The Fundamental Theorem of Calculus

Quick answer

The Fundamental Theorem of Calculus joins the two halves of calculus. Part 1: the accumulation function F(x) = ∫ₐˣ f(t) dt, the area so far, has derivative f(x) — adding area at x happens at the rate f(x). Part 2: to evaluate a definite integral, find any antiderivative F of f and compute F(b) − F(a). So ∫₀¹ x² dx = [x³/3]₀¹ = 1/3, with no limit of sums at all. Read as net change, Part 2 says integrating a rate recovers the total change.

What you'll learn

Interpret and differentiate accumulation functions
Apply Part 1, including with the chain rule
Evaluate definite integrals with antiderivatives (Part 2)
Use the net change theorem

Accumulation functions

Fix a starting point $a$ and let the upper limit vary. The result is a new function:

F(x) = \int_a^x f(t)\,dt

$F(x)$ is the signed area under $f$ from $a$ up to $x$ — the amount accumulated so far. (The variable inside is called $t$ so that $x$ can mean the moving endpoint.) $F$ increases where $f > 0$ , since area is being added, and decreases where $f < 0$ .

Part 1: the derivative of an integral

Fundamental Theorem of Calculus, Part 1

If $f$ is continuous on an interval containing $a$ , then $F(x) = \int_a^x f(t)\,dt$ is differentiable and

\frac{d}{dx}\int_a^x f(t)\,dt = f(x)

Why the rate of accumulation is f(x)

Push $x$ a little to the right, to $x + h$ . The accumulated area grows by a thin strip from $x$ to $x + h$ . The strip is about $h$ wide and $f(x)$ tall:

F(x + h) - F(x) \approx f(x)\,h \quad\Longrightarrow\quad \frac{F(x + h) - F(x)}{h} \approx f(x)

y = f(t)

Area so far, and the strip added next

As $h \to 0$ the approximation becomes exact, because $f$ is continuous and the strip’s top flattens to the height $f(x)$ . Area accumulates at a rate equal to the height of the curve at the moving edge. Integration and differentiation undo each other.

Part 2: evaluating integrals

An antiderivative of $f$ is a function $F$ with $F' = f$ .

Fundamental Theorem of Calculus, Part 2

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ , then

\int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b

This follows from Part 1: $G(x) = \int_a^x f(t)\,dt$ is one antiderivative, and any other differs from it by a constant $C$ , which cancels in $F(b) - F(a)$ . Since $G(a) = 0$ , the difference is $G(b)$ , the integral.

Read in terms of rates, Part 2 is the net change theorem: integrating a rate of change $F'$ over $[a, b]$ gives the total change $F(b) - F(a)$ . Integrating a velocity, for example, gives the displacement.

Worked examples

Common mistakes

Practice problems

Evaluate $\int_1^3 2x\,dx$ .

Answer

$8$

Full solution

$\Big[x^2\Big]_1^3 = 9 - 1 = 8$ .
Evaluate $\int_0^2 (3x^2 - 4x + 1)\,dx$ .

Answer

$2$

Full solution

$\Big[x^3 - 2x^2 + x\Big]_0^2 = (8 - 8 + 2) - 0 = 2$ .
Evaluate $\int_1^e \tfrac{1}{x}\,dx$ .

Answer

$1$

Full solution

$\Big[\ln x\Big]_1^e = \ln e - \ln 1 = 1 - 0 = 1$ .
Evaluate $\int_0^{\ln 2} e^x\,dx$ .

Answer

$1$

Full solution

$\Big[e^x\Big]_0^{\ln 2} = 2 - 1 = 1$ .
Find $\tfrac{d}{dx}\int_2^x \tfrac{t}{t^2 + 1}\,dt$ .

Answer

$\tfrac{x}{x^2 + 1}$

Full solution

Part 1: the integrand evaluated at the upper limit.
Find $\tfrac{d}{dx}\int_0^{3x} \cos(t^2)\,dt$ .

Answer

$3\cos(9x^2)$

Full solution

$\cos\big((3x)^2\big) \cdot 3 = 3\cos(9x^2)$ .
Find $\tfrac{d}{dx}\int_x^5 \sqrt{t}\,dt$ .

Answer

$-\sqrt{x}$

Full solution

Reverse the limits: $\int_x^5 \sqrt{t}\,dt = -\int_5^x \sqrt{t}\,dt$ , whose derivative is $-\sqrt{x}$ .
A particle’s velocity is $v(t) = 3t^2 - 2$ m/s. Find its displacement from $t = 0$ to $t = 2$ .

Answer

$4$ m

Full solution

$\int_0^2 (3t^2 - 2)\,dt = \Big[t^3 - 2t\Big]_0^2 = 8 - 4 = 4$ .
Let $F(x) = \int_0^x (t^2 - 4)\,dt$ . Where is $F$ increasing?

Answer

Where $x < -2$ or $x > 2$

Full solution

$F'(x) = x^2 - 4$ by Part 1, which is positive when $|x| > 2$ .
A student evaluates $\int_0^{\pi/2} \cos x\,dx$ as $\Big[\sin x\Big]$ at $\tfrac{\pi}{2}$ only, getting $1$ , then says the method works because the answer is right. What is the flaw?

Hint

What is $\sin 0$ ?

Answer

The lower term was skipped; it happens to be $\sin 0 = 0$ here, but it is not $0$ in general.

Full solution

The correct computation is $\sin\tfrac{\pi}{2} - \sin 0 = 1 - 0 = 1$ .

Skipping $F(a)$ gives the right answer only when $F(a) = 0$ . For $\int_0^{\pi} \sin x\,dx$ the same shortcut gives $1$ instead of $2$ .

Frequently asked questions

What does the Fundamental Theorem of Calculus say?

Part 1: the derivative of ∫ from a to x of f(t) dt is f(x). Part 2: ∫ from a to b of f(x) dx = F(b) − F(a), where F is any antiderivative of f.

What is an accumulation function?

A function defined by an integral with a variable upper limit, F(x) = ∫ from a to x of f(t) dt. It gives the accumulated signed area from a up to x.

How do I differentiate ∫ from 0 to x² of f(t) dt?

Use Part 1 with the chain rule: f(x²) · 2x.

Does it matter which antiderivative I use in Part 2?

No. Two antiderivatives differ by a constant, and it cancels in F(b) − F(a).

What is the net change theorem?

The integral of a rate of change over [a, b] is the total change: ∫ from a to b of F′(x) dx = F(b) − F(a).

What to learn next

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Accumulation functions

Part 1: the derivative of an integral

Why the rate of accumulation is f(x)

Part 2: evaluating integrals

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What does the Fundamental Theorem of Calculus say?

What is an accumulation function?

How do I differentiate ∫ from 0 to x² of f(t) dt?

Does it matter which antiderivative I use in Part 2?

What is the net change theorem?

What to learn next

More Calculus lessons