Difficulty: Intermediate | Prerequisites: Sections 1–2 of Chapter 4 (Riemann sums and the definite integral), derivative rules from Chapter 3
The Fundamental Theorem of Calculus is the bridge between differential and integral calculus. It says that differentiation and integration are inverse operations, and it gives you a practical shortcut: instead of grinding through Riemann-sum limits, you find an antiderivative and evaluate it at two points. The Net Change Theorem extends this idea to real-world problems where you recover a total quantity from its rate of change.
The FTC has two parts. Part 1 says the derivative of an integral-defined function is just the integrand evaluated at the upper limit. Part 2 says you can evaluate any definite integral by finding an antiderivative F and computing F(b) − F(a). The Net Change Theorem applies Part 2 to rates of change: integrate the rate to get the total change.
Antiderivative
A function F(x) whose derivative is f(x), i.e. F'(x) = f(x). In simple terms, it is the function you get when you "undo" a derivative.
Indefinite integral
The family of all antiderivatives of f, written ∫ f(x) dx = F(x) + C. The result is a function (plus a constant), not a number.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It accounts for the fact that infinitely many functions share the same derivative (they differ only by a vertical shift).
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and g(x) = ∫ₐˣ f(t) dt, then g'(x) = f(x). In simple terms, differentiating an integral gives you the original function back.
Fundamental Theorem of Calculus, Part 2 (FTC2)
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a). This is the practical evaluation tool.
Evaluation bar notation
F(x) |ₐᵇ or [F(x)]ₐᵇ means F(b) − F(a). A shorthand used when applying FTC2.
Net Change Theorem
The integral of a rate of change F'(x) from a to b equals the net change F(b) − F(a). Think of it as: "add up all the little changes and you get the total change."
Displacement
The net change in position: ∫ₐᵇ v(t) dt = s(b) − s(a). Can be zero or negative if the object reverses direction.
Total distance travelled
The integral of the absolute value of velocity: ∫ₐᵇ |v(t)| dt. Always non-negative, because it counts all motion regardless of direction.
If f is continuous on [a, b], define g(x) = ∫ₐˣ f(t) dt for a ≤ x ≤ b. Then g is differentiable and g'(x) = f(x).
The intuition: g(x) accumulates area as x moves to the right. The rate at which that area grows is exactly the height of the function at x.
If f(x) = ∫ₓ² √(t² + 3) dt, flip the limits to get −∫₂ˣ √(t² + 3) dt, then differentiate. Result: f'(x) = −√(x² + 3).
When the upper limit is a function of x rather than x itself, apply the chain rule.
H(x) = ∫₋₁ˣ³ sin(2t²) dt. Let u = x³. By FTC1 + chain rule: H'(x) = sin(2(x³)²) · 3x² = 3x² sin(2x⁶).
F(x) = ∫₃ˣ² cos(t)/(1 + t⁵) dt. Let u = x². Then F'(x) = [cos(x²)/(1 + x¹⁰)] · 2x.
Split the integral at any constant c:
J(x) = ∫₂ₓ³ˣ⁺¹ sin(t⁴) dt = ∫₀³ˣ⁺¹ sin(t⁴) dt − ∫₀²ˣ sin(t⁴) dt.
Differentiate each piece with the chain rule and combine:
J'(x) = sin((3x+1)⁴) · 3 − sin((2x)⁴) · 2.
If f is continuous on [a, b] and F is any antiderivative of f (meaning F' = f), then:
∫ₐᵇ f(x) dx = F(b) − F(a)
This eliminates the need to evaluate limits of Riemann sums. Find one antiderivative, plug in the endpoints, subtract.
F(x)|ₐᵇ means "evaluate F at b, then subtract F at a." Some textbooks write [F(x)]ₐᵇ instead.
∫₁⁹ √x dx: antiderivative of x^(1/2) is (2/3)x^(3/2). Evaluate: (2/3)(27) − (2/3)(1) = 52/3.
∫₋₁⁵ (u+1)(u−2) du: expand to u² − u − 2, antiderivative is u³/3 − u²/2 − 2u. Evaluate at 5 and −1, subtract.
∫₁² (x⁴ + 1)/x² dx: rewrite as x² + x⁻², antiderivative is x³/3 − 1/x. Evaluate at 2 and 1, subtract.
The two parts of the FTC together establish that differentiation and integration are inverse processes. Part 1 says "the derivative of an integral gives back the integrand." Part 2 says "the integral of a derivative gives back the net change." Part 2 is the workhorse for computing definite integrals in practice.
The indefinite integral ∫ f(x) dx represents the most general antiderivative of f. It is a family of functions F(x) + C, where C is the constant of integration.
Linearity properties:
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
∫ k · f(x) dx = k · ∫ f(x) dx
If F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a). This is a restatement of FTC2, framed around rates of change.
The theorem says: if you sum up all the instantaneous rates of change from a to b, you get the total (net) change. In the form the course emphasises:
F(b) = F(a) + ∫ₐᵇ F'(x) dx
This is useful when you know the starting value and the rate, and want the ending value.
Displacement (net change in position): ∫ₐᵇ v(t) dt. This can be positive, negative, or zero.
Total distance travelled: ∫ₐᵇ |v(t)| dt. This is always non-negative.
The difference matters whenever the velocity changes sign (the object reverses direction).
Strategy for ∫ₐᵇ |f(x)| dx:
Find where f(x) = 0 on [a, b]. These are the split points.
Break [a, b] into sub-intervals at the split points.
On each sub-interval, determine whether f is positive or negative. Where f < 0, replace |f(x)| with −f(x).
Evaluate the integral on each piece and add the results.
Example: ∫₀² |x − 1| dx. The function x − 1 = 0 at x = 1. On [0, 1], x − 1 < 0, so |x − 1| = −(x − 1) = 1 − x. On [1, 2], x − 1 > 0. Evaluate each piece: ∫₀¹ (1 − x) dx + ∫₁² (x − 1) dx = 1/2 + 1/2 = 1.
Contrast with ∫₀² (x − 1) dx = 0 (the net area, where the negative and positive parts cancel).
\frac{d}{dx}\int_a^x f(t)\,dt = f(x) \quad \text{(FTC Part 1)}\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\cdot u'(x) \quad \text{(FTC1 + chain rule)}\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F' = f \quad \text{(FTC Part 2)}F(b) = F(a) + \int_a^b F'(x)\,dx \quad \text{(Net Change Theorem)}\text{Displacement} = \int_a^b v(t)\,dt\text{Total distance} = \int_a^b |v(t)|\,dtBasic antiderivatives to know:
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫ cos x dx = sin x + C
∫ sin x dx = −cos x + C
∫ sec²x dx = tan x + C
∫ sec x tan x dx = sec x + C
∫ csc²x dx = −cot x + C
∫ csc x cot x dx = −csc x + C
If v(t) = 8t − t² is an object's velocity and it starts at position 5 at t = 0, the Net Change Theorem gives its position at t = 6 as: s(6) = 5 + ∫₀⁶ (8t − t²) dt. Compute the integral, add the starting value, and you have the answer without ever solving a differential equation.
If h'(t) is the growth rate of a tree in inches per year, then ∫₅⁸ h'(t) dt is the total number of inches the tree grew between years 5 and 8.
Water flows from a tank at r(t) = 10 − 2t litres per minute. The integral ∫₁³ r(t) dt tells you how many litres accumulated in a bucket placed under the tank from minute 1 to minute 3.
Students often forget the chain rule when differentiating ∫ₐᵘ⁽ˣ⁾ f(t) dt. You must multiply by u'(x). Omitting it is one of the most common errors on exams.
Confusing displacement with total distance is a frequent mistake. If velocity is sometimes negative, ∫ₐᵇ v(t) dt and ∫ₐᵇ |v(t)| dt give different answers.
When the variable of integration is in the lower limit (∫ₓᵇ f(t) dt), students forget to flip the sign before differentiating.
Students sometimes write the indefinite integral without + C. This loses marks on nearly every assignment and exam.
⚠️ FTC1 + chain rule problems ("find the derivative of ∫ₐᵘ⁽ˣ⁾ f(t) dt") appear on virtually every MTH 132 exam.
⚠️ Be ready to evaluate definite integrals involving absolute values by splitting at zeros.
⚠️ "Given s(0) and v(t), find s(T)" problems test the Net Change Theorem directly. Write out s(T) = s(0) + ∫₀ᵀ v(t) dt.
True or false: d/dx [∫₁ˣ t³ dt] = x³. (True, by FTC1.)
Fill in the blank: ∫₂⁵ 3 dx = ___. (9.)
True or false: ∫₀² (x − 1) dx gives the total area between the curve and the x-axis on [0, 2]. (False; it gives the net area, which is 0. The total area is 1.)
Fill in the blank: If F'(x) = f(x) and F(1) = 4, then F(3) = 4 + ___. (∫₁³ f(x) dx.)
True or false: d/dx [∫₀ˣ² sin(t) dt] = sin(x²). (False; by the chain rule, the answer is sin(x²) · 2x.)
Q: Find the derivative of H(x) = ∫ₓ¹ (1 − t²)/(1 + t⁴) dt.
A: Flip the limits: H(x) = −∫₁ˣ (1 − t²)/(1 + t⁴) dt. By FTC1: H'(x) = −(1 − x²)/(1 + x⁴).
Q: Evaluate ∫₀³ (t² − 4) dt.
A: Antiderivative: t³/3 − 4t. At t = 3: 9 − 12 = −3. At t = 0: 0. Result: −3.
Q: Evaluate ∫₀³ |t² − 4| dt.
A: t² − 4 = 0 at t = 2 (on [0, 3]). On [0, 2], t² − 4 < 0, so |t² − 4| = 4 − t². On [2, 3], t² − 4 > 0. ∫₀² (4 − t²) dt + ∫₂³ (t² − 4) dt = 16/3 + 7/3 = 23/3.
Q: An object has velocity v(t) = 8t − t² and starts at position 5 at t = 0. What is its position at t = 6?
A: s(6) = 5 + ∫₀⁶ (8t − t²) dt = 5 + [4t² − t³/3]₀⁶ = 5 + 144 − 72 = 77.
Q: Find d/dx [∫ₓˣ² 5 sin²(t) dt].
A: Split at 0: ∫₀ˣ² 5 sin²(t) dt − ∫₀ˣ 5 sin²(t) dt. Differentiate each with chain rule: 5 sin²(x²) · 2x − 5 sin²(x) · 1 = 10x sin²(x²) − 5 sin²(x).
The FTC closes the loop opened in Sections 1 and 2: you no longer need Riemann sums to compute definite integrals. The Substitution Rule (Chapter 4, Section 5) extends the antiderivative toolkit so you can handle more complicated integrands. In Calculus II, the ideas here generalise to integration by parts, partial fractions, and improper integrals. The Net Change Theorem reappears in differential equations and physics whenever you recover a quantity from its rate.
Fundamental Theorem of Calculus, FTC, FTC Part 1, FTC Part 2, antiderivative, indefinite integral, constant of integration, net change theorem, displacement, total distance travelled, absolute value integral, evaluation bar notation, chain rule with integrals, MTH 132, Calculus I, Michigan State University