Difficulty: Intermediate | Prerequisites: Riemann sums, antiderivatives, basic integration rules
The Fundamental Theorem of Calculus (FTC) is the bridge between differential calculus and integral calculus. It says two things: first, that accumulating a quantity via integration can be "undone" by differentiation; second, that you can evaluate a definite integral by finding any antiderivative and plugging in the endpoints. If you have been treating derivatives and integrals as separate toolkits, this is where they snap together. You should already be comfortable with antiderivative rules and the meaning of a definite integral as a signed area.
The FTC connects derivatives and integrals. Part 1 says that differentiating an accumulation function gives back the original integrand. Part 2 says you can evaluate a definite integral using any antiderivative evaluated at the endpoints. The average value formula and net-change theorem are direct consequences.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b], then the function g(x) = integral from a to x of f(t) dt is continuous on [a, b], differentiable on (a, b), and g'(x) = f(x).
In simple terms, if you define a function by accumulating area under f from a fixed starting point to x, the rate of change of that accumulated area at any point is just f(x) itself.
Fundamental Theorem of Calculus, Part 2 (FTC2) / Evaluation Theorem
If f is continuous on [a, b] and F is any antiderivative of f, then the integral from a to b of f(x) dx = F(b) - F(a).
Think of it as: to compute a definite integral, find any antiderivative, evaluate it at the top and bottom limits, and subtract.
Average value of a function
The average value of f on [a, b] is (1/(b - a)) times the integral from a to b of f(x) dx.
In simple terms, it is the single constant height that would produce the same total area over the interval. The "mean value" of the function on that interval.
Net change theorem
The integral from a to b of F'(x) dx = F(b) - F(a). Integrating a rate of change over an interval gives the net change in the original quantity.
Think of it as: if you know the speed at every instant, integrating the speed gives you the total displacement (net change in position).
Accumulation function
A function defined by g(x) = integral from a to x of f(t) dt. The output at each x is the signed area under f from a to x. Sometimes called an "area-so-far" function.
FTC1 tells you that differentiation and integration are inverse operations. If you build a function by integrating, then differentiating it returns you to the original integrand.
FTC2 gives you a computational shortcut. Instead of computing limits of Riemann sums, you find an antiderivative and subtract endpoint values.
Together they say: integration accumulates, differentiation recovers the rate, and antiderivatives let you evaluate the accumulation exactly.
Formula: f_avg = (1/(b - a)) * integral from a to b of f(x) dx.
Rearranging: integral from a to b of f(x) dx = f_avg * (b - a). This is why, if the average value is 10 on [a, b], the integral equals 10(b - a).
Computed with a calculator when the antiderivative is not elementary (e.g. f(x) = cos x on [1, 5]).
If you know f'(x) and f at one point, you can find f at another point: f(b) = f(a) + integral from a to b of f'(x) dx.
Example pattern: given f'(x) = ln(2 + sin x) and f(3) = 4, find f(5). Write f(5) = f(3) + integral from 3 to 5 of ln(2 + sin x) dx = 4 + (numerical value). This is a direct application of FTC2 rewritten as a net-change equation.
The expression lim (h approaches 0) of (1/h) * integral from x to x+h of f(t) dt equals f(x).
This is the derivative definition applied to the accumulation function g(x) = integral from a to x of f(t) dt, because g'(x) = lim (h approaches 0) of [g(x+h) - g(x)] / h = lim (h approaches 0) of (1/h) * integral from x to x+h of f(t) dt.
The answer is f(x), not f'(x) and not 0. This is FTC1 in disguise.
FTC1: d/dx [integral from a to x of f(t) dt] = f(x)
FTC2: integral from a to b of f(x) dx = F(b) - F(a), where F' = f
Average value: f_avg = (1/(b - a)) * integral from a to b of f(x) dx
Net change: f(b) = f(a) + integral from a to b of f'(x) dx
The net-change theorem is how engineers compute total displacement from a velocity function or total charge from a current function. The average value formula appears in physics when calculating the mean temperature of an object over time, which is precisely what Newton's Law of Cooling problems ask for (e.g. average coffee temperature over 30 minutes).
Students often confuse the average value formula with averaging the endpoint values, (f(a) + f(b)) / 2. That is the trapezoidal approximation for a linear function, not the average value formula. They are different things.
The limit expression (1/h) * integral from x to x+h of f(t) dt as h approaches 0 is frequently mistaken to equal f'(x) or 0. It equals f(x). The integral over a shrinking interval divided by the width of that interval returns the integrand's value at that point.
When using the net change theorem, students sometimes write f(5) = integral from 3 to 5 of f'(x) dx, forgetting to add the initial value f(3). The integral gives the change, not the final value.
Students sometimes think any antiderivative will do and forget that +C cancels in a definite integral. That is correct, but they then mistakenly add +C when evaluating definite integrals, which is unnecessary and causes confusion.
⚠️ The average value formula is heavily tested. Know the formula cold and be able to rearrange it: if average value and interval are given, find the integral; if the integral and interval are given, find the average value.
⚠️ "Construct and evaluate an integral equation" problems (given dy/dx and an initial condition, find y at another point) appear regularly. The setup is always y(b) = y(a) + integral from a to b of (dy/dx) dx.
⚠️ The limit form of FTC1 (lim as h approaches 0 of (1/h) * integral from x to x+h of f(t) dt = f(x)) is a popular multiple-choice distractor test. Know it evaluates to f(x).
True or false: The average value of f on [a, b] is always equal to (f(a) + f(b)) / 2.
False. That is only true for linear functions.
Fill in the blank: If the average value of f on [a, b] is 10, then the integral from a to b of f(x) dx = ______.
10(b - a).
True or false: lim (h approaches 0) of (1/h) * integral from x to x+h of f(t) dt = f'(x).
False. It equals f(x).
Q: (Calculator) What is the average value of f(x) = cos x on the interval [1, 5]?
A: f_avg = (1/(5 - 1)) * integral from 1 to 5 of cos x dx = (1/4)(sin 5 - sin 1) ≈ (1/4)(-0.9589 - 0.8415) ≈ -0.450. Answer: (B) -0.450.
Q: If the average value of f on [a, b] is 10, then the integral from a to b of f(x) dx equals what?
A: Rearrange f_avg = (1/(b - a)) * integral. So integral = 10(b - a) = 10b - 10a. Answer: (C) 10b - 10a.
Q: (Calculator) Given f'(x) = ln(2 + sin x) and f(3) = 4, find f(5).
A: f(5) = f(3) + integral from 3 to 5 of ln(2 + sin x) dx = 4 + integral from 3 to 5 of ln(2 + sin x) dx ≈ 4 + 0.555 ≈ 4.555. Answer: (D) 4.555.
Q: What is lim (h approaches 0) of (1/h) * integral from x to x+h of f(t) dt?
A: f(x). This is FTC1 applied to the derivative definition of the accumulation function. Answer: (D) f(x).
This material connects directly to Riemann sums (Chapter 4.2/4.3), since FTC2 replaces the limit-of-sums process with antiderivative evaluation. It also feeds into differential equations: the net-change theorem is the simplest case of solving a separable DE with an initial condition. Later topics on area between curves, volumes of revolution, and arc length all rely on setting up and evaluating definite integrals using FTC2.
Related Terms / Search Tags: FTC, fundamental theorem of calculus, FTC part 1, FTC part 2, evaluation theorem, average value of a function, mean value theorem for integrals, net change theorem, accumulation function, antiderivative evaluation, calculus maximus WS 4.3, AP Calculus FTC, integral initial value problem