Calculus Practice: Solving Definite and Indefinite Integrals

1. Integration of Polynomials

Let f(x) = -x² + 2x + 3. The integral is:

• a) ∫(-x² + 2x + 3) dx = -x³/3 + x² + 3x + C
• b) [-x³/3 + x² + 3x] from 1 to 2 = 11/3

2. Finding f(x) from f'(x)

Let f'(x) = 3x² - 3. Given f(2) = 1, find f(x):

f(x) = x³ - 3x + C. Substituting f(2) = 1: 8 - 6 + C = 1, so C = -1. Thus, f(x) = x³ - 3x - 1.

3. Solving for f(x) with Initial Conditions

Let f'(x) = 6x² + 2x - 1. Given f(2) = 5:

f(x) = 2x³ + x² - x + C. Substituting f(2) = 5: 2(8) + 4 - 2 + C = 5, so 18 + C = 5, C = -13. Thus, f(x) = 2x³ + x² - x - 13.

4. Polynomial Integration

Let f(x) = 3x² - 4x:

• a) ∫f(x) dx = x³ - 2x² + C
• b) Evaluation results in 32.

5. Definite Integral Properties

Consider ∫₁⁵ f(x) dx = 6:

• a) ∫₁⁵ 2f(x) dx = 2 ∫₁⁵ f(x) dx = 2(6) = 12
• b) ∫₁³ 3 dx = [3x]₁³ = 3(3) - 3(1) = 6

6. Integration using Substitution

Let f(x) = ∫ (8 / (2x - 1)) dx. Using u = 2x - 1, du = 2 dx:

∫ (8/u) * (du/2) = 4 ln|2x - 1| + C. Given point (1, 5): 4 ln(1) + C = 5, so C = 5. f(x) = 4 ln|2x - 1| + 5.

7. Exponential Function Integration

Given f'(x) = 9e^3x passing through (ln 2, 15):

f(x) = 3e^3x + C. 3e^{3(ln 2)} + C = 15 → 3(8) + C = 15 → C = -9. f(x) = 3e^3x - 9.

8. Trigonometric Integration

Given f'(x) = 2 cos(2x) passing through (π/12, 2):

f(x) = sin(2x) + C. sin(2 * π/12) + C = 2 → sin(π/6) + C = 2 → 0.5 + C = 2 → C = 1.5. f(x) = sin(2x) + 1.5.

9. Integration with Radicals

Let f(x) = 20x / (5x² + 4)^3/2:

• a) ∫ f(x) dx = -4 / √(5x² + 4) + C
• b) Definite integral evaluation yields 10/7.

10. Solving for Constant k

Given ∫₀^{e^k} (2 / (1 + 2x)) dx = ln 3:

[ln(1 + 2x)]₀^{e^k} = ln(1 + 2e^k) - ln(1) = ln 3. Therefore, 1 + 2e^k = 3, 2e^k = 2, e^k = 1, so k = 0.

11. Exponential Integration

Find ∫ 4xe^{x^2-9} dx:

Using u = x² - 9, du = 2x dx: 2 ∫ e^u du = 2e^{x^2-9} + C. Given f(3) = 7: 2e⁰ + C = 7, C = 5. f(x) = 2e^{x^2-9} + 5.

12. Derivative Analysis

Given f'(x) = 6x² - 24x:

• a) f(x) = 2x³ - 12x² + C. With f(1) = 0, C = 10. f(x) = 2x³ - 12x² + 10.
• b) Critical point at x = 2.
• c) Concavity: f''(x) = 12x - 24. Concave up for x > 2.

15. Logarithmic Integration

Given f'(x) = 8x / (x² + 5) passing through (2, 9 ln 3):

f(x) = 4 ln|x² + 5| + C. 4 ln(9) + C = 9 ln 3 → 8 ln 3 + C = 9 ln 3 → C = ln 3. f(x) = 4 ln|x² + 5| + ln 3.