Application of Integrals — Class 12 MCQs with Answers

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

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

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

   A∫ₐᵇ f(x) dxBf(b)−f(a)Cf'(x)D(b−a)·f(a)
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   Answer: ∫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. Q2 of 6Easy

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

   A∫_a^b f(x) dxB∫ f(x) dxCf(b) − f(a)Db − a
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   Answer: ∫_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. Q3 of 6Medium

   Area between y = x² and y = 4x:

   

   A4B8C32/3D16
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   Answer: 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.

4. Q4 of 6Medium

   Assertion (A): The area of the region in the first quadrant bounded by the parabola y = x², the y-axis and the line y = 4 is 8/3 square units. Reason (R): In the first quadrant, the parabola y = x² can be written as x = √y.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: R is true, since x ≥ 0 in the first quadrant. The region lies between the y-axis and the curve for 0 ≤ y ≤ 4, so its area is ∫₀⁴ √y dy = (2/3) × 8 = 16/3, not 8/3. So A is false (8/3 is the area between the curve and the x-axis from x = 0 to x = 2).

5. Q5 of 6Hard

   The area bounded by the parabola y = x² and the line y = 4 is:

   

   A8/3B32/3C64/3D16/3
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   Answer: 32/3

   Explanation: The curves meet at x = ±2, and the integral of 4 − x² between them is 32/3 square units.

6. Q6 of 6Hard

   Assertion (A): The area of the region bounded by the ellipse x²/a² + y²/b² = 1 (a, b > 0) is b/a times the area of the region bounded by the circle x² + y² = a². Reason (R): For every x in [−a, a], the ordinate (b/a)√(a² − x²) of the ellipse is b/a times the ordinate √(a² − x²) of the circle, so each vertical strip of the ellipse is b/a times the matching strip of the circle.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: A is true: πab = (b/a) × πa². R is true: solving the ellipse for y ≥ 0 gives y = (b/a)√(a² − x²). Area is the integral of these ordinates over the same interval, so multiplying every ordinate by b/a multiplies the area by b/a; R explains A.

Application of Integrals notes: key concepts (Class 12 Mathematics)

Definite integrals measure plane areas. The chapter finds the area under a curve as ∫y dx, the area about the y-axis as ∫x dy, the area between two curves as ∫(upper − lower), and the standard areas of circles, semicircles, parabolas and ellipses — after sketching the region and locating limits from intersection points.

Area under a curve

For y = f(x) ≥ 0 between x = a and x = b, the area is ∫ₐᵇ y dx; the definite integral is the region's area.

Area about the y-axis

When integrating along y, area = ∫ x dy taken from y = c to y = d, expressing x as a function of y between the y-limits.

Region below the x-axis

There ∫y dx is negative, so the geometric area is its absolute value; split the interval at points where f changes sign.

Area between two curves

Between x = a and x = b with f ≥ g, area = ∫ₐᵇ (f(x) − g(x)) dx, the upper curve minus the lower.

Area between curve and line

Solve the curve and line together for the intersection points, which become the limits, then integrate the difference.

Finding the limits

Intersection points are found by solving the two equations simultaneously; they set the bounds of every area integral.

Area of a circle

For x² + y² = r², area = πr²; a quarter comes from ∫₀ʳ √(r²−x²)dx = πr²/4 and a semicircle is ½πr².

Area of an ellipse

The region x²/a² + y²/b² = 1 has area πab; one quadrant is πab/4, obtained by integrating y = b√(1−x²/a²).

Area under standard curves

Areas under y = sin x, y = eˣ, y = √x and parabolas y² = 4ax follow directly from their definite integrals.

Parabola and latus rectum

The region bounded by y² = 4ax and its latus rectum x = a has area 8a²/3, a common board result.

Using symmetry

For symmetric regions (circle, ellipse) compute one quadrant and multiply, cutting the algebra.