The Fundamental Theorem of Calculus

Learn evaluating with antiderivatives, derivatives of accumulation.

Lesson 3 of 32 parts Practice Flashcards

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Evaluating with Antiderivatives

Objective: Explain evaluating with antiderivatives, apply the stated method, and check its conditions.

If f is continuous on [a,b] and F′=f, then ∫ₐᵇ f(x)dx=F(b)−F(a). This Fundamental Theorem links accumulation to differentiation. An antiderivative’s additive constant cancels in the endpoint difference. Evaluate upper endpoint minus lower endpoint.

Formulas

$∫ₐᵇ f=F(b)−F(a)$

Worked examples

Try itEvaluate ∫₀^7 6x dx.Show answer

Step 1: An antiderivative is F(x)=(6/2)x². Step 2: Use F(7)−F(0). Step 3: The integral is 147.

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Evaluating with Antiderivatives

Objective: Explain evaluating with antiderivatives, apply the stated method, and check its conditions.

Formulas

$∫ₐᵇ f=F(b)−F(a)$

Worked examples

Try itEvaluate ∫₀^7 6x dx.Show answer

Step 1: An antiderivative is F(x)=(6/2)x². Step 2: Use F(7)−F(0). Step 3: The integral is 147.

Derivatives of Accumulation

Objective: Explain derivatives of accumulation, apply the stated method, and check its conditions.

If f is continuous, G(x)=∫ₐˣ f(t)dt has derivative G′(x)=f(x). The upper limit varies with x. A nontrivial upper limit g(x) would introduce a chain factor g′(x); the integration variable t is a dummy variable. A changing upper bound g(x) requires the chain rule.

Formulas

$d/dx ∫ₐˣ f(t)dt=f(x)$

Worked examples

Try itFor G(x)=∫₀^x (6t²+7) dt, find G′(7).Show answer

Step 1: The integrand is continuous, so FTC gives G′(x)=ax²+b. Step 2: Evaluate 6(7)²+7. Step 3: G′(7)=301.

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