Difficulty: Intermediate | Prerequisites: Familiarity with derivatives, basic antiderivatives, and Riemann sums (Sections 5.1 and 5.2).
The Fundamental Theorems of Calculus connect two ideas that seem unrelated on the surface: the derivative (rate of change) and the definite integral (accumulated area under a curve). This section sits at the pivot point of the entire calculus sequence. Everything before it, limits, derivatives, Riemann sums, builds toward this result, and everything after it, integration techniques, applications, differential equations, depends on it. If you missed the last few weeks, you will want a solid grip on antiderivatives and the definition of the definite integral before reading on.
FTC Part 2 lets you evaluate a definite integral by finding an antiderivative and plugging in the endpoints. FTC Part 1 says that if you define a function as the integral of a continuous function with a variable upper limit, its derivative gives you back the original function. The Generalized FTC 1 extends this to cases where the limits of integration are themselves functions, using the chain rule.
Antiderivative (indefinite integral)
A function F(x) whose derivative equals f(x), i.e. F'(x) = f(x). Think of it as "undoing" differentiation. If the derivative tells you the speed, the antiderivative tells you the position.
Definite integral
The value ∫ₐᵇ f(x) dx, representing the net signed area between f(x) and the x-axis from x = a to x = b. In simple terms, this means the total accumulated quantity over an interval.
Accumulation function
A function defined as g(x) = ∫ₐˣ f(t) dt, where the upper limit varies. Think of it as a running total: as x moves to the right, the function accumulates more area under f.
Continuity
A function f is continuous on [a, b] if it has no breaks, jumps, or holes on that interval. Both parts of the FTC require f to be continuous. In simple terms, you can draw the graph without lifting your pen.
Constant of integration (C)
The arbitrary constant added to an indefinite integral, reflecting the fact that antiderivatives are only unique up to a vertical shift. Think of it as the "unknown starting point" that disappears when you evaluate a definite integral.
If F is continuous on [a, b] and F'(x) = f(x), then:
\int_a^b f(x)\,dx = F(b) - F(a)Equivalently: ∫ₐᵇ F'(x) dx = F(b) – F(a).
This is the theorem that makes definite integrals practical. Rather than computing a limit of Riemann sums (which is tedious), you find any antiderivative F of the integrand, evaluate it at the two endpoints, and subtract.
Evaluate ∫₀¹ x dx.
The antiderivative of x is x²/2.
Evaluate at the endpoints: [x²/2] from 0 to 1 = 1²/2 – 0²/2 = 1/2.
Geometrically, this is the area of the triangle under y = x from 0 to 1, which checks out: (1/2)(base)(height) = (1/2)(1)(1) = 1/2.
If f is continuous on an interval containing a, then:
\frac{d}{dx}\int_a^x f(t)\,dt = f(x)In words: if you define a function by integrating f from a fixed lower limit to a variable upper limit x, then the derivative of that function is simply f evaluated at x. Integration and differentiation are inverse operations.
Let g(x) = ∫₁ˣ t dt. Find g'(x).
By FTC 1, g'(x) = x. Done.
To verify this directly:
Evaluate the integral: ∫₁ˣ t dt = [t²/2] from 1 to x = x²/2 – 1/2.
Differentiate: g'(x) = d/dx (x²/2 – 1/2) = 2x/2 = x.
Both approaches agree, confirming FTC 1.
If f is continuous and g(x) and h(x) are differentiable, then:
\frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x)This handles the case where both the upper and lower limits of integration are functions of x, not just constants. It combines FTC 1 with the chain rule.
When the lower limit is a constant (say 0), the formula simplifies to f(h(x)) · h'(x), because the derivative of the constant lower limit is zero.
Let F(x) = ∫₀^(x²) √s ds. Find F'(x).
The upper limit is h(x) = x², so h'(x) = 2x.
The lower limit is 0 (a constant), so that term drops out.
Apply the formula: F'(x) = √(x²) · 2x = x · 2x = 2x².
Let g(x) = ∫₁^(sin x) [t⁷ – 6 ln|t|] / [4 cos(t)] dt. Find g'(x).
The upper limit is h(x) = sin x, so h'(x) = cos x.
Apply the generalized FTC 1: g'(x) = [(sin x)⁷ – 6 ln|sin x|] / [4 cos(sin x)] · cos x.
These appear repeatedly in FTC problems. Memorise them.
Integral | Result |
|---|---|
∫ sin(x) dx | –cos(x) + C |
∫ cos(x) dx | sin(x) + C |
∫ tan(x) dx | –ln |
∫ cot(x) dx | ln |
∫ sec²(x) dx | tan(x) + C |
∫ csc²(x) dx | –cot(x) + C |
∫ sec(x) tan(x) dx | sec(x) + C |
∫ (1/x) dx | ln |
FTC Part 2 is why engineers can compute the total distance a rocket travels by integrating its velocity function, rather than adding up tiny increments by hand. FTC Part 1 underpins how computers calculate accumulated quantities in real time: a sensor feeding a continuous stream of temperature readings can be modelled as an accumulation function whose derivative at any moment is just the current reading.
Students often forget to apply the chain rule in the Generalized FTC 1. If the upper limit is x², you must multiply by d/dx(x²) = 2x. Leaving out the 2x is one of the most common errors on exams.
Students sometimes include the "+C" constant when evaluating a definite integral. The constant of integration cancels out when you subtract F(a) from F(b), so it never appears in a definite integral answer.
Students sometimes confuse FTC 1 and FTC 2. FTC 2 evaluates a definite integral using antiderivatives. FTC 1 differentiates an accumulation function. They are related but distinct operations.
Students often assume the variable of integration (the "t" in ∫ f(t) dt) matters. It does not. The dummy variable can be t, s, u, or anything else; it disappears after integration.
⚠️ Expect at least one problem asking you to differentiate an integral with a function (not just x) as the upper limit. The chain rule step is what separates full marks from partial credit.
⚠️ Problems that define g(x) = ∫₀ˣ f(t) dt and then ask for critical points of g are classic exam material. You set g'(x) = f(x) = 0 and use the first derivative test.
⚠️ The common integrals table is assumed knowledge. Exam problems will not remind you that ∫ sec²(x) dx = tan(x) + C.
⚠️ Initial-value problems (finding g(π) when given g'(z) and g(1) = 1) combine antidifferentiation with solving for C. These appear regularly.
True or false: d/dx ∫₂ˣ cos(t) dt = cos(x). True.
Fill in the blank: ∫₀³ 2x dx = ___. 9 (since [x²] from 0 to 3 = 9 – 0).
True or false: When evaluating a definite integral, you must include +C in your answer. False. The constant cancels.
Fill in the blank: If F(x) = ∫₀^(x³) sin(t) dt, then F'(x) = ___. sin(x³) · 3x² (chain rule).
True or false: FTC 1 requires f to be continuous. True.
Q: Evaluate ∫₀¹ (y^(π⁴) + sin(y)) dy.
A: y^(π⁴+1) / (π⁴+1) – cos(y), evaluated from 0 to 1, gives 1/(π⁴+1) – cos(1) + 1.
Q: Given g'(z) = (2z²+1)/z + sec²(z) and g(1) = 1, find g(π).
A: Rewrite as 2z + 1/z + sec²(z). Integrate to get z² + ln|z| + tan(z) + C. Use g(1) = 1 to find C = –tan(1). Then g(π) = π² + ln(π) – tan(1).
Q: Let g(x) = ∫₀^(x²) (3√t – 1) dt. Find all critical points of g and classify them.
A: First evaluate the integral: g(x) = [2t^(3/2) – t] from 0 to x² = 2(x²)^(3/2) – x² = 2x³ – x². Then g'(x) = 6x² – 2x = 2x(3x – 1). Setting g'(x) = 0 gives x = 0 and x = 1/3. Using a sign chart: g has a local maximum at x = 0 and a local minimum at x = 1/3.
Q: If g(x) = ∫₁^(sin x) [t⁷ – 6 ln|t|] / [4 cos(t)] dt, what is g'(x)?
A: By the Generalized FTC 1: g'(x) = [(sin x)⁷ – 6 ln|sin x|] / [4 cos(sin x)] · cos x.
This connects directly to integration techniques (u-substitution, integration by parts) because those methods rely on FTC 2 to evaluate the results. The Generalized FTC 1 reappears in multivariable calculus when working with line integrals and path-dependent accumulation functions. The idea of "differentiation undoes integration" also grounds the study of differential equations, where you start from a derivative relationship and work backward to find the original function.
Fundamental Theorem of Calculus, FTC, FTC 1, FTC 2, evaluation theorem, accumulation function, antiderivative, indefinite integral, definite integral, net change theorem, integral calculus, chain rule with integrals, generalized FTC, variable limits of integration, Newton-Leibniz formula, area under a curve, MATH 101, Section 5.3, integral calculus UT Austin