Application of Integrals

Previous Years' CBSE Board Questions

8.2 Area under Simple Curves

SAI (2 marks)

Sketch the region bounded by the lines <math>2x + y = 8</math>, y = 2, y = 4 and the y-axis. Hence, obtain its area using integration. (2023)

Using integration, find the area bounded by the curve <math>y^2 = 4x</math>, y-axis and y = 3.

(2021C) Ev (2020, NCERT, 2018, Delhi 2014) [Fr]

1. Using integration, find the area of the region bounded by the line <math>2y = -x + 8</math>, x-axis, <math>x = 2</math> and <math>x = 4</math>. (2021C) Ev

SAII (3 marks)

Find the area of the following region using integration. <math>\{(x, y): y^2 \le 2x \text{ and } y \ge x - 4\}</math> (2023)

1. Using integration, find the area of the region bounded by <math>y = mx (m > 0)</math>, <math>x = 1</math>, <math>x = 2</math> and the x-axis.

(2023)

1. Using integration, find the area of the region <math>\{(x, y) : y^2 \le x \le y\}.</math> (Term II, 2021-22) Ev

LAI (4 marks)

1. Using integration, find the area of the region <math>\{(x, y): 4x^2 + 9y^2 \le 36, 2x + 3y \ge 6\}.</math> (Term II, 2021-22) [FV]
2. Using integration, find the area of the region bounded by lines <math>x - y + 1 = 0</math>, <math>x = -2</math>, <math>x = 3</math> and x-axis. (Term II, 2021-22) [EV]
3. If the area of the region bounded by the curve <math>y^2 = 4ax</math> and the line <math>x = 4a</math> is <math>\frac{256}{3}</math> sq. units, then using integration find the value of a, where <math>a > 0</math>.

(Term II, 2021-22) [iv

1. Find the area of the region bounded by curve <math>4x^2 = y</math> and the line <math>y = 8x + 12</math>, using integration.

(Term II, 2021-22) [EV]

Find the area bounded by the curves y = |x - 1| and y = 1, using integration. (Term II, 2021-22) [Cr]

LAII (5/6 marks)

1. Using integration, find the area of the region bounded by the circle <math>x^2 + y^2 = 16</math>, line <math>y = x</math> and
2. axis, but lying in the 1st quadrant.

1. Find the area of the following region using integration: <math>\{(x, y): y < |x| + 2, y > x^2\}</math> (2020) Ev
2. Using integration, find the area of a triangle whose vertices are (1, 0), (2, 2) and (3, 1). (2020) Ev

Using integration, find the smaller area enclosed by

the circle <math>x^2 + y^2 = 4</math> and the line <math>x + y = 2</math>. (2020) Ev

Using integration, find the area of the region in the

first quadrant enclosed by the x-axis, the line <math>y = x</math> and

the circle <math>x^2 + y^2 = 32</math>.

1. If the area between the curves <math>x = y^2</math> and <math>x = 4</math> is divided into two equal parts by the line <math>x = a</math>, then find the value of a using integration. (2020C) Ev
2. Using the method of integration, find the area of the region bounded by the lines <math>3x - 2y + 1 = 0</math>, <math>2x + 3y - 21 = 0</math> and <math>x - 5y + 9 = 0</math>. (2019) Ev
3. Find the area bounded by the circle <math>x^2 + y^2 = 16</math> and the line <math>\sqrt{3}y = x</math> in the first quadrant, using integration. (Delhi 2017) [Cr]
4. Find the area enclosed between the parabola <math>4y = 3x^2</math> and the straight line <math>3x - 2y + 12 = 0</math>.

(Al 2017, 2015C)

1. Using integration, find the area of the region bounded by the line <math>x - y + 2 = 0</math>, the curve <math>x = \sqrt{y}</math> and y - axis. (Foreign 2015)
2. Find the area of the region in the first quadrant enclosed by the y-axis, the line <math>y = x</math> and the circle <math>x^2 + y^2 = 32</math>, using integration. (NCERT, Delhi 2015C)
3. Find the area of the smaller region bounded by the ellipse <math>\frac{x^2}{9} + \frac{y^2}{4} = 1</math> and the line <math>\frac{x}{3} + \frac{y}{2} = 1</math>.

(Foreign 2014) (cr

1. Using integration, find the area of the region bounded by the curves: <math>y = |x + 1| + 1, x = -3, x = 3, y = 0</math> (Delhi 2014C)
2. Using integration, find the area bounded by the curve <math>x^2 = 4y</math> and the line <math>x = 4y - 2</math>. (Delhi 2014C)
3. Using integration, find the area of the region in the first quadrant enclosed by the x – axis, the line y = x and the circle <math>x^2 + y^2 = 18</math>.

(Al 2014C)

CBSE Sample Questions

8.2 Area under Simple Curves

VSA (1 mark)

Find the area bounded by <math>y = x^2</math>, the x- axis and the lines <math>x = -1</math> and <math>x = 1</math>. (2020-21) Ev

(2 marks)

1. Find the area of the region bounded by the parabola <math>y^2 = 8x</math> and the line <math>x = 2</math>. (2020-21) Ev

(3 marks)

1. Find the area of the region bounded by the curves <math>x^2 + y^2 = 4</math>, <math>y = \sqrt{3}x</math> and x-axis in the first quadrant. (2020-21)
2. Find the area of the ellipse <math>x^2 + 9y^2 = 36</math> using integration. (2020-21) Cr

(4/5 marks)

1. Make a rough sketch of the region <math>\{(x, y): 0 \le y \le x^2,</math> <math>0 \le y \le x</math>, <math>0 \le x \le 2</math> and find the area of the region using integration. (2022-23) EV
2. Using integration, find the area of the region in the first quadrant enclosed by the line <math>x + y = 2</math>, the parabola <math>y^2 = x</math> and the x-axis. (Term II, 2021-22)
3. Using integration, find the area of the region <math>\{(x,y): 0 \le y \le \sqrt{3}x, x^2 + y^2 \le 4\}</math>. (Term II, 2021-22) Cr

Detailed SOLUTIONS

Previous Years' CBSE Board Questions

From the graph, ABCD is the required region.

(2, 3)

(4, 2) 2 6 8 6

7+ 6 D <math>\Rightarrow v = 4</math> 3 > y=2 We have, <math>y^2 \le 2x</math> ... (i)

1 В ... (ii)

≥x. 2 3 <math display="block">2x + y = 8</math> (8, 4) <math>7y = x - 4</math> 2 <math>x' \in</math> (4, 0) <math>\rightarrow x</math>

0<br>-2 A

<math>=\frac{1}{2}\left[8y-\frac{y^2}{2}\right]^4=\frac{1}{2}\left[\left(32-\frac{16}{2}\right)-\left(16-\frac{4}{2}\right)\right]</math> <math>y^2 = 2x</math>

<math>=\frac{1}{2}\times10=5</math> sq.units <math>y^2 = 4x</math>

Now, area = <math>\int_{0}^{4} \left(\frac{8-y}{2}\right) dy = \frac{1}{2} \int_{0}^{4} (8-y) dy</math>

y =3 <del>≪</del> (2.25,3) X ► X 0

2. Required area = <math>\int_{A}^{3} \frac{y^2}{4} dy</math> <math>=\left[\frac{y^3}{12}\right]_0^3 = \frac{9}{4}</math> sq. units.

3. Required area

<math>=\int_{1}^{4} \left(\frac{-x+8}{2}\right) dx = \left[\frac{-x^2}{4} + 4x\right]_{2}^{4} = 5 \text{ sq. units}</math>

and <math>y \ge x - 4</math>

Here, the shaded area represents the required area.

Required area = <math>\int_{-2}^{2} (y+4) dy - \int_{2}^{4} \frac{y^2}{2} dy</math>

<math>=\left[\frac{y^2}{2}+4y\right]_{3}^{4}-\frac{1}{2}\left[\frac{y^3}{3}\right]_{3}^{4}</math>

<math>=\left[\left(\frac{16}{2}+16\right)-\left(\frac{4}{2}-8\right)\right]-\frac{1}{2}\left[\frac{64}{3}-\left(\frac{-8}{3}\right)\right]</math>

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