Class 12 Maths Chapter 8 Application of Integrals MCQs

Class 12 Maths Chapter 8 Application of Integrals MCQs are provided to help students improve their test scores. Answers and full explanations are provided for these multiple-choice questions. The questions are written in accordance with NCERT guidelines and the CBSE syllabus (2022–2023). Click here to access all chapter-by-chapter MCQs for class 12 Maths.

MCQs on Class 12 Maths Chapter 8 Application of Integrals

Check out the Class 12 Maths Chapter 8 Application of Integrals multiple-choice questions. Each MCQ has four options, but only one of them is correct. Students must select the appropriate option and compare their results to the solutions on this page. Also, check important questions for class 12 Maths.

Download PDF – Chapter 8 Applications of Integrals

1) The area bounded by the curves y² = 4x and y = x is equal to

1. 1/3
2. 8/3
3. 35/6
4. None of these

Answer: (b) 8/3

Explanation:

For the given curves, y² = 4x and y = x, the intersection points are (0, 0) and (4, 4).

Therefore, the area bounded by the curves,

A =₀∫⁴ [√(4x)-x] dx [since y² = 4x, y = √(4x) ]

A = 2. ₀∫⁴√x dx – ₀∫⁴ x dx

Integrate the function and apply the limits, we get

A = (4/3)(8) – 8

A = (32-24)/3 = 8/3.

Hence, option (b) is the correct answer.

2) The area of the figure bounded by the curve y = log_ex, the x-axis and the straight line x = e is

1. 5-e
2. 3+e
3. 1
4. None of these

Answer: (c) 1

Explanation:

At, x= 1, y = log_e (1) = 0

At, x = e, y = log_e (e) = 1

Therefore, A = ₁∫^e log_ex dx

Using integration by parts,

A = [x log_e x – x]₁^e

Now, apply the limits, we get

A = [e-e-0+1]

A = 1

3) The area of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is

1. ⅜ sq. units
2. ⅝ sq. units
3. ⅞ sq. units
4. 9/8 sq. units

Answer: (d) 9/8 sq. units

Explanation:

For the curves x² = y and x = 4y-2, the points of intersection are x = -1 and x = 2.

Hence, the required area, A = _-1∫² {[(x+2)/4]- [x²/4] } dx

Now, integrate the function and apply the limits, we get

A = (¼)[(10/3)-(-7/6)]

A = (¼)(9/2) = 9/8 sq. units

Hence, the correct answer is option (d) 9/8 sq. units

4) The area enclosed between the graph of y = x³ and the lines x = 0, y = 1, y = 8 is

1. 7
2. 14
3. 45/4
4. None of these

Answer: (c) 45/4

Explanation:

Given curve, y=x³ or x = y^1/3.

Hence, the required area, A = ₁∫⁸ y^1/3 dy

A = [(y^4/3)/(4/3)]₁⁸

Now, apply the limits, we get

A = (¾)(16-1)

A = (¾)(15) = 45/4.

Hence, option (c) 45/4 is the correct answer.

5) The area of the region bounded by the curve y² = x, the y-axis and between y = 2 and y = 4 is

1. 52/3 sq. units
2. 54/3 sq. units
3. 56/3 sq. units
4. None of these

Answer: (c) 56/3

Explanation:

Given: y² = x

Hence, the required area, A = ₂∫⁴ y² dy

A = [y³/3]₂⁴

A = (4³/3) – (2³/3)

A = (64/3) – (8/3)

A = 56/3 sq. units.

6) Area of the region bounded by the curve y = cos x between x = 0 and x = π is

1. 1 sq. units
2. 2 sq. units
3. 3 sq. units
4. 4 sq. units

Answer: (b) 2 sq. units

Explanation:

Given: y= cos x and also provided that x= 0 and x = π

Hence, the required area, A =₀∫^π |cos x| dx

It can also be written as,

A = 2₀∫^π/2 cos x dx

Now, integrate the function, we get

A = 2[sin x]₀^π/2

Now, apply the limits we get

A = 2 sq. units

Hence, the correct answer is option (b) 2 sq. units.

7) Area of the region bounded by the curve x = 2y + 3, the y-axis and between y = -1 and y = 1 is

1. 6 sq. units
2. 4 sq. units
3. 8 sq. units
4. 3/2 sq. units

Answer: (a) 6 sq. units

Explanation:

Required Area =_-1∫¹(2y+3)dy

A=[(2y²/2)+3y]_-1¹

Now, apply the limits, we get

A = 1+3-1+3

A = 6 sq. units.

8) The area bounded by the curve y = x³, the x-axis and two ordinates x = 1 and x = 2 is

1. 15/2 sq. units
2. 15/4 sq. units
3. 17/2 sq. units
4. 17/4 sq. units

Answer: (b) 15/4 sq. units

Explanation:

Required Area = ₁∫² x³ dx

A = [x⁴/4]₁²

Now, apply the limits, we get

A = [(2⁴/4) – (¼)]

A = (16/4) – (¼)

A = 15/4

Hence, the required area is 15/4 is the correct answer.

9) The area of the region bounded by the circle x² + y² = 1 is

1. 2π sq. units
2. 3π sq. units
3. 4π sq. units
4. 1π sq. units

Answer: (d) 1π sq. units

Explanation: Given:

Given circle equation is x²+y² =1, whose centre is (0, 0) and radius is 1.

Therefore, y² = 1-x²

y=√(1-x²)

Hence, the required area, A = 4₀∫¹√(1-x²) dx

Now, integrate the function and apply limits, we get

A = 4(½)(π/2)

A = π sq. units

Hence, option (d) 1π sq. unit is the correct answer.

10) Area bounded by the curve y = sin x and the x-axis between x = 0 and x = 2π is

1. 2 sq. units
2. 3 sq. units
3. 4 sq. units
4. None of these

Answer: (c) 4 sq. units

Explanation:

Required Area, A = ₀∫^2π |sin x| dx

A = ₀∫^π sin x dx + _π∫^2π(-sin x) dx

Now, substitute the limits, we get

A= 4 sq. units.

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