How does the Fundamental Theorem of Calculus connect derivatives and integrals?
Quick answer
If f is continuous throughout the interval you are considering, define A(x) = ∫ₐˣ f(t) dt. The Fundamental Theorem of Calculus says A′(x) = f(x): the derivative of the accumulated amount gives the integrand’s value at x. For example, A(x) = ∫₀ˣ t² dt = x³/3, so A′(x) = x², the original integrand. [S1]
The question, the details, the next step.
I know derivatives describe change and integrals can describe accumulation, but how does the Fundamental Theorem connect them?
Think of an integral with a variable upper limit as an accumulation function. For a continuous function f on the interval, set A(x) = ∫ₐˣ f(t) dt. As x moves, the accumulated amount changes. The theorem says its instantaneous rate of change is A′(x) = f(x). In other words, differentiating the accumulation function recovers the value of the function being accumulated. [S1]
The continuity condition matters: f should have no breaks across the interval you are considering. It is not enough to check only the value at x. [S1]
Can you work through a simple example and show what each symbol is doing?
Take f(t) = t² and define A(x) = ∫₀ˣ t² dt. Here, t is the integration variable, 0 is the fixed starting point, and x is the moving endpoint. Evaluating the integral gives A(x) = x³/3. Differentiate that expression: A′(x) = x². That matches the original integrand, t², with its input written as x. So the integral builds A, and differentiation takes you back to f. [S1]
For a quick check, choose x = 2: A′(2) = 4, which is also f(2) = 2². This follows from the same relationship. [S1]
Why does the integral use t, but the derivative gives f(x)? I keep mixing up those letters.
The t inside the integral is a temporary label for the input that ranges through the accumulation. The x outside the integral is the endpoint you can vary. You could write A(x) = ∫₀ˣ u² du instead; the accumulated function would still be x³/3. The letter inside the integral does not determine the result’s variable—the moving endpoint does.
A useful reading is: “A of x is the accumulated value of f from a to x.” Then the theorem tells you that the rate at which this total changes at x is f(x). The continuity requirement applies across the interval in question. [S1]
What steps should I use on a homework problem, and what if I can’t confirm continuity?
Use this short routine:
- Identify the integrand f and the variable endpoint x.
- Check whether f is continuous throughout the interval from the fixed starting point to the x-values you are considering.
- Match the problem to A(x) = ∫ₐˣ f(t) dt.
- If the condition holds, write A′(x) = f(x).
- If useful, evaluate the integral separately to check the relationship, as in the t² example.
If continuity is uncertain, first inspect the given function and interval rather than assuming the rule applies. The cited theorem establishes the result under continuity; it does not settle every possible case where continuity fails. For those cases, use the definitions and results your course has covered or ask your instructor which theorem applies. [S1]
A practical checklist
Write down the integrand and the variable upper endpoint.
State the interval being considered.
Check that the integrand is continuous throughout that interval.
Apply A′(x) = f(x) only when the stated condition holds.
Keep the dummy integration variable separate from the endpoint variable.
A few more useful answers
Does the integration variable have to be t?
No. It is a temporary label, so you can use another letter consistently. The variable upper endpoint determines the variable in the accumulation function.
Does the theorem say every integral is the derivative of its integrand?
The result here applies to an accumulation function with a variable upper endpoint when the integrand is continuous throughout the interval being considered. It does not establish that conclusion for every integral or every discontinuous integrand. [S1]
What is the main idea to remember?
Accumulating f from a fixed starting point to x creates A(x); differentiating A with respect to x recovers f(x), provided the continuity condition holds. [S1]
Sources
References used for this guide.
- [S1] OpenStax Calculus Volume 2, Section 1.3: The Fundamental Theorem of Calculus openstax.org · Checked