrt{25-x^2} + \frac{25}{2}\sin^{-1}(\frac{x}{5})]_0^5 = 2 (\frac{25}{2} \cdot \frac{\pi}{2}) = \frac{25\pi}{2}$ sq. units.
25π/2 sq. units
Q13
2019 Board
00:00
Find the area bounded by the curve $y = \cos x$ from $x = 0$ to $x = \pi/2$. 2 Marks
Step 1: Set up Integral
Area $= \int_0^{\pi/2} \cos x \, dx$.
Step 2: Evaluate
$= [\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$ sq. unit.
1 sq. unit
Q14
2015 Board
00:00
Find the area enclosed by the parabola $y^2 = 4ax$ and the line $x = a$. 2 Marks
Step 1: Set up Integral
Area $= 2 \int_0^a \sqrt{4ax} \, dx = 4\sqrt{a} \int_0^a \sqrt{x} \, dx$.
Step 2: Evaluate
$= 4\sqrt{a} [\frac{2}{3}x^{3/2}]_0^a = \frac{8}{3}a^2$ sq. units.
8/3 a² sq. units
Q15
2012 Board
00:00
Find the area bounded by the parabola $y = 4 - x^2$ and the x-axis. 3 Marks
Step 1: Find Intersection with X-axis
$4 - x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Step 2: Set up Integral
Area $= \int_{-2}^2 (4 - x^2) \, dx$.
Step 3: Evaluate
$= [4x - \frac{x^3}{3}]_{-2}^2 = (8 - 8/3) - (-8 + 8/3) = 16 - 16/3 = 32/3$ sq. units.
32/3 sq. units