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Calculus II (Integration & Series) Unit 1: Review: FTC & substitution

Calculus II builds on two ideas from Calculus I: the Fundamental Theorem and u-substitution. Get these automatic before new techniques.

Covers: Fundamental Theorem, substitution, accumulation functions

Study guide

Calculus 2 builds on the Fundamental Theorem of Calculus. Part 2 evaluates definite integrals with antiderivatives: ∫ₐᵇ f(x) dx = F(b) − F(a). Part 1 says accumulation functions g(x) = ∫ₐˣ f(t) dt are antiderivatives of f, so g′(x) = f(x). With a variable upper limit, apply the chain rule: d/dx ∫ₐ^u(x) f(t) dt = f(u(x))·u′(x). Accumulation functions are increasing where f is positive and have extrema where f changes sign.

U-substitution is the main technique for reversing the chain rule. Pick u to be an inner function whose derivative (up to a constant factor) also appears in the integrand, such as u = x² in ∫ 2x e^(x²) dx. Rewrite everything in terms of u and du, integrate, and substitute back.

For definite integrals, either substitute back to x at the end or change the limits: plug the original limits into u = g(x) to get new u-limits, then evaluate in u without returning to x. Mixing the two methods is a common error.

Basic antiderivatives to know cold include ∫ eᵘ du = eᵘ + C, ∫ 1/u du = ln|u| + C, ∫ sin u du = −cos u + C, ∫ sec²u du = tan u + C, and ∫ 1/(1 + u²) du = arctan u + C.

Also know: ∫ sec u tan u du = sec u + C, ∫ tan u du = ln|sec u| + C, ∫ sec u du = ln|sec u + tan u| + C, and ∫ aᵘ du = aᵘ/ln a + C. In general, ∫ 1/(a² + u²) du = (1/a)arctan(u/a) + C and ∫ 1/√(a² − u²) du = arcsin(u/a) + C. Completing the square turns integrands like 1/(x² + 4x + 13) into this form, because x² + 4x + 13 = (x + 2)² + 9.

If the numerator is a constant multiple of the denominator's derivative, substitute u = denominator: ∫ g′(x)/g(x) dx = ln|g(x)| + C. If both limits vary, d/dx ∫_h(x)^g(x) f(t) dt = f(g(x))·g′(x) − f(h(x))·h′(x).

Key ideas

• FTC part 2 evaluates definite integrals with antiderivatives.
• FTC part 1 with the chain rule: d/dx ∫ₐ^g(x) f(t) dt = f(g(x))·g′(x).
• u-substitution: choose u as an inner function whose derivative appears in the integrand.
• For definite integrals, change the bounds to u-values or switch back to x before evaluating.

Formulas and rules

Common mistakes

• Forgetting the chain-rule factor g′(x) in FTC part 1.
• Mixing x-bounds with a u-integrand.

Words to know

FTC with a variable upper limit

d/dx ∫ₐ^g(x) f(t) dt = f(g(x))·g′(x)

Choosing u (substitution)

Pick an inner function whose derivative also appears: ∫ 2x·eˣ² dx → u = x².

∫ eᵘ du

eᵘ + C

Changing limits in substitution

New limits come from plugging the old limits into u.

FTC part 2

∫ₐᵇ f(x) dx = F(b) − F(a)

FTC part 1

d/dx ∫ₐˣ f(t) dt = f(x)

Accumulation function

g(x) = ∫ₐˣ f(t) dt; increasing where f > 0.

∫ 1/u du

ln|u| + C

∫ sin u du

−cos u + C

∫ 1/(1 + u²) du

arctan u + C

Sample practice questions

33 more questions like these, from level 1 to 5, are in the practice set.