Calculus Practice: Solving Definite and Indefinite Integrals
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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