Application of Integrals — Class 12 MCQs with Answers

Application of Integrals — important questions & MCQs with answers (Class 12 Mathematics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Area under y=f(x) from a to b (f≥0):

   A.∫ₐᵇ f(x) dx✓ Correct

   B.f(b)−f(a)

   C.f'(x)

   D.(b−a)·f(a)

   Answer: A. ∫ₐᵇ f(x) dx

   Explanation: ∫ₐᵇ f(x) dx sums thin vertical strips f(x)·dx across the interval — that IS the area. Subtracting the two endpoint values of f itself confuses the function with its antiderivative and gives a different number.

2. Q2Easy

   Area under y = f(x) from x = a to x = b (f ≥ 0):

   A.∫_a^b f(x) dx✓ Correct

   B.∫ f(x) dx

   C.f(b) − f(a)

   D.b − a

   Answer: A. ∫_a^b f(x) dx

   Explanation: A definite integral with both limits, ∫_a^b f(x) dx, gives the exact area under a nonnegative curve; dropping the limits leaves only a family of antiderivatives, and b − a alone ignores the curve's height entirely.

3. Q3Easy

   Area between y = f(x) and y = g(x) with f ≥ g:

   A.∫_a^b (f(x) − g(x)) dx✓ Correct

   B.∫ f(x) − g(x)

   C.f(a) − g(a)

   D.f(b) − g(b)

   Answer: A. ∫_a^b (f(x) − g(x)) dx

   Explanation: With f above g on [a,b], area = ∫_a^b (f(x) − g(x)) dx — always upper curve minus lower curve, integrated over the whole interval, not just evaluated at the endpoints.

4. Q4Easy

   Area of circle x² + y² = r²:

   A.4πr²

   B.2πr

   C.r²

   D.πr²✓ Correct

   Answer: D. πr²

   Explanation: Same standard result as any circle: enclosed area is πr², independent of how the boundary is traced. Confusing this with circumference 2πr is the usual slip.

5. Q5Easy

   Area of upper semicircle of x² + y² = 4:

   A.π

   B.4π

   C.2π✓ Correct

   D.8π

   Answer: C. 2π

   Explanation: The upper semicircle of x²+y²=4 has radius 2, so its area is half the full circle: ½×π×2² = 2π. Forgetting the ½ gives the tempting full-circle value 4π instead.

6. Q6Easy

   Area enclosed by y² = 4ax and its latus rectum (x = a):

   A.4a²

   B.(8a²)/3✓ Correct

   C.a²

   D.4a²/3

   Answer: B. (8a²)/3

   Explanation: For y²=4ax, the upper half has height 2√(ax); doubling for the lower half and integrating, area = 2∫₀ᵃ2√(ax)dx = (8/3)a². This latus-rectum area is a standard result worth memorising directly.

7. Q7Medium

   Area of circle x²+y²=r²:

   A.4πr²

   B.2πr

   C.r²

   D.πr²✓ Correct

   Answer: D. πr²

   Explanation: A circle's enclosed area is πr² — memorise it as the base geometric result; integrating 2∫₀ʳ√(r²−x²)dx reproduces the same value but is unnecessary busywork for a full circle.

8. Q8Medium

   Area bounded by y = x, x-axis, and x = 4 equals:

   A.16

   B.4

   C.2

   D.8✓ Correct

   Answer: D. 8

   Explanation: The region under y=x from 0 to 4 is a right triangle with base 4 and height 4, so area = ½×4×4 = 8. Using the full rectangle base×height instead of halving it gives the distractor 16.

9. Q9Medium

   Area between y = x² and y = 4x:

   A.4

   B.8

   C.32/3✓ Correct

   D.16

   Answer: C. 32/3

   Explanation: Between x=0 and 4, the line y=4x lies above the parabola y=x², so area = ∫₀⁴(4x−x²)dx = [2x²−x³/3]₀⁴ = 32−64/3 = 32/3. Subtracting the curves in the wrong order flips the sign entirely.

10. Q10Medium

    The area of the region enclosed by the circle x² + y² = 16 is:

    A.256π

    B.16π✓ Correct

    C.4π

    D.8π

    Answer: B. 16π

    Explanation: The radius is 4, so the area is π × 16 = 16π square units.

11. Q11Medium

    Area enclosed by y = √(4 − x²) (upper semicircle):

    A.4

    B.4π

    C.π

    D.2π✓ Correct

    Answer: D. 2π

    Explanation: y=√(4−x²) traces the upper semicircle of radius 2, so its area is half the full circle: ½×π×2² = 2π. Squaring the radius before taking half is where the 4π distractor sneaks in.

12. Q12Hard

    If part of curve is below x-axis, total area is found by:

    A.Ignoring the part

    B.Splitting and taking absolute values✓ Correct

    C.Negating once

    D.Doubling

    Answer: B. Splitting and taking absolute values

    Explanation: Where the curve dips below the x-axis, the definite integral there comes out negative even though the enclosed region has real area, so you must split at every x-axis crossing and add the absolute value of each piece. Skipping the split lets pieces cancel.

13. Q13Hard

    The area under y = √x from x = 0 to x = 4 is:

    A.8/3

    B.4

    C.32/3

    D.16/3✓ Correct

    Answer: D. 16/3

    Explanation: The integral of √x is 2x^(3/2)/3, which at 4 gives 16/3 square units.

14. Q14Hard

    The area between the curves y = x² and y = x³ from x = 0 to x = 1 is:

    A.1/12✓ Correct

    B.1/6

    C.1/4

    D.1/3

    Answer: A. 1/12

    Explanation: On this interval x² lies above x³, and the integral of x² − x³ is 1/3 − 1/4 = 1/12.

Application of Integrals — FAQs

More Class 12 Mathematics chapters

• 01Relations and Functions
• 02Inverse Trigonometric Functions
• 03Matrices
• 04Determinants
• 05Continuity and Differentiability
• 06Application of Derivatives

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