AP Calculus: Fundamental Theorem of Calculus (FTC) – Part 1 and Part 2

Fundamental Theorem of Calculus (FTC) – Part 1 and Part 2

Concept Summary

• Fundamental Theorem of Calculus (FTC) Part 1: Connects differentiation and integration by stating that if ( f ) is continuous on ([a, b]), then the function ( F(x) = \int_a^x f(t) \, dt ) is differentiable on ((a, b)) and ( F'(x) = f(x) ). Significance: Shows integration and differentiation are inverse processes.
• FTC Part 2: Evaluates definite integrals using antiderivatives: ( \int_a^b f(x) \, dx = F(b) - F(a) ), where ( F ) is any antiderivative of ( f ). Significance: Provides a practical method to compute area under curves.
• Antiderivative: A function ( F ) such that ( F'(x) = f(x) ). Significance: Required for applying FTC Part 2.
• Continuity Requirement: ( f ) must be continuous on the interval for FTC Part 1 to apply. Significance: Ensures ( F(x) ) is differentiable and ( F'(x) = f(x) ).
• Variable Upper Limit: In FTC Part 1, the upper limit of integration must be a variable (e.g., ( x )) for ( F(x) ) to be a function. Significance: Differentiating with respect to this variable yields ( f(x) ).

Core Questions

WHAT (definitional)

Q: What is the Fundamental Theorem of Calculus Part 1? A: If ( f ) is continuous on ([a, b]), then ( F(x) = \int_a^x f(t) \, dt ) is differentiable on ((a, b)) and ( F'(x) = f(x) ). Trap/Clarification: The derivative of the integral is the original function only if the upper limit is the variable of differentiation (e.g., ( x )).

Q: What is the Fundamental Theorem of Calculus Part 2? A: If ( F ) is an antiderivative of ( f ) on ([a, b]), then ( \int_a^b f(x) \, dx = F(b) - F(a) ). Trap/Clarification: The antiderivative ( F ) is not unique (any constant ( +C ) works), but the difference ( F(b) - F(a) ) cancels ( C ).

WHY (causal/explanatory)

Q: Why does FTC Part 1 show differentiation and integration are inverse processes? A: Because differentiating the integral of ( f ) (with a variable upper limit) returns ( f ), reversing the integration. Trap/Clarification: This does not mean ( \int f'(x) \, dx = f(x) ) unless ( f ) is continuous (e.g., ( f(x) = |x| ) fails at ( x = 0 )).

Q: Why is FTC Part 2 important for computing definite integrals? A: It reduces the problem of finding the area under ( f ) to evaluating an antiderivative at two points, avoiding Riemann sums. Trap/Clarification: The antiderivative must be continuous on ([a, b]); discontinuities (e.g., ( \int_{-1}^1 \frac{1}{x} \, dx )) violate FTC Part 2.

HOW (process/application)

Q: How do you apply FTC Part 1 to find ( \frac{d}{dx} \int_2^x \sin(t^2) \, dt )? A: Directly differentiate: ( \frac{d}{dx} \int_2^x \sin(t^2) \, dt = \sin(x^2) ). Trap/Clarification: If the upper limit is a function (e.g., ( \int_2^{x^2} \sin(t) \, dt )), use the Chain Rule: ( 2x \sin(x^2) ).

Q: How do you compute ( \int_1^3 2x \, dx ) using FTC Part 2? A: Find an antiderivative ( F(x) = x^2 ), then evaluate ( F(3) - F(1) = 9 - 1 = 8 ). Trap/Clarification: Forgetting to evaluate at both limits (e.g., writing ( F(3) ) only) is a common error.

CAN (conditions/possibilities)

Q: Can FTC Part 1 be applied if ( f ) is discontinuous at a point in ([a, b])? A: No; ( f ) must be continuous on the entire interval for ( F(x) ) to be differentiable. Trap/Clarification: A single discontinuity (e.g., ( f(x) = \frac{1}{x} ) at ( x = 0 )) invalidates FTC Part 1.

Q: Under what conditions can FTC Part 2 be used to evaluate ( \int_a^b f(x) \, dx )? A: ( f ) must be continuous on ([a, b]), and an antiderivative ( F ) must exist on ([a, b]). Trap/Clarification: Piecewise functions (e.g., ( f(x) = 1 ) for ( x \leq 0 ), ( f(x) = 2 ) for ( x > 0 )) require splitting the integral at discontinuities.

Quick Facts & Traps

• Fact: FTC Part 1 requires the upper limit to be a variable (e.g., ( x )) for differentiation to yield ( f(x) ).
• Trap: Differentiating ( \int_a^b f(x) \, dx ) (constant limits)-Reality: The result is 0, not ( f(x) ).
• Fact: FTC Part 2 works for any antiderivative ( F ) (e.g., ( F(x) + C )), since ( C ) cancels in ( F(b) - F(a) ).
• Trap: Using ( F(x) = \ln|x| ) for ( \int_{-1}^1 \frac{1}{x} \, dx )-Reality: ( \ln|x| ) is not continuous on ([-1, 1]), so FTC Part 2 fails.
• Fact: For ( \int_a^g(x) f(t) \, dt ), the derivative is ( f(g(x)) \cdot g'(x) ) (Chain Rule).
• Trap: Ignoring the Chain Rule when the upper limit is a function-Reality: Leads to incorrect derivatives (e.g., ( \frac{d}{dx} \int_0^{x^2} f(t) \, dt = f(x^2) ), not ( f(x^2) \cdot 2x )).

Rapid-Fire True/False

• Statement: If ( F(x) = \int_0^x f(t) \, dt ), then ( F'(x) = f(x) ) even if ( f ) is discontinuous at ( x = 1 ). Answer: FALSE Why the common mistake happens: Overlooking the continuity requirement for FTC Part 1.

• Statement: ( \int_0^2 (3x^2) \, dx = x^3 \big|_0^2 = 8 ) is correct. Answer: TRUE Why the common mistake happens: Forgetting to evaluate at both limits (e.g., writing ( x^3 ) without substitution).

• Statement: The derivative of ( \int_1^{x^3} \cos(t) \, dt ) is ( \cos(x^3) ). Answer: FALSE Why the common mistake happens: Ignoring the Chain Rule (correct answer: ( 3x^2 \cos(x^3) )).