Fundamental Theorem of Calculus

Fundamental Theorem of Calculus

The theorem that establishes the connection between derivatives, antiderivatives, and definite integrals. The fundamental theorem of calculus is typically given in two parts.

Two-part Fundamental Theorem of Calculus: Part 1 defines g(x)=∫f(t)dt, continuous and differentiable with g'(x)=f(x); Part 2:...

Key Formula

\textbf{Part 1: } \frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x) \qquad \textbf{Part 2: } \int_{a}^{b} f(x)\,dx = F(b) - F(a)

Where:

$f(x)$ = A continuous function on the interval [a, b]
$F(x)$ = Any antiderivative of f(x), meaning F'(x) = f(x)
$a$ = The lower limit of integration
$b$ = The upper limit of integration
$t$ = A dummy variable of integration (placeholder inside the integral)

Worked Example

Problem: Use Part 2 of the Fundamental Theorem of Calculus to evaluate the definite integral ∫ from 1 to 4 of 2x dx.

Step 1: Identify the integrand and find its antiderivative. The integrand is f(x) = 2x. An antiderivative of 2x is F(x) = x².

F(x) = x^2 \quad \text{since } F'(x) = 2x

Step 2: Apply the Fundamental Theorem (Part 2). Evaluate F at the upper limit b = 4 and subtract F at the lower limit a = 1.

\int_{1}^{4} 2x\,dx = F(4) - F(1)

Step 3: Compute F(4) and F(1).

F(4) = 4^2 = 16, \quad F(1) = 1^2 = 1

Step 4: Subtract to get the final value of the definite integral.

16 - 1 = 15

Answer: The definite integral equals 15.

Another Example

This example uses Part 1 of the theorem (differentiating an integral) rather than Part 2 (evaluating a definite integral). It shows how integration and differentiation undo each other.

Problem: Use Part 1 of the Fundamental Theorem of Calculus to find g'(x) if g(x) = ∫ from 2 to x of (3t² + 1) dt.

Step 1: Recognize the structure. The function g(x) is defined as an accumulation function — an integral with a variable upper limit.

g(x) = \int_{2}^{x}(3t^2 + 1)\,dt

Step 2: Apply Part 1 of the Fundamental Theorem directly. Since the upper limit is simply x (not a more complicated expression), the derivative equals the integrand evaluated at x.

g'(x) = \frac{d}{dx}\int_{2}^{x}(3t^2 + 1)\,dt = 3x^2 + 1

Step 3: Verify by thinking about it: Part 1 tells you that integrating a function and then differentiating gives you back the original function. Replace t with x in the integrand.

g'(x) = 3x^2 + 1

Answer: g'(x) = 3x² + 1.

Frequently Asked Questions

What is the difference between Part 1 and Part 2 of the Fundamental Theorem of Calculus?

Part 1 tells you that differentiation undoes integration: if you define a function by integrating f from a constant to x, then its derivative is f(x). Part 2 tells you how to evaluate a definite integral numerically: find any antiderivative F, then compute F(b) − F(a). Part 1 is about the relationship between the two operations, while Part 2 is the practical computation tool you use most often.

Why is the Fundamental Theorem of Calculus important?

Before this theorem, computing the area under a curve required summing infinitely many thin rectangles (Riemann sums), which is extremely tedious. The theorem provides a shortcut: just find an antiderivative and plug in the limits. It also reveals the deep fact that the two central ideas of calculus — rates of change (derivatives) and accumulation (integrals) — are inverse processes.

Does the Fundamental Theorem of Calculus work for all functions?

The standard version requires the function f to be continuous on the closed interval [a, b]. If f has a discontinuity within the interval, you may need to split the integral at the point of discontinuity and handle each piece separately. There are more advanced versions of the theorem that relax the continuity requirement, but the basic form you learn in a first calculus course assumes continuity.

FTC Part 1 vs. FTC Part 2

	FTC Part 1	FTC Part 2
What it states	d/dx of ∫ from a to x of f(t) dt equals f(x)	∫ from a to b of f(x) dx equals F(b) − F(a)
Purpose	Shows differentiation reverses integration	Provides a method to evaluate definite integrals
Output	A function (the derivative of the accumulation function)	A number (the net signed area under the curve)
When you use it	When asked to differentiate an integral with a variable limit	When asked to compute the value of a definite integral

Why It Matters

The Fundamental Theorem of Calculus appears in virtually every calculus course and is tested on standardized exams like the AP Calculus AB and BC exams. In physics, it connects velocity (the derivative of position) with displacement (the integral of velocity), letting you move between these quantities. Nearly every technique you learn later in calculus — u-substitution, integration by parts, improper integrals — relies on Part 2 of this theorem to produce a final numerical answer.

Common Mistakes

Mistake: Forgetting to subtract F(a) and only computing F(b).

Correction: Part 2 always requires the difference F(b) − F(a). Even if the lower limit is 0, you must compute F(0) and subtract it — it is not always zero.

Mistake: Using Part 1 incorrectly when the upper limit is a function of x, such as x², instead of just x.

Correction: If g(x) = ∫ from a to h(x) of f(t) dt, you must apply the chain rule: g'(x) = f(h(x)) · h'(x). For example, if the upper limit is x², multiply by the derivative 2x.

Related Terms

Definite Integral — Evaluated using Part 2 of the FTC
Antiderivative of a Function — The function F used in the FTC formula
Derivative — Inverse operation to integration per Part 1
Theorem — The FTC is a foundational theorem of calculus
Integration by Substitution — A technique for finding antiderivatives to use with the FTC
Riemann Sum — The limit definition the FTC replaces for evaluation
Chain Rule — Needed when applying Part 1 with a composite upper limit
Continuous Function — Continuity of f is a requirement of the FTC