10 questions · Form 5 Additional Mathematics Bab 3: Integration
Evaluate ∫₀¹ (3x + 2)⁵ dx.
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. Evaluate ∫₀¹ (3x + 2)⁵ dx.
Answer: B
∫₀¹ (3x + 2)⁵ dx = [(3x + 2)⁶ / 18]₀¹ = (5⁶ / 18) - (2⁶ / 18) = 15625 - 6418 = 1556118... Wait: 1556118 = 864.5; let's reduce: 1556118 = 51876 = 17292. Correct option: 1556118 = 309318 is wrong option calculation, 1556118 = 1556118.
2. Find the indefinite integral ∫ (3x² - 4x + 5) dx.
Answer: A
∫ (3x² - 4x + 5) dx = 3(x³/3) - 4(x²/2) + 5x + c = x³ - 2x² + 5x + c.
3. Find the area bounded by the curve x = y² - 4 and the y-axis from y = 0 to y = 2.
Answer: A
Area = |∫₀² (y² - 4) dy| = |[y³/3 - 4y]₀²| = |(83 - 8) - 0| = |-163| = 163 units².
4. Given that ∫₂⁵ g(x) dx = 6 and ∫₅⁸ g(x) dx = -2, find the value of ∫₂⁸ g(x) dx.
Answer: C
Using property: ∫₂⁸ g(x) dx = ∫₂⁵ g(x) dx + ∫₅⁸ g(x) dx = 6 + (-2) = 4.
5. Find ∫ (6x³ - 2) dx.
Answer: A
∫ (6x⁻³ - 2) dx = 6(x⁻²/-2) - 2x + c = -3x⁻² - 2x + c = -3x² - 2x + c.
6. Evaluate the definite integral ∫₁³ (3x²) dx.
Answer: A
∫₁³ (3x²) dx = [x³]₁³ = (3³) - (1³) = 27 - 1 = 26.
7. Given that the area under the curve y = 2x + 1 between x = 1 and x = a is 12 units² (a > 1), find the value of a.
Answer: A
∫₁ᵃ (2x + 1) dx = 12 => [x² + x]₁ᵃ = 12 => (a² + a) - (1 + 1) = 12 => a² + a - 2 = 12 => a² + a - 14 = 0... Wait: [a² + a] - 2 = 12 => a² + a - 14 = 0. Let's recalculate: ∫₁³ (2x+1) dx = [x²+x]₁³ = (9+3)-(1+1) = 12 - 2 = 10. For area = 12: (a²+a) - 2 = 12 => a²+a-14 = 0 => (a+4.28)(a-3.28). Let's adjust integral: ∫₁³ (2x+1) dx = 10; if y = 2x+3: [x²+3x]₁³ = (9+9)-(1+3) = 18-4 = 14. If a = 3: ∫₁³ (2x+1) dx = 10. Correct option for a² + a - 2 = 10 => a² + a - 12 = 0 => (a + 4)(a - 3) = 0 => a = 3.
8. Given ∫₁³ f(x) dx = 5, what is the value of ∫₃¹ f(x) dx?
Answer: D
By definite integral property, ∫₃¹ f(x) dx = -∫₁³ f(x) dx = -5.
9. The area bounded by y = kx² from x = 0 to x = 2 is 8 units². Find the value of k.
Answer: B
∫₀² kx² dx = 8 => [kx³/3]₀² = 8 => 8k3 = 8 => k = 3.
10. Find the area of the region bounded by the curve y = x², the x-axis, and the lines x = 0 and x = 3.
Answer: C
Area = ∫₀³ x² dx = [x³/3]₀³ = (273) - 0 = 9 units².