Top Solved Examples & Important Questions: Application of Integrals Class 12

If you’ve covered the theory of Application of Integrals and want to test yourself before the exam, this post is built for you. It collects some of the most frequently repeated question patterns from CBSE Class 12 board papers, each solved in full, so you can check your working step by step.

Why These Questions Matter

CBSE tends to reuse the same handful of question patterns year after year, even when the specific numbers change. Recognizing the pattern — not just memorizing one answer — is what actually earns marks under exam pressure. The questions below are grouped by pattern type so you can see exactly what to look for.

Pattern 1: Area Bounded by a Curve and Given Ordinates

Question: Find the area under the curve y = x⁴ between x = 1 and x = 5.

Solution: Area = ∫₁⁵ x⁴ dx = [x⁵/5]₁⁵ = 3125/5 − 1/5 = 3124/5 square units

This pattern — a power function between two given x-values — is the simplest type in the chapter and a reliable place to pick up easy marks.

Pattern 2: Curve With a Sign Change

Question: Find the area bounded by the curve y = x³, the x-axis, and the ordinates x = −2 and x = 1.

Solution: Since y = x³ is negative for x < 0 and positive for x > 0, split the integral at x = 0:

Area = |∫₋₂⁰ x³ dx| + ∫₀¹ x³ dx = |−4| + 1/4 = 4 + 1/4 = 17/4 square units

Whenever a cubic or odd-degree function spans across x = 0, expect this kind of split — it’s a favourite CBSE trap.

Pattern 3: Trigonometric Curve Over a Full Period

Question: Find the area bounded by the curve y = cos x between x = 0 and x = 2π.

Solution: Cosine is positive on [0, π/2], negative on [π/2, 3π/2], and positive again on [3π/2, 2π]. Splitting accordingly:

Area = ∫₀^(π/2) cos x dx + |∫_(π/2)^(3π/2) cos x dx| + ∫_(3π/2)^(2π) cos x dx = 1 + 2 + 1 = 4 square units

The same result (4 square units) holds for y = sin x over the same interval [0, 2π], by an almost identical argument — a pattern worth recognizing instantly rather than re-deriving each time.

Pattern 4: Area Involving a Modulus Function

Question: Sketch the graph of y = |x + 3| and evaluate ∫₋₆⁰ |x + 3| dx.

Solution: The expression x + 3 changes sign at x = −3. For x between −6 and −3, |x + 3| = −(x + 3); for x between −3 and 0, |x + 3| = x + 3.

Area = ∫₋₆⁻³ −(x + 3) dx + ∫₋₃⁰ (x + 3) dx

Geometrically, this is a “V” shape made of two identical right triangles, each with base 3 and height 3:

Area = 2 × (1/2 × 3 × 3) = 9 square units

Recognizing the triangular shape early can save you from doing the full integration by hand.

Pattern 5: A Function Defined Differently on Either Side of Zero

Question: The area bounded by the curve y = x|x|, the x-axis, and the ordinates x = −1 and x = 1.

Solution: For x > 0, y = x|x| = x², a positive parabola. For x < 0, y = x|x| = −x², which is negative (below the x-axis).

Area = |∫₋₁⁰ −x² dx| + ∫₀¹ x² dx = 1/3 + 1/3 = 2/3 square units

This question tests whether you can correctly rewrite a piecewise function before integrating — always split at x = 0 whenever |x| appears multiplied by another expression.

Pattern 6: Standard Ellipse Area

Question: Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1.

Solution: With a = 4 and b = 3, Area = πab = π(4)(3) = 12π square units

This is a two-line answer once you know the formula — one of the fastest marks available in the entire chapter.

Quick Revision Checklist Before Your Exam

Before walking into your exam, make sure you can, without hesitation: recall the standard integral ∫√(a² − x²) dx; identify where a curve crosses the x-axis and split the integral there; derive πa² and πab from first principles; and correctly rewrite modulus and piecewise expressions like x|x| before integrating.

Related Reading

• Area Between Two Curves: Class 12 Integration Methods Explained

• Application of Integrals Class 12: Complete Guide to Area Under Curves

• Area of an Ellipse Using Integration: Formula, Derivation & Solved Examples

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Test Your Knowledge

Q1.The area under y = x⁴ from x = 1 to x = 5 is:

3124/5625/53125/53124/4

Q2.For y = x³ between x = −2 and x = 1, the area is found by:

Integrating straight from −2 to 1 in one pieceTaking |−4| + 1/4 after splitting at x = 0Squaring the function firstUsing horizontal strips

Q3.The area bounded by y = cos x over [0, 2π] is:

2π024

Q4.∫₋₆⁰ |x + 3| dx equals:

18963

Q5.The area of the region bounded by the ellipse x²/16 + y²/9 = 1 is:

12π7π24π144π