Source: MATH 104 Calculus 1, Penn State
Tags: FTC, fundamental theorem of calculus, first fundamental theorem, second fundamental theorem, antiderivative, definite integral, integration, differentiation
Difficulty: Intermediate | Prerequisites: Comfortable with limits, continuity, and basic differentiation. You should know what a derivative is and have worked with indefinite integrals.
The Fundamental Theorems of Calculus sit at the centre of MATH 104 and connect the two main operations you have been learning, differentiation and integration, into a single coherent framework. Up to this point, integrals and derivatives may have felt like separate techniques. The FTC shows they are inverses of each other, which is what makes evaluating definite integrals practical rather than requiring you to compute limits of Riemann sums every time. You need a solid grasp of antiderivatives and continuity before this material will click.
The First Fundamental Theorem says that if you differentiate the integral of a continuous function, you get the original function back. The Second Fundamental Theorem says you can evaluate any definite integral by finding an antiderivative and plugging in the limits. Together, they turn integration from a limit-of-sums problem into an antiderivative problem.
Antiderivative
A function F(x) whose derivative equals f(x), i.e. F'(x) = f(x). Think of it as the reverse of taking a derivative.
Definite integral
The signed area under the curve f(x) from x = a to x = b, written ∫ₐᵇ f(x) dx. In simple terms, it adds up infinitely many tiny slices of area between two bounds.
Continuous function
A function with no breaks, jumps, or holes in its graph on a given interval. Think of it as a function you can draw without lifting your pen.
Fundamental Theorem of Calculus (FTC)
The pair of theorems linking differentiation and integration. In simple terms, they say that derivatives and integrals undo each other.
Riemann sum
An approximation of a definite integral using a finite number of rectangles. Think of it as the "rough draft" version of an integral before you take the limit.
States that differentiation and integration are inverse operations
If f is continuous on [a, b], define F(x) = ∫ₐˣ f(t) dt. Then F is differentiable on (a, b) and F'(x) = f(x)
In plain terms: the derivative of the "area so far" function gives you back the original function
This guarantees that every continuous function has an antiderivative
The key requirement is continuity. If f has a discontinuity, the theorem does not apply at that point
Provides the practical evaluation method for definite integrals
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a)
"Any" antiderivative works because all antiderivatives of f differ by a constant, and the constant cancels when you subtract
This is the theorem you use most often in practice: find an antiderivative, plug in the upper and lower bounds, subtract
Part 1 tells you that the process works (integration followed by differentiation returns you to the start)
Part 2 tells you how to compute the answer (find F, evaluate F(b) − F(a))
Together they eliminate the need to evaluate definite integrals via Riemann sum limits
FTC Part 1
If f is continuous on [a, b], then for x in (a, b):
d/dx [ ∫ₐˣ f(t) dt ] = f(x)
FTC Part 2
If F'(x) = f(x) and f is continuous on [a, b], then:
∫ₐᵇ f(x) dx = F(b) − F(a)
The expression F(b) − F(a) is often written as [F(x)]ₐᵇ or F(x) |ₐᵇ.
The FTC is the reason engineers can calculate total work, total distance, or accumulated charge by finding antiderivatives instead of summing infinitely many tiny pieces by hand. Any time a physicist writes "integrate the force over the path to get work done," they are relying on Part 2 to turn that integral into a subtraction of two antiderivative values.
Students often think Part 1 and Part 2 say the same thing. They do not. Part 1 is about differentiating an integral; Part 2 is about evaluating one.
Students sometimes forget the continuity requirement. The FTC requires f to be continuous on the interval. If there is a discontinuity inside [a, b], you cannot apply it blindly; you need to split the integral at the discontinuity.
A common error is writing F(a) − F(b) instead of F(b) − F(a). The order matters: upper bound minus lower bound.
Students occasionally confuse "an antiderivative" with "the antiderivative." There are infinitely many antiderivatives (they differ by a constant), but the constant cancels in Part 2, so any one of them works.
⚠️ Expect at least one problem requiring you to apply FTC Part 1 with a chain rule twist, e.g. d/dx [ ∫ₐ^(x²) f(t) dt ]. You need to multiply by the derivative of the upper limit.
⚠️ Evaluation problems using Part 2 are near-guaranteed. Know your basic antiderivatives cold: power rule, trig functions, exponentials.
⚠️ Conceptual questions may ask you to explain the relationship between the two parts, or to state what conditions are required for the FTC to hold.
True or false: The First Fundamental Theorem guarantees that every continuous function has an antiderivative.
Fill in the blank: ∫ₐᵇ f(x) dx = F() − F(), where F is any antiderivative of f.
True or false: If f has a discontinuity at c inside [a, b], you can still apply FTC Part 2 directly to ∫ₐᵇ f(x) dx.
Fill in the blank: d/dx [ ∫₂ˣ sin(t) dt ] = ___.
True or false: All antiderivatives of a given function differ by a constant.
Answers: 1. True. 2. F(b) − F(a). 3. False (split the integral at c). 4. sin(x). 5. True.
Q: Evaluate ∫₁³ (2x + 1) dx using the Second Fundamental Theorem.
A: The antiderivative is F(x) = x² + x. Then F(3) − F(1) = (9 + 3) − (1 + 1) = 12 − 2 = 10.
Q: Let G(x) = ∫₀ˣ cos(t) dt. Find G'(x).
A: By FTC Part 1, G'(x) = cos(x).
Q: Find d/dx [ ∫₁^(x³) e^t dt ].
A: Let u = x³. By FTC Part 1 combined with the chain rule: the answer is e^(x³) · 3x².
Q: Why does the "+C" constant from antidifferentiation not appear when you evaluate a definite integral?
A: Because definite integration subtracts F(a) from F(b). Any constant C would appear in both terms and cancel: (F(b) + C) − (F(a) + C) = F(b) − F(a).
Q: State the conditions required for FTC Part 2 to apply.
A: f must be continuous on the closed interval [a, b], and F must be an antiderivative of f on that interval.
The FTC connects directly to the derivative rules (product rule, chain rule) because FTC Part 1 problems with variable upper limits often require the chain rule. It also sets up the substitution method (u-substitution), which is the integration counterpart of the chain rule. If you go on to study improper integrals, you will extend Part 2 to cases where one of the bounds is infinite.
FTC, fundamental theorem of calculus, first fundamental theorem, second fundamental theorem, antiderivative, definite integral, indefinite integral, Riemann sum, area under a curve, net change theorem, accumulation function, evaluation theorem, integration and differentiation inverse, MATH 104, calculus 1, Penn State