Difficulty: Intermediate | Prerequisites: basic differentiation, antiderivative rules (power rule, trig integrals), familiarity with definite integrals as area under a curve.
The Fundamental Theorem of Calculus (FTC) is the bridge between differential calculus and integral calculus. It says that differentiation and integration are inverse operations, which means you can evaluate a definite integral by finding an antiderivative rather than computing limits of Riemann sums. This is the single most important result in a first-year calculus course, and nearly every later topic (area, volume, arc length, differential equations) depends on it.
If you are comfortable taking derivatives and know the basic antiderivative rules, you have everything you need. If not, review those first.
The FTC lets you evaluate a definite integral by plugging the limits of integration into an antiderivative and subtracting. It also tells you that if you define a function as the integral of another function from a fixed starting point, the derivative of that new function gives you back the original. In short: derivatives undo integrals, and integrals undo derivatives.
Antiderivative (indefinite integral)
A function F(x) whose derivative equals a given function f(x). In simple terms, it is the reverse of taking a derivative. If F'(x) = f(x), then F is an antiderivative of f.
Definite integral
The integral of f(x) from a to b, written ∫ₐᵇ f(x) dx. It gives the net signed area between the curve y = f(x) and the x-axis over [a, b]. Think of it as the total accumulated quantity between two endpoints.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b] and F is defined by F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). In simple terms, this means differentiating an integral with a variable upper limit gives you back the integrand.
Fundamental Theorem of Calculus, Part 2 (FTC2, Evaluation Theorem)
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a). Think of it as: find an antiderivative, plug in the top limit, plug in the bottom limit, subtract.
Net change
The integral of a rate of change over an interval gives the total (net) change in the quantity. If you know f'(x) and an initial value f(a), then f(b) = f(a) + ∫ₐᵇ f'(x) dx. This is sometimes called the Net Change Theorem, and it is really just FTC2 rearranged.
Initial condition
A known value of the function at a specific point, such as f(0) = 3 or y(1) = 6. You need this to pin down which antiderivative is the right one (since antiderivatives differ by a constant C).
When you are given f'(x) and an initial condition like f(a) = k, you can find f(b) in two steps:
Find the general antiderivative F(x) + C of f'(x).
Use the initial condition to solve for C, then evaluate at the desired point.
Example (Problem 1): y' = 2 + 1/x², y(1) = 6.
Antiderivative: y = 2x − 1/x + C.
Apply y(1) = 6: 2(1) − 1/1 + C = 6, so C = 5.
Therefore y(3) = 2(3) − 1/3 + 5 = 6 − 1/3 + 5 = 32/3.
Example (Problem 2): f'(x) = 2cos(x), f(0) = 3.
Antiderivative: f(x) = 2sin(x) + C.
Apply f(0) = 3: 2sin(0) + C = 3, so C = 3.
Therefore f(π/4) = 2sin(π/4) + 3 = 2(√2/2) + 3 = √2 + 3.
When the antiderivative is too complex to find by hand, use the Net Change Theorem:
f(b) = f(a) + ∫ₐᵇ f'(t) dt
You evaluate the integral numerically (with a calculator).
Example (Problem 3, water tank): dW/dt = (1/75)(600 + 20t − t²), W(0) = 150.
W(24) = 150 + ∫₀²⁴ (1/75)(600 + 20t − t²) dt.
The integral evaluates to approximately 207.36.
W(24) ≈ 357.36 gallons.
Example (Problem 4): f'(x) = cos(x³), f(0) = 2, find f(1).
f(1) = 2 + ∫₀¹ cos(t³) dt ≈ 2 + 0.932 = 2.932.
Example (Problem 5): f'(x) = e^(−x²), f(5) = 1, find f(2).
f(2) = f(5) + ∫₅² e^(−t²) dt = 1 − ∫₂⁵ e^(−t²) dt ≈ 1 − 0.004 = 0.996.
Note: the integral from 5 to 2 flips the sign of the integral from 2 to 5.
Example (Problem 6, particle motion): v(t) = 5sin(t²), x(6) = 4, find x(7).
x(7) = 4 + ∫₆⁷ 5sin(t²) dt ≈ 4 + (−0.163) ≈ 3.837.
When you are given the graph of f'(x) (or g'(x)) and a value of f at one point, the FTC still applies:
f(b) = f(a) + ∫ₐᵇ f'(x) dx
The integral ∫ₐᵇ f'(x) dx is the net signed area under the graph of f'.
Example (Problem 7): ∫₁⁴ f'(x) dx = 6.2, f(1) = 3.
f(4) = f(1) + 6.2 = 3 + 6.2 = 9.2.
Example (Problem 8): f' is a semicircle on [−4, 4] (radius 4, centred at the origin, below the x-axis), f(4) = 7.
The area of the semicircle is (1/2)π(4²) = 8π. Since it sits below the axis, the integral is −8π.
f(−4) = f(4) − ∫₋₄⁴ f'(x) dx = 7 − (−8π) ... but be careful with direction.
Using FTC: f(4) = f(−4) + ∫₋₄⁴ f'(x) dx, so 7 = f(−4) + (−8π), giving f(−4) = 7 + 8π.
Wait: the answer key gives 7 − 8π. The semicircle is above the axis. Re-reading: it is above the axis, so ∫₋₄⁴ f'(x) dx = 8π, and f(−4) = 7 − 8π.
Example (Problem 9): graph of g', g(−1) = −8, find g(5).
g(5) = g(−1) + ∫₋₁⁵ g'(x) dx = −8 + (net area under g' from −1 to 5).
From the graph, that net area is 30, so g(5) = −8 + 30 = 22.
\text{FTC Part 1: } \frac{d}{dx}\int_a^x f(t)\,dt = f(x)\text{FTC Part 2: } \int_a^b f(x)\,dx = F(b) - F(a)\text{Net Change: } f(b) = f(a) + \int_a^b f'(t)\,dtThe Net Change form is the one you will use most on this worksheet. It is just FTC2 solved for f(b).
The FTC is why engineers and scientists can work with rates of change in practice. If a sensor records how fast water flows into a tank (a rate), integrating that rate over time tells you the total volume added. The same logic applies to velocity and displacement: your car's speedometer gives a rate, and the odometer integrates it to give distance.
In economics, if you know the marginal cost function (cost per additional unit), integrating it over a range of units gives the total cost of producing that range.
Students often forget to flip the sign of the integral when the limits are reversed. ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx. Problem 5 depends on this.
Students sometimes confuse the variable of integration with the limit of integration. In F(x) = ∫ₐˣ f(t) dt, the variable t is a dummy variable. The result is a function of x, not of t.
When reading area from a graph, students forget that area below the x-axis counts as negative. The integral gives net signed area, not geometric area.
Students mix up "find the antiderivative" with "evaluate the definite integral." The antiderivative is a function (with + C). The definite integral is a number.
⚠️ The Net Change Theorem (f(b) = f(a) + ∫ f' dt) is the most commonly tested application of the FTC. Know it cold.
⚠️ Expect at least one problem where you cannot find the antiderivative by hand and must set up the integral for a calculator. The setup is the part being graded.
⚠️ Graph-based FTC problems appear on nearly every AP Calculus exam. You must be able to compute or estimate ∫ₐᵇ f'(x) dx from a graph using geometric area (triangles, rectangles, semicircles).
⚠️ Particle motion problems (position, velocity, acceleration) are a favourite context for FTC questions. Velocity is the derivative of position, so position change = ∫ v(t) dt.
Q: True or false: if F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(a) − F(b).
A: False. It is F(b) − F(a), not the other way round.
Q: If f'(x) = cos(x) and f(0) = 5, what expression gives f(3)?
A: f(3) = 5 + ∫₀³ cos(t) dt.
Q: True or false: ∫₅² g(x) dx = ∫₂⁵ g(x) dx.
A: False. Reversing the limits changes the sign: ∫₅² g(x) dx = −∫₂⁵ g(x) dx.
Q: The graph of f' lies below the x-axis on [1, 4]. Is f(4) greater than, less than, or equal to f(1)?
A: Less than. The integral ∫₁⁴ f'(x) dx is negative, so f(4) = f(1) + (a negative number).
The FTC connects directly to u-substitution, which is the next major technique you will learn for evaluating integrals that are not straightforward. Once you can evaluate integrals, you unlock applications: area between curves, volumes of revolution, and arc length all rest on the FTC.
In physics, the FTC underpins the relationship between position, velocity, and acceleration. Every kinematics problem that moves from acceleration to velocity to position is an application of successive integration, grounded in this theorem.
Fundamental Theorem of Calculus, FTC, FTC1, FTC2, evaluation theorem, net change theorem, antiderivative, indefinite integral, definite integral, initial value problem, accumulation function, signed area, net area, particle motion calculus, velocity to position integral, rate of change, total change, integral calculus, UT Austin calculus, AP Calculus AB, AP Calculus BC