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Generate Fundamental Theorem of Calculus Worksheets

Explore the crucial connection between differentiation and integration. Use it to evaluate definite integrals.

Connecting Two Big Ideas: Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is the bridge between differentiation and integration, proving that these two operations are inverses of each other. Part 1 says that integrating a function and then differentiating gives back the original function. Part 2 says that a definite integral can be evaluated by finding an antiderivative. This theorem turns area calculations from tedious summations into straightforward algebra.

Components of the Fundamental Theorem

This section covers both parts of the theorem:

  • FTC Part 1: If F(x) = ∫ from a to x of f(t) dt, then F'(x) = f(x). Differentiation undoes integration.
  • FTC Part 2: ∫ from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f. This provides a practical method for computing definite integrals.
  • Net Change Theorem: The integral of a rate of change over an interval gives the net change: ∫ from a to b of f'(x) dx = f(b) - f(a).
  • Evaluating with FTC: The step-by-step process of finding the antiderivative, substituting the bounds, and computing the difference.

Examples of the Fundamental Theorem

FTC Part 1 Examples

  • If F(x) = ∫ from 0 to x of t² dt, then F'(x) = x² — the derivative of the integral returns the original function.
  • If F(x) = ∫ from 1 to x of (3t + 5) dt, then F'(x) = 3x + 5.
  • If F(x) = ∫ from 2 to x of cos(t) dt, then F'(x) = cos(x).

FTC Part 2 Examples

  • Evaluate ∫ from 1 to 3 of 2x dx: Antiderivative is x². F(3) - F(1) = 9 - 1 = 8.
  • Evaluate ∫ from 0 to 2 of (3x² + 1) dx: Antiderivative is x³ + x. F(2) - F(0) = (8 + 2) - (0 + 0) = 10.
  • Evaluate ∫ from 1 to 4 of √x dx: Antiderivative is 2x^(3/2)/3. F(4) - F(1) = 2(8)/3 - 2(1)/3 = 16/3 - 2/3 = 14/3 ≈ 4.67.

Net Change Examples

  • A population grows at a rate of P'(t) = 100 + 2t people per year. The population gain from year 0 to year 5 is ∫ from 0 to 5 of (100 + 2t) dt = [100t + t²] from 0 to 5 = 500 + 25 = 525 people.
  • Water flows into a tank at 4t liters per minute. Total water from t = 0 to t = 3 is ∫ 4t dt = 2t², evaluated as 2(9) - 0 = 18 liters.
  • A car's velocity is v(t) = 3t m/s. Distance from t = 2 to t = 6 is ∫ 3t dt = 3t²/2, evaluated as 3(36)/2 - 3(4)/2 = 54 - 6 = 48 meters.

Step-by-Step Evaluation Examples

  • Evaluate ∫ from -1 to 2 of (x² - 1) dx: Step 1: Antiderivative is x³/3 - x. Step 2: F(2) = 8/3 - 2 = 2/3. Step 3: F(-1) = -1/3 + 1 = 2/3. Step 4: 2/3 - 2/3 = 0.
  • Evaluate ∫ from 0 to Ï€ of sin(x) dx: Antiderivative is -cos(x). Evaluate: -cos(Ï€) - (-cos(0)) = -(-1) - (-1) = 1 + 1 = 2.
  • Evaluate ∫ from 1 to e of 1/x dx: Antiderivative is ln(x). Evaluate: ln(e) - ln(1) = 1 - 0 = 1.