Definite Integration — Chapter Test | Class 12 IMO

Q1

What is the area of the circle x² + y² = a² using definite integration in the first quadrant?

The area in the first quadrant is ∫[0 to a] √(a² - x²) dx. Using the standard formula, it evaluates to (πa²)/4.

Q2

Evaluate lim n→∞ (1/n) [ (1/n)² + (2/n)² + ... + (n/n)² ].

This converts to ∫[0 to 1] x² dx. Evaluating this gives [x³/3] from 0 to 1 = 1/3.

Q3

If ∫₀¹ (x² + kx + 1) dx = 1, then k equals:

∫₀¹ (x² + kx + 1) dx = [x³/3 + kx²/2 + x]₀¹ = 1/3 + k/2 + 1 = 4/3 + k/2 = 1 → k/2 = −1/3 → k = −2/3.

Q4

Evaluate ∫₀¹ x (1 − x)⁹ dx.

Beta function: ∫₀¹ x^(m−1) (1−x)^(n−1) dx = B(m,n) = Γ(m)Γ(n)/Γ(m+n). Here x = x¹ so m−1=1 → m=2. (1−x)⁹ so n−1=9 → n=10. B(2,10) = Γ(2)Γ(10)/Γ(12) = 1! 9! / 11! = 1/(10×11) = 1/110.

Q5

Determine the area of the region bounded by the curve y = 2x - x² and the x-axis.

The curve intersects the x-axis at x = 0 and x = 2. Area = ∫[0 to 2] (2x - x²) dx = [x² - x³/3] from 0 to 2 = (4 - 8/3) - 0 = 4/3.

Q6

A decorative garden patch in Kerala is shaped as the area bounded by y = sin x, x = 0, x = π, and the x-axis. What is the area of this patch in square units?

Area = ∫[0 to π] sin x dx = [-cos x] from 0 to π = -(-1) - (-1) = 1 + 1 = 2 square units.

Q7

A sweet shop sells boxes of laddoos. The demand function is p(x) = 100 - x², where x is the quantity. If the equilibrium quantity is 6 boxes, what is the Consumer Surplus? (Assume equilibrium price is p(6))

Equilibrium price = 100 - 36 = 64. Consumer Surplus = ∫[0 to 6] (100 - x² - 64) dx = ∫[0 to 6] (36 - x²) dx = [36x - x³/3] from 0 to 6 = 216 - 72 = 144.

Q8

Evaluate the integral involving the greatest integer function: ∫[0 to 1.5] [x] dx.

Split the integral where the integer value changes: ∫[0 to 1] 0 dx + ∫[1 to 1.5] 1 dx = 0 + [x] from 1 to 1.5 = 1.5 - 1 = 0.5.

Q9

∫₀^π |cos x| dx equals:

|cos x| is positive. ∫₀^π |cos x| dx = ∫₀^(π/2) cos x dx + ∫_(π/2)^π (−cos x) dx = [sin x]₀^(π/2) + [−sin x]_(π/2)^π = (1−0) + (0−(−1)) = 2.

Q10

Using the property of definite integrals, evaluate ∫[0 to π/2] (sin x)/(sin x + cos x) dx.

Let I = ∫[0 to π/2] (sin x)/(sin x + cos x) dx. Using property ∫[0 to a] f(x) dx = ∫[0 to a] f(a-x) dx, I = ∫[0 to π/2] (cos x)/(cos x + sin x) dx. Adding both gives 2I = ∫[0 to π/2] 1 dx = π/2. So, I = π/4.

Q11

Evaluate ∫[-1 to 2] |x| dx.

Split at 0: ∫[-1 to 0] -x dx + ∫[0 to 2] x dx = [-x²/2] from -1 to 0 + [x²/2] from 0 to 2 = (0 - (-1/2)) + (2 - 0) = 1/2 + 2 = 5/2.

Q12

If ∫₀¹ xˣ dx = ∫₀¹ (1 − x)ˣ dx, then the value of α is:

Let I = ∫₀¹ xˣ dx. Put x = 1 − t, dx = −dt. Limits: 0→1, 1→0. I = ∫₁⁰ (1−t)ˣ (−dt) = ∫₀¹ (1−t)ˣ dt. So equality holds for any α. The variable of integration is dummy; the integral of xˣ from 0 to 1 equals integral of (1−x)ˣ from 0 to 1 by substitution, regardless of α.

Q13

If ∫[0 to 1] f(x) dx = 5 and ∫[0 to 1] g(x) dx = 2, find ∫[0 to 1] [3f(x) - 2g(x)] dx.

Using the linearity property: 3∫f(x)dx - 2∫g(x)dx = 3(5) - 2(2) = 15 - 4 = 11.

Q14

If ∫₀ᵏ (2x + 3) dx = 10, then k equals:

∫₀ᵏ (2x+3) dx = [x² + 3x]₀ᵏ = k² + 3k = 10 → k² + 3k − 10 = 0 → (k+5)(k−2) = 0 → k = 2 (positive).

Q15

Water flows into an overhead tank in Mumbai at a rate of R(t) = 10e^(0.2t) liters/minute. How much water is added to the tank in the first 5 minutes?

Volume = ∫[0 to 5] 10e^(0.2t) dt = [10/0.2 e^(0.2t)] from 0 to 5 = 50[e¹ - e⁰] = 50(e - 1) liters.