Welcome back from the midterm. Week 7 left the derivative and the definite integral looking like two unrelated limits — one about tangent slopes, one about area. This week proves they are inverse processes, which is the single most useful fact in the whole course: it replaces the Riemann-sum limit of Week 7 with ordinary antidifferentiation for nearly every integral you will ever compute by hand.
9.1 The Fundamental Theorem, Part 1
Define F(x) = ∫ax f(t) dt — the integral from a fixed point a to a variable upper limit x. If f is continuous, then
In words: differentiating an integral with respect to its upper limit gives back the original integrand. The rate at which accumulated area grows, at x, is exactly the height of the curve there — the derivative undoes the integral.
Example 9.1 — Applying FTC Part 1 directly
Find d/dx ∫2x √(1+t3) dt.
Step 1 — Recognise the form of Eq. 9.1
Here f(t) = √(1+t3), a continuous function, and the upper limit is x itself.
Step 2 — Apply the theorem
d/dx ∫2x √(1+t3) dt = √(1+x3) — no antiderivative was ever computed.
When the upper limit is itself a function g(x) rather than plain x, the Chain Rule (Week 4 §4.10) attaches automatically:
9.2 The Fundamental Theorem, Part 2
If f is continuous on [a,b] and F is any antiderivative of f (meaning F′=f), then
written F(x)]ab. This is the practical half: it replaces the Riemann-sum limit of Week 7 §7.5 with "find any antiderivative, then subtract" — no limits, no sums.
Example 9.2 — Recomputing Week 7's integral instantly
Evaluate ∫02 x2 dx using FTC Part 2, and compare to the limit computation in Week 7, Example 7.2.
Step 1 — Find any antiderivative of x²
F(x) = x3/3 works, since F′(x) = x2.
Step 2 — Apply Eq. 9.3
F(2) − F(0) = 8/3 − 0
= 8/3, matching the limit-of-Riemann-sums answer from Week 7 exactly — without a single sum or limit computed this time.
9.3 Why the Two Parts Fit Together
Part 1 says the derivative of an integral (with a variable upper limit) recovers the integrand. Part 2 says the integral can be evaluated using any antiderivative. Together they say differentiation and integration are inverse operations — the same relationship as f and f−1 undoing each other, but for the two central operations of calculus rather than for numbers.
9.4 Indefinite Integrals and Antiderivative Families
The indefinite integral ∫ f(x) dx denotes the entire family of antiderivatives of f, not a single function or a number:
where F′=f and C is an arbitrary constant. This is exactly the consequence of the Mean Value Theorem from Week 6 §6.5: any two antiderivatives of the same function differ only by a constant, so once one is found, all of them are known.
9.5 Basic Antiderivative Formulas
Each differentiation formula from Weeks 4 and 4.9 reads backwards as an antiderivative formula:
The restriction n ≠ −1 in Eq. 9.5a leaves a gap — ∫ x−1 dx — that Week 11 fills once the natural logarithm is introduced.
9.6 The Substitution Method for Indefinite Integrals
Substitution is the antiderivative counterpart of the Chain Rule, and is often called u-substitution. To integrate a composite structure, name the inner function u = g(x), compute du = g′(x)dx, and rewrite the whole integral in terms of u alone.
Example 9.3 — Substitution, indefinite integral
Evaluate ∫ 2x cos(x2) dx.
Step 1 — Choose u to be the inner function
u = x2, so du = 2x dx — which is exactly the factor sitting next to cos(x2).
Step 2 — Rewrite entirely in terms of u
∫ 2x cos(x2) dx = ∫ cos u du
Step 3 — Integrate in u, then substitute back
sin u + C
= sin(x2) + C
Substitution succeeds exactly when, after choosing u, every remaining piece of the integral — not just some of it — can be rewritten purely in terms of u (including dx itself, via du). If a leftover x cannot be eliminated, the chosen u was wrong, or the integral needs a different technique (Week 12).
9.7 Substitution in Definite Integrals
For a definite integral, substitution has an extra bookkeeping step: the limits of integration must also convert from x-values to the corresponding u-values, so that the final evaluation never needs to substitute back to x at all.
Example 9.4 — Substitution, definite integral
Evaluate ∫01 x√(x2+1) dx.
Step 1 — Substitute and convert the limits
u = x2+1, du = 2x dx, so x dx = du/2. At x=0, u=1; at x=1, u=2.
Step 2 — Rewrite the whole integral in u, including the limits
∫12 √u (du/2) = (1/2)∫12 u1/2 du
Step 3 — Integrate and evaluate directly in u (no back-substitution needed)
(1/2)·(2/3)u3/2]12 = (1/3)(23/2−1)
= (2√2 − 1)/3
9.8 The Area Between Two Curves
If f(x) ≥ g(x) on [a,b], the area between the curves is
— always the top curve minus the bottom curve, so the integrand stays non-negative and the answer is a genuine (unsigned) area, even when part of the region dips below the x-axis.
Example 9.5 — Area between two curves
Find the area of the region bounded by y=x2 and y=x+2.
Step 1 — Find the intersection points (the limits of integration)
x2 = x+2 ⇒ x2−x−2=0 ⇒ (x−2)(x+1)=0 ⇒ x=−1, 2.
Step 2 — Determine which curve is on top
At x=0 (inside the interval): line gives 2, parabola gives 0, so the line is on top.
Step 3 — Apply Eq. 9.6 and evaluate
A = ∫−12 [(x+2) − x2] dx = [x2/2 + 2x − x3/3]−12
At x=2: 2+4−8/3 = 10/3. At x=−1: 1/2−2+1/3 = −7/6.
A = 10/3 − (−7/6) = 20/6+7/6 = 27/6 = 9/2
9.9 Worked Capstone Example
Example 9.6 — Combining FTC Part 1, the Chain Rule, and substitution
Find h′(x) for h(x) = ∫1x2 sin(√t) dt, then verify the underlying integral could also be attacked with substitution.
Step 1 — Apply Eq. 9.2 (FTC Part 1 with a function upper limit)
h′(x) = sin(√x2)·d/dx[x2] = sin(x)·2x, using √x2=x for x ≥ 0.
h′(x) = 2x sin x
Step 2 — Note the substitution structure inside the integral itself
Inside ∫ sin(√t) dt, the substitution u=√t (so t=u2, dt=2u du) converts it into ∫ 2u sin u du — not elementary by the formulas in §9.5 alone (it needs integration by parts, Week 12), which is exactly why FTC Part 1 was the efficient route here: it never required finding this antiderivative at all.
9.10 Nomenclature
| Symbol | Meaning |
|---|---|
| FTC | Fundamental Theorem of Calculus |
| ∫ f(x)dx | Indefinite integral: the family of antiderivatives of f |
| C | Constant of integration |
| u-substitution | Rewriting an integral in a new variable u=g(x) to reverse the Chain Rule |
| F(x)]ab | Shorthand for F(b) − F(a) |
9.11 Problems
- FTC Part 1. Find d/dx ∫0x t3 dt.
- FTC Part 1 with the Chain Rule. Find d/dx ∫0sin x √(1+t2) dt.
- FTC Part 2. Evaluate ∫14 (3x2−2x) dx.
- Indefinite integral. Find ∫ (4x3 − sec2x) dx.
- Substitution, indefinite. Evaluate ∫ x2(x3+1)4 dx.
- Substitution, trig. Evaluate ∫ sin3x cos x dx.
- Substitution, definite. Evaluate ∫01 3x2√(x3+1) dx.
- Area between curves, i. Find the area bounded by y=x2 and y=4.
- Area between curves, ii. Find the area bounded by y=x3 and y=x for x∈[0,1].
- Combined. Find the area bounded by y=√x and y=x2, using substitution if needed during the integration step.
9.12 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §5.4 (the Fundamental Theorem of Calculus), §5.5 (indefinite integrals and the substitution method), §5.6 (definite integral substitutions and the area between curves).
- Pearson MyLab Math — the Week 9 homework set is assigned directly from the sections above; see the Calendar for the due date.