Short on time? This page is the recap. For explanations, the figure and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: ∫ (a to b) f(x) dx is the definite integral from a to b; [F(x)] (a to b) means F(b) − F(a).
Recap
- An antiderivative of f is any F with F′ = f. Antiderivatives of the same f differ only by a constant.
- If f is continuous, the accumulation function ∫ (a to x) f(t) dt is one antiderivative (Topic 6.4).
- Theorem: if f is continuous on [a, b] and F is any antiderivative of f, then ∫ (a to b) f(x) dx = F(b) − F(a).
- No + C is needed: it cancels.
- The answer is a signed quantity. It can be negative or zero.
Key relationships
| Idea | Statement |
|---|---|
| Antiderivative | F′(x) = f(x) |
| Evaluation | ∫ (a to b) f(x) dx = [F(x)] (a to b) = F(b) − F(a) |
| Net change | ∫ (a to b) F′(x) dx = F(b) − F(a) |
| Final amount | F(b) = F(a) + ∫ (a to b) F′(x) dx |
| Riemann sum link | lim Σ f(xᵢ) Δx = ∫ (a to b) f(x) dx, then evaluate with F |
Basic antiderivatives: xⁿ → xⁿ⁺¹/(n + 1) for n ≠ −1; 1/x → ln x (x > 0); eˣ → eˣ; cos x → sin x; sin x → −cos x; sec²x → tan x.
The five-step method
- Check that f is continuous on the whole of [a, b].
- Rewrite roots and fractions as powers.
- Find F term by term, and check F′ = f.
- Write [F(x)] (a to b).
- Substitute the upper limit, then subtract the bracketed value at the lower limit.
Assumptions behind the method
- f must be continuous on the closed interval [a, b]. A single point where f is undefined inside the interval breaks the theorem.
- F must be an antiderivative on the whole interval, not just at the endpoints.
Mistakes to avoid
- F(a) − F(b) instead of F(b) − F(a).
- No brackets around F(a), so only its first term is subtracted.
- Writing f(b) − f(a): substituting into the integrand instead of the antiderivative.
- Sign slips with trig: the antiderivative of sin x is −cos x.
- Using the theorem across a discontinuity, for example ∫ (−1 to 2) (3/x⁴) dx.
- Calling net change “distance”. ∫ v(t) dt is net change in position.
Quick self-check
- Evaluate ∫ (1 to 9) √x dx. (52/3: [(2/3)x^(3/2)] (1 to 9) = 18 − 2/3)
- Evaluate ∫ (0 to π) cos x dx. (0: [sin x] (0 to π) = 0 − 0. The areas above and below the axis cancel.)
- F′ = f, f is continuous, F(2) = 7 and F(6) = 1. Find ∫ (2 to 6) f(x) dx. (1 − 7 = −6)
Next: practice questions.