Application of Integrals: CBSE Class 12 Previous Year Questions

14 different questions from Application of Integrals (NCERT Chapter 8) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions5 questions 4 marks questions3 questions 5 marks questions6 questions

Most asked Application of Integrals questions

Q101 markMCQApplication of Integrals CBSE 2026 · 65/1/1

The area bounded by the curve $y = x ∣ x ∣$ , $x$ -axis and the ordinates $x = - 1$ and $x = 1$ is given by

(A)0
(B) $\frac{1}{3}$
(C) $\frac{2}{3}$
(D) $\frac{4}{3}$

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Answer: (C)

\frac{2}{3}

For $x \geq 0$ , $y = x^{2}$ ; for $x < 0$ , $y = - x^{2}$ . The curve is symmetric about the origin.
Area $= 2 \int_{0}^{1} x^{2} d x = 2 \times \frac{1}{3} = \frac{2}{3}$ sq unit.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q101 markMCQApplication of Integrals CBSE 2026 · 65/2/1

Which of the following expressions will give the area of region bounded by the curve $y = x^{2}$ and line $y = 16$ ?

(A) $\int_{0}^{4} x^{2} d x$
(B) $2 \int_{0}^{4} x^{2} d x$
(C) $\int_{0}^{16} y d y$
(D) $2 \int_{0}^{16} y d y$

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Answer: (D)

2 \int_{0}^{16} y d y

The region lies between $y = 0$ and $y = 16$ , with $x$ from $- y$ to $y$ .
It is symmetric about the y-axis, so area $= 2 \int_{0}^{16} x d y = 2 \int_{0}^{16} y d y$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q121 markMCQApplication of Integrals CBSE 2026 · 65/3/1

The area of the shaded region of the circle given below is equal to :

Diagram for CBSE 2026 Class 12 Maths question 12

(A) $\int_{1}^{3} 9 - y^{2} d y$
(B) $2 \int_{1}^{3} 9 - y^{2} d y$
(C) $\int_{0}^{3} 9 - x^{2} d x$
(D) $2 \int_{0}^{3} 9 - x^{2} d x$

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Answer: (B)

2 \int_{1}^{3} 9 - y^{2} d y

The shaded part is the part of the circle $x^{2} + y^{2} = 9$ above the line $y = 1$ .
Taking horizontal strips, x runs from $- 9 - y^{2}$ to $9 - y^{2}$ , for y from 1 to 3.
Area $= \int_{1}^{3} 2 9 - y^{2} d y = 2 \int_{1}^{3} 9 - y^{2} d y$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q111 markMCQApplication of Integrals CBSE 2026 · 65/4/1

The area of the region bounded by the curve y = x and x-axis, between x = 0 and x = 2 is :

(A)2 sq. units
(B) $\frac{1}{2}$ sq. unit
(C)1 sq. unit
(D)4 sq. units

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Answer: (A) 2 sq. units

Area $= \int_{0}^{2} x d x = [\frac{x ^{2}}{2}]_{0}^{2}$ .
$= 2$ sq. units.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q121 markMCQApplication of Integrals CBSE 2026 · 65/5/1

An ant is observed crawling on a sheet of paper along a straight line given by equation $y = 2 x - 4$ . Area of the surface covered by the ant bounded by y-axis, x-axis and $x = 1$ is :

(A)1 sq. unit
(B)3 sq. units
(C)2 sq. units
(D)4 sq. units

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Answer: (B) 3 sq. units

For $0 \leq x \leq 1$ , $y = 2 x - 4 < 0$ , so the region lies below the x-axis.
Area $= \int_{0}^{1} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{1} = 3$ sq. units.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q384 marksCase StudyApplication of Integrals CBSE 2026 · 65/1/1

Roundabouts are often made on busy roads to ease the traffic and avoid red lights.
One such round-about is made such that equation representing its boundary is given by $C_{1}; x^{2} + y^{2} = 64$ .
There is a circular pond with a fountain in the middle of the roundabout whose equation is given by $C_{2} : x^{2} + y^{2} = 4$ .
Based on the given information, answer the following questions :
(i) Represent the given equations $C_{1}$ and $C_{2}$ with the help of a diagram. (1)
(ii) Express y as a function of $x$ , (y = f(x)), for both $C_{1}$ an $C_{2}$ . (1)
(iii) Using integration find the area of region covered by the roundabout. (2)
OR (iii) Using integration, find the area of region covered by circular pond. (2)

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Answer: (i) Two concentric circles with centre O(0, 0) and radii 8 and 2 (ii)

C_{1}

y = \pm 64 - x^{2}

;

C_{2}

y = \pm 4 - x^{2}

(iii)

64 π

sq units; OR

4 π

sq units

(i) $C_{1}$ is a circle with centre (0, 0) and radius 8; $C_{2}$ is a circle with centre (0, 0) and radius 2, inside $C_{1}$ .
(ii) $C_{1}$ : $y = \pm 64 - x^{2}$ , $- 8 \leq x \leq 8$ ; $C_{2}$ : $y = \pm 4 - x^{2}$ , $- 2 \leq x \leq 2$ .
(iii) By symmetry, area $= 4 \int_{0}^{8} 64 - x^{2} d x = 4 [\frac{x}{2} 64 - x^{2} + \frac{64}{2} sin^{- 1} \frac{x}{8}]_{0}^{8} = 4 \times 32 \times \frac{π}{2} = 64 π$ sq units.
OR (iii) Area $= 4 \int_{0}^{2} 4 - x^{2} d x = 4 [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{2} = 4 \times 2 \times \frac{π}{2} = 4 π$ sq units.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q374 marksCase StudyApplication of Integrals CBSE 2026 · 65/2/1

A racing track is build around an elliptical ground whose equation is given by $9 x^{2} + 16 y^{2} = 144$ . The width of the track is 3 m as shown below :
Based on given information, answer the following questions :
(i) Express y as a function of $x$ from the given equation of ellipse. (1)
(ii) Integrate the function obtained in (i) with respect to $x$ . (1)
(iii) (a) Find the area of the region enclosed within the elliptical ground excluding the track using integration. (2)
OR (iii) (b) Write the co-ordinates of the points P and Q where the outer edge of the track cuts $x$ axis and y axis in first quadrant and find the area of the triangle formed by points P, O, Q using integration. (2)

Diagram for CBSE 2026 Class 12 Maths question 37

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Answer: (i)

y = \pm \frac{3}{4} 16 - x^{2}

(ii)

\frac{3}{8} x 16 - x^{2} + 6 sin^{- 1} \frac{x}{4} + C

(iii)(a)

12 π

sq m OR (iii)(b) P(7, 0), Q(0, 6); area 21 sq m

(i) $16 y^{2} = 144 - 9 x^{2} \Rightarrow y = \pm \frac{3}{4} 16 - x^{2}$ (upper half: $y = \frac{3}{4} 16 - x^{2}$ ).
(ii) $\int \frac{3}{4} 16 - x^{2} d x = \frac{3}{4} [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}] + C = \frac{3}{8} x 16 - x^{2} + 6 sin^{- 1} \frac{x}{4} + C$ .
(iii)(a) Area $= 4 \int_{0}^{4} \frac{3}{4} 16 - x^{2} d x = 3 [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{0}^{4} = 3 (8 \cdot \frac{π}{2}) = 12 π$ sq m.
(iii)(b) Outer edge has semi-axes $4 + 3 = 7$ and $3 + 3 = 6$ : P(7, 0), Q(0, 6).
Line PQ: $\frac{x}{7} + \frac{y}{6} = 1 \Rightarrow y = 6 - \frac{6 x}{7}$ .
Area of $△ P O Q = \int_{0}^{7} (6 - \frac{6 x}{7}) d x = 42 - 21 = 21$ sq m.

Also asked in: 2026 65/2/2, 2026 65/2/3

Q384 marksCase StudyApplication of Integrals CBSE 2026 · 65/3/1

There is a triangular park in the society. The park is divided into two sections as shown in the figure.
In the region OAC, children are allowed to play games like cricket, football, while in the region AOB, activities which involve running are not allowed.
The vertices of the triangular park ABC are A(0, 4), B(– 2, 0) and C(3, 0).
Based on the above information, answer the following questions :
(i) Write the equation of the boundary line AB of the park. (1)
(ii) Write the equation of the boundary line AC of the park. (1)
(iii) (a) Using integration, find the area of region OAC, in which children are allowed to play cricket, football. (2)
OR (iii) (b) Using integration, find the area of region AOB. (2)

Diagram for CBSE 2026 Class 12 Maths question 38

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Answer: (i)

y = 2 x + 4

(ii)

4 x + 3 y = 12

(iii)(a) 6 sq units (iii)(b) 4 sq units

(i) Slope of AB $= \frac{4 - 0}{0 + 2} = 2$ , so $y = 2 x + 4$ .
(ii) Slope of AC $= \frac{0 - 4}{3 - 0} = - \frac{4}{3}$ , so $y = 4 - \frac{4 x}{3}$ , i.e. $4 x + 3 y = 12$ .
(iii)(a) Area OAC $= \int_{0}^{3} (4 - \frac{4 x}{3}) d x = [4 x - \frac{2 x ^{2}}{3}]_{0}^{3} = 12 - 6 = 6$ sq units.
(iii)(b) Area AOB $= \int_{- 2}^{0} (2 x + 4) d x = [x^{2} + 4 x]_{- 2}^{0} = 0 - (4 - 8) = 4$ sq units.

Also asked in: 2026 65/3/2, 2026 65/3/3

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