What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus is the statement that differentiation and integration are inverse operations. It has two parts: the first says that the derivative of an accumulator function recovers the integrand, and the second says that a definite integral can be evaluated by finding any antiderivative and subtracting its values at the two endpoints.

What does F prime of x equals f of x mean geometrically?

It means the slope of the accumulator curve F at any point x is exactly the height of the integrand curve f at that same x. Where f is positive, F climbs. Where f is zero, F is flat. Where f is large, F climbs steeply. The two curves are linked: one is the running area, the other is the rate at which that area is being added.

How do you compute a definite integral using the FTC?

Find any antiderivative G of the integrand f, then evaluate G at the upper bound and subtract its value at the lower bound. The result is the definite integral from the lower bound to the upper bound. For example, to integrate t squared from zero to two, take the antiderivative t cubed over three and compute eight over three minus zero, which gives eight over three.

What is the difference between Part 1 and Part 2 of the FTC?

Part 1 builds a new function as the accumulated integral and proves that its derivative is the original integrand. Part 2 goes the other direction: it uses Part 1 to compute the value of a specific definite integral by evaluating any antiderivative at the two endpoints. Part 1 is the theoretical link; Part 2 is the practical evaluation tool.

Why is the FTC important?

Without the FTC, computing a definite integral would require summing infinitely many thin rectangles, which is impractical. The FTC reduces this to finding a single antiderivative and evaluating at two points. It also explains why integration and differentiation are studied together: they are not separate topics but two sides of one operation.

Fundamental Theorem of Calculus

How to use

The Function picker on the left lists six functions in two groups, Polynomial and Transcendental; tapping one switches $f$ and its accumulator $F$ and resets $a$ to $0$ and $x$ to $2$ . Learn more about the function families
The Display checkboxes f(t), F(x) and area show or hide the solid integrand, the dashed accumulator and the shading; the legend under the plot lists only what is visible. Learn more about the display toggles
The Parameters sliders set lower bound a and upper bound x, the moving point, both from $-5$ to $5$ ; Reset puts them back at $a = 0$ , $x = 2$ . Learn more about the a and x sliders
Accent color under Appearance recolours the sliders, the card and the Applied chips; the Reset next to it returns to blue. Learn more about the accent color
The S, M, L and XL buttons above the plot set its size. Learn more about plot size
The shaded region runs under $f$ from $a$ to $x$ ; it counts negative where $f$ is below the t-axis and flips sign when $x$ is left of $a$ . Learn more about the shaded area
The At the moving point x card shows Area, F(x) and f(x) = F'(x); the first two always match, and the third is the slope of $F$ at $x$ . Learn more about the at-the-point card
The info panel has two tabs: Explanation gives the chosen function, its accumulator and the current values; Concepts walks through both parts of the theorem. Learn more about the info panel

Quadraticf(t) = t²F(x) = x³/3 − a³/3

880 × 460

f(t) = t² (integrand)F(x) = ∫ₐˣ f(t) dt (accumulator)shaded area

At the moving point xx = 2 (a = 0)

Area ∫ₐˣ f(t) dt

2.667

shaded region under f

F(x)

2.667

height on F

f(x) = F'(x)

4

slope of F at x

The shaded area from a to x equals the height of F at x; and the slope of F at x equals f(x).FTC

Applieda=0x=2F'(x) = f(x)

Quadratic: accumulator and integrand

The shaded region is the area under f from a to x. Sliding x grows or shrinks that area, and the height of F tracks it exactly.

Integrand ·

f(t) = t²

Accumulator ·

F(x) = x³/3 − a³/3

Right now

a =

0

, x =

2

. Shaded area =

2.667

, F(x) =

2.667

(they match by construction). f(x) =

4

, which is also

F'(x)

— the slope of the F curve at x.

The two halves of the theorem

Part 1 (the link). Define

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

— the accumulated area. Then F is differentiable and

F'(x) = f(x)

. Differentiation undoes integration.

Part 2 (evaluation). If G is any antiderivative of f, then

\int_a^b f(t)\,dt = G(b) - G(a)

. That's how you compute definite integrals without summing rectangles forever.

Where this leads

f(x) is a height and F(x) an accumulated area; they share an x and nothing else. See the quadratic family or compare all six families.

Key Terms

Getting Started

The Function Families

Identity: Area That Grows Like a Triangle

Quadratic: the Tool's Opening State

Cubic: a Faster Integrand, a Faster Accumulator

Sine: an Accumulator That Turns Around

Cosine: Where the Two Curves Swap Roles

Exponential: Accumulator and Integrand Almost Coincide

The a and x Sliders

The Shaded Area

Reading the At-The-Point Card

Display Toggles

Plot Size and the Info Panel

What Is the FTC

Part 1 - The Accumulator

Part 2 - Evaluation

Related Concepts

Key Terms

Fundamental Theorem of Calculus (FTC) — the statement that differentiation and integration are inverse operations. Has two parts: Part 1 says the derivative of an accumulator function equals the integrand; Part 2 gives a way to compute definite integrals using antiderivatives.

Integrand — the function

f(t)

being integrated. The curve whose area you are accumulating.

Accumulator function — the function

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

. It returns the signed area under

f

from a fixed lower bound

a

to the moving upper bound

x

.

Antiderivative — any function

G

with

G'(x) = f(x)

. Antiderivatives differ by a constant; the accumulator is one specific antiderivative pinned by

F(a) = 0

.

Definite integral — the signed area

\int_a^b f(t)\, dt

. Computed via Part 2 as

G(b) - G(a)

for any antiderivative

G

.

Signed area — area counted positively where

f > 0

and negatively where

f < 0

Getting Started

The page opens with the Quadratic family

f(t) = t^2

loaded, the lower bound

a

parked at

0

, and the upper bound

x

2

. You see three things together on the graph:

• The solid blue curve of

f

(the integrand).

• The dashed deep-blue curve of

F

(the accumulator).

• The shaded region between

f

and the t-axis from

a

x

.

Two thin vertical lines mark

a

and

x

on the t-axis. Marker dots sit on

f

and

F

at the upper bound

x

.

The boxed card below the graph displays three numbers: the shaded area, the height

F(x)

, and the value

f(x) = F'(x)

. The first two always match. The third is the slope of

F

x

.

To explore quickly, drag the upper bound x slider and watch every part of the picture update at once.

The Function Families

DemoThe function families

Step 0 of 5

Six families are organized into two groups in the left panel, each shipped with a closed-form antiderivative.

Polynomial:

• Identity —

f(t) = t

F(x) = x^2/2 - a^2/2

. The simplest case where signed area depends on whether

x > a

x < a

.

• Quadratic —

f(t) = t^2

F(x) = x^3/3 - a^3/3

. The power rule in its cleanest form.

• Cubic —

f(t) = t^3

F(x) = x^4/4 - a^4/4

. Odd integrand; useful for seeing how negative regions cancel positive ones.

Transcendental:

• Sine —

f(t) = \sin t

F(x) = -\cos x + \cos a

. Periodic accumulator with bounded swings.

• Cosine —

f(t) = \cos t

F(x) = \sin x - \sin a

. Same period; accumulator is a sine.

• Exponential —

f(t) = e^t

F(x) = e^x - e^a

. The accumulator grows the same way the integrand does.

Identity: Area That Grows Like a Triangle

With

f(t) = t

a = 0

and

x = 2

, the shaded region is a right triangle of base

2

and height

2

, so its area is

2

— and the accumulator reads

F(2) = 2

to match.

Geometry gives the whole answer here without any integration:

F(x) = \frac{1}{2} \cdot x \cdot x = \frac{x^2}{2}

, a parabola. The dashed F curve on the graph is exactly that parabola.

Identity, frozen at a = 0, x = 2

The shaded region is a triangle of area 2, and the dashed accumulator passes through height 2 at x = 2. A straight integrand gives a parabolic F.

This family is the cleanest illustration of Part 1. Differentiate the area function and you get

\frac{d}{dx}\left(\frac{x^2}{2}\right) = x

, which is the integrand you started with. The slope of the dashed curve at any

x

equals the height of the solid one at that same

x

.

It also shows why a straight integrand produces a curved accumulator. Area accrues faster as

x

grows, because each new sliver is taller than the last — so F bends upward even though f does not.

Quadratic: the Tool's Opening State

f(t) = t^2

from

a = 0

x = 2

. The shaded area works out to

\int_0^2 t^2,dt = \frac{2^3}{3} = \frac{8}{3} \approx 2.667

and

F(2)

reads the same. Note that

f(2) = 4

while

F(2) = 2.667

: the two markers sit at different heights, and they are measuring different things.

Quadratic, frozen at a = 0, x = 2

Two markers at different heights on the same vertical: f(2) = 4 is a height, F(2) = 2.667 is an accumulated area. The slope of the dashed curve there is 4, not 2.667.

That difference is the point most worth pinning down.

f(x)

is a height — how tall the integrand is at

x

F(x)

is an accumulated area — everything collected between

a

and

x

. They share an

x

and nothing else.

What connects them is the slope. The dashed F curve is steepest where the solid f curve is tallest, and

F'(2) = f(2) = 4

exactly. Reading the two curves that way — heights on one, slopes on the other — is what makes Part 1 visible rather than symbolic.

Cubic: a Faster Integrand, a Faster Accumulator

f(t) = t^3

[0, 2]

gives

\int_0^2 t^3,dt = \frac{2^4}{4} = 4

with

f(2) = 8

. The accumulator is

F(x) = x^4/4

, which climbs steeply once past

x = 1

Cubic, frozen at a = 0, x = 2

f(2) = 8 with an accumulated area of 4. F(x) = x⁴/4 climbs sharply past x = 1, and stays positive on the left of the axis even though the integrand is negative there.

Integrating raises the degree by one, every time:

t

accumulates to

x^2/2

t^2

x^3/3

t^3

x^4/4

. That is the power rule running backwards, and the three polynomial families lay it out in order.

There is a detail the negative side of the graph makes visible. For

x < 0

the cubic is below the axis, so area accrues negatively — yet

F(x) = x^4/4

is positive there. No contradiction: with

x < 0 = a

the integral runs backwards, and the two sign reversals cancel. The signed area discussion covers the bookkeeping.

Sine: an Accumulator That Turns Around

f(t) = \sin(t)

from

0

2

. The area comes to

\int_0^2 \sin(t),dt = -\cos(2) + \cos(0) \approx 1.416

with

f(2) \approx 0.909

. Here

F(x) = 1 - \cos(x)

, a shifted cosine wave.

Sine, frozen at a = 0, x = 2

Area 1.416 collected so far, with the integrand at 0.909 and falling. Past x = π the shaded slivers go negative and the dashed accumulator turns back down.

The interesting behaviour is just past

x = \pi

. There the sine drops below the axis, each new sliver of area is negative, and the accumulator stops rising and begins to fall — even though it is still positive.

That turning point is Part 1 again, read in reverse: F has a maximum exactly where

f

crosses zero going downward, because

F' = f

and a maximum needs

F' = 0

. Every zero of the integrand is a critical point of the accumulator.

Cosine: Where the Two Curves Swap Roles

f(t) = \cos(t)

[0, 2]

accumulates to

\int_0^2 \cos(t),dt = \sin(2) - \sin(0) \approx 0.909

while

f(2) = \cos(2) \approx -0.416

— the integrand has already gone negative, though the accumulated total is still comfortably positive.

The accumulator is

F(x) = \sin(x)

, so this family draws a cosine and a sine on the same axes.

Cosine, frozen at a = 0, x = 2

The integrand has already gone negative at -0.416 while the running total is still +0.909. The accumulator here is exactly sin(x) - the two curves are the familiar pair, swapped.

Compare this state against the sine family and the pairing becomes obvious: there the integrand was

\sin

and the accumulator a shifted

\cos

; here the integrand is

\cos

and the accumulator is

\sin

. Differentiation and integration walk the same cycle in opposite directions.

The negative readings also make the sign convention concrete. Between

\pi/2

and

2

the shaded region lies below the axis and is subtracted, which is why the total of

0.909

is smaller than the area of the visible region above the axis alone.

Exponential: Accumulator and Integrand Almost Coincide

f(t) = e^t

from

0

2

gives

\int_0^2 e^t,dt = e^2 - 1 \approx 6.389

against

f(2) = e^2 \approx 7.389

. The two differ by exactly

1

, and they will differ by exactly

1

at every

x

Exponential, frozen at a = 0, x = 2

f(2) = 7.389 against F(2) = 6.389 - a gap of exactly 1, and the same gap at every x, because F(x) = eˣ - 1 is the integrand shifted down one unit.

That is the signature of the exponential. Since

F(x) = e^x - e^0 = e^x - 1

, the accumulator is the integrand shifted down by one unit — the two curves on screen are parallel in the vertical sense, never converging or diverging.

It follows directly from

f' = f

. The only functions whose accumulated area matches their own height, up to a constant, are the exponentials, and the constant here is fixed by the starting point:

F(a) = 0

always, so at

a = 0

the offset must be

e^0 = 1

The a and x Sliders

DemoThe a and x sliders

Step 0 of 5

The lower bound a slider sets the fixed starting point of integration. By construction

F(a) = 0

— the accumulator passes through zero exactly at

x = a

. Move

a

and the whole

F

curve shifts vertically; the integrand

f

is unaffected.

The upper bound x slider sweeps the moving endpoint:

• The right edge of the shaded region tracks

x

in real time.

• The marker dots on both curves slide to the new

x

.

• The numeric card updates with the new area, the new

F(x)

, and the new

f(x)

.

Try setting

a = 0

and dragging

x

from

0

rightward. The shaded area grows as

x

grows, and the height of the

F

curve climbs by exactly the same amount.

The Reset button next to the Parameters label restores the family's default

a

and

x

The Shaded Area

The shaded region between the integrand

f

and the t-axis from

a

x

is the visual heart of the tool. Its signed area is what the definite integral

\int_a^x f(t)\, dt

computes.

A few things to notice as you slide

x

:

• When

f > 0

on the interval, the shaded area is positive and grows.

• When

f < 0

, the shaded area is negative — the integrand sits below the t-axis, and the integral subtracts that contribution.

• When the integrand crosses zero, the accumulator

F

momentarily stops growing. A flat spot of

F

— and a local maximum or minimum — sits exactly there.

• When

a = x

, the shaded region collapses to a single point. Area is zero;

F(a) = 0

by definition.

• Sliding

x

to the left of

a

flips the orientation, and the area enters with the opposite sign.

Reading the At-The-Point Card

DemoReading the card: area, F(x), slope

Step 0 of 5

The boxed card below the graph displays the three quantities at the moving point

x

:

• Area — the signed area of the shaded region from

a

x

. Computed directly as

\int_a^x f(t)\, dt

.

• F(x) — the height of the accumulator curve at

x

. Computed from the closed-form antiderivative.

• f(x) = F'(x) — the value of the integrand at

x

, equal to the slope of

F

x

.

The first two numbers are always equal by construction — that's how

F

is defined.

The third number is the punchline of Part 1 of the FTC: the slope of the accumulator at

x

is exactly the integrand evaluated at

x

. Differentiation undoes integration.

Read the row as you slide: left tells you the area piling up, middle tells you the same area as a height, right tells you how fast that pile is growing.

Display Toggles

DemoDisplay, plot size and info panel

Step 0 of 5

The Display section in the left panel hides individual layers when one is in the way:

• f(t) — toggles the integrand curve. Off, you see just the accumulator and the shaded area outline.

• F(x) — toggles the dashed accumulator curve. Off, the picture reduces to a classical area under f view without the FTC overlay.

• area — toggles the shaded fill. Off, the two function curves remain but the area visualization is gone.

The legend below the graph updates to show only the visible layers.

The Accent color picker at the bottom recolors the highlight throughout — slider track, verdict border, key labels — useful for screenshots or personal preference.

Plot Size and the Info Panel

The row of buttons above the plot sets its size: S is

500 \times 400

, M is

700 \times 550

, L is

900 \times 850

and XL is

1100 \times 850

pixels. The page opens at

880 \times 460

, and the current size is printed next to the buttons. A new size redraws the same curves, shading and marker dots; the numbers in the card do not change.

The info panel on the right has two tabs.

• Explanation follows the controls. It names the function, gives the integrand

f(t)

and the accumulator

F(x)

as formulas, then under Right now lists the current

a

and

x

, the shaded area,

F(x)

and

f(x)

, and closes with the two halves of the theorem. A short note at the end links to the matching family section on this page. The tab rewrites itself whenever you pick a function or move a slider.

• Concepts does not change with the controls. It explains the accumulator

F

, what

F'(x)

is, why Part 2 lets you evaluate an integral with any antiderivative, and what the picture shows.

For the three numbers in the card, see the at-the-point card.

What Is the FTC

The Fundamental Theorem of Calculus is the bridge between the two central operations of calculus: differentiation (slope of a curve) and integration (area under a curve). It says the two operations are inverses of each other.

The theorem has two parts, both essential.

Part 1 establishes that if you define a function by integrating

f

from a fixed lower bound up to a moving upper bound, the derivative of that function recovers

f

. Symbolically, if

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

, then

F'(x) = f(x)

.

Part 2 uses Part 1 to give a practical way to compute definite integrals: pick any antiderivative

G

f

, and the integral

\int_a^b f(t)\, dt

equals

G(b) - G(a)

.

For full coverage of proofs and applications, see the fundamental theorem of calculus page.

Part 1 - The Accumulator

FTC Part 1. If

f

is continuous on

[a, b]

and

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

for

x

[a, b]

, then

F

is differentiable on

(a, b)

and

F'(x) = f(x).

The intuition: push

x

a little further right by

\Delta x

. The shaded area grows by a thin strip of width

\Delta x

and height roughly

f(x)

. So

\Delta F \approx f(x) \cdot \Delta x

, which means

F'(x) = f(x)

in the limit.

This is the reason the accumulator is sometimes called the integral function — it shows that integration produces a function whose derivative is the original integrand. Differentiation undoes integration, point by point.

In the tool, you can verify this visually. Pick any

x

, note the slope of the

F

curve there (use the rate at which the dot is climbing), and compare it to the height of

f

at the same

x

. They match.

Part 2 - Evaluation

FTC Part 2. If

f

is continuous on

[a, b]

and

G

is any antiderivative of

f

(so

G'(x) = f(x)

everywhere on the interval), then

\int_a^b f(t)\, dt = G(b) - G(a).

This is the practical workhorse of integration. Without it, computing a definite integral would mean summing the areas of infinitely many thin rectangles. With it, you find an antiderivative — usually by reversing a differentiation rule — and evaluate at the endpoints.

For example, to compute

\int_0^2 t^2\, dt

:

• Find an antiderivative:

G(t) = t^3/3

works.

• Evaluate at the endpoints:

G(2) - G(0) = 8/3 - 0 = 8/3

.

That's exactly what the tool reports when you load the Quadratic family with

a = 0

and

x = 2

.

For deeper coverage of integration techniques and antiderivative tables, see the integration techniques page.