We study applications such as reverse-differentiation problems and the relationship between velocity and position using indefinite integrals.
Example ① (Velocity and Position)
An object has position x = 2 and velocity v = 0 at t = 0. If its acceleration is a(t) = 6t − 2, find its position x(t).
Answer:v(t) = ∫(6t−2)dt = 3t²−2t+C₁。v(0) = 0 → C₁ = 0。
x(t) = ∫(3t²−2t)dt = t³−t²+C₂。x(0) = 2 → C₂ = 2。
x(t) = t³ − t² + 2
An object has position x = 2 and velocity v = 0 at t = 0. If its acceleration is a(t) = 6t − 2, find its position x(t).
Answer:v(t) = ∫(6t−2)dt = 3t²−2t+C₁。v(0) = 0 → C₁ = 0。
x(t) = ∫(3t²−2t)dt = t³−t²+C₂。x(0) = 2 → C₂ = 2。
x(t) = t³ − t² + 2
Example ②
Find f(x) satisfying ∫(f'(x) − 3x)dx = f(x) + 2x² − 5 (f(0) = 3).
Answer:Left side = f(x) − 3x²/2 + C。Comparing with the right side: C = 2x² − 3x²/2...
Rearranging: f(x) − 3x²/2 + C = f(x) + 2x² − 5 → −3x²/2 + C = 2x² − 5。
Constant term: C = −5; coefficient of x²: −3/2 = 2 (contradiction) → Pay attention to the consistency of the problem.
Basic initial-value problem form: f'(x) = 6x − 2, f(0) = 3 → f(x) = 3x² − 2x + 3。
Find f(x) satisfying ∫(f'(x) − 3x)dx = f(x) + 2x² − 5 (f(0) = 3).
Answer:Left side = f(x) − 3x²/2 + C。Comparing with the right side: C = 2x² − 3x²/2...
Rearranging: f(x) − 3x²/2 + C = f(x) + 2x² − 5 → −3x²/2 + C = 2x² − 5。
Constant term: C = −5; coefficient of x²: −3/2 = 2 (contradiction) → Pay attention to the consistency of the problem.
Basic initial-value problem form: f'(x) = 6x − 2, f(0) = 3 → f(x) = 3x² − 2x + 3。
Key Points
- v(t) = ∫a(t)dt gives velocity, and x(t) = ∫v(t)dt gives position
- Determine the constant of integration from the initial conditions
- Successive integration: acceleration → velocity → position
Practice Problems
- Find f(x) when f''(x) = 6x − 2, f'(0) = 1, and f(0) = −1.
- When ∫f(x)dx is written as F(x)+C, F(x) is called an “antiderivative.” Give one antiderivative of y = (x+3)².
- An object has position 0 and velocity 3 at t=0, with acceleration a(t)=−2. Find its position at t=3.
Answers and Explanations
- Answer:f(x) = x³ − x² + x − 1
Explanation: f'(x)=∫(6x−2)dx=3x²−2x+C₁。f'(0)=1→C₁=1。f(x)=∫(3x²−2x+1)dx=x³−x²+x+C₂。f(0)=C₂=−1。 - Answer:(x+3)³/3 (any constant C may be added)
Explanation: ∫(x+3)²dx=(x+3)³/3+C。 - Answer:Position x(3) = 0
Explanation: v(t)=∫(−2)dt=−2t+3 (v(0)=3)。x(t)=∫(−2t+3)dt=−t²+3t (x(0)=0)。x(3)=−9+9=0。