Fundamental Theorem of Calculus

Equivalent forms

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

\int_a^b F'(x)\,dx = F(b)-F(a)

Two of mathematics' great machines — the derivative and the integral — turn out to be one machine viewed from opposite ends.

Symbol	Name	SI unit	Dimension	Range
$b$	Upper limit Right endpoint of integration	1	1	0 – 6.283
$a$	Lower limit Left endpoint of integration	1	1	0 – 3
$k$	Frequency Shape parameter of the integrand	1	1	0.5 – 3
$A$	Accumulated areaoutput Signed area from a to b	1	1	-5 – 5

Unit systems

Symbol	SI	CGS	Imperial
$b$	1	1	1

Where it holds

Requires f continuous on [a,b] (Part 2 needs F an antiderivative). Extends to piecewise-continuous f via the Lebesgue integral.

Dimensional analysis

\int f dx (dim f \times dim x)

F(b)-F(a) (dim of F)

F is an antiderivative of f, so

[F] = [f][x]

; both sides share that dimension.

\checkmark

Discovery

Newton & Leibniz · 1670

Isaac Barrow glimpsed it; Newton and Leibniz independently forged it into calculus around 1665–1675, igniting the priority dispute that split British and Continental mathematics for a century. It is the theorem that made physics computable.

Try this

Differentiation and integration look like opposite chores. The Fundamental Theorem says they are the same act run backwards — the area you sweep out is the antiderivative you differentiate away.

Slide the upper limit and watch the shaded area grow; the rate it grows at any instant is exactly the height of the curve there.

Research status: stable

Real-world applications

Recovering position from acceleration in kinematics
Work $= \int F\cdot dx$ and impulse $= \int F dt$
Computing total charge from current, or energy from power
The basis of every numerical-integration scheme

Common misconceptions

The integral is 'the area' — it is signed area; regions below the axis count negative.
You need the limits to differentiate an integral — Part 1 differentiates a variable-limit integral directly.
FTC needs f differentiable — it only needs f continuous.

Experimental verification

Every time a physicist integrates acceleration to get velocity, or power over time to get energy, the answer matches measurement — the theorem is used, not tested, but its consequences are checked continuously.

Derivation

Define A

(x)=\int

_a^x f.

Then A(x+h)-A

(x)=\int _x^{x+h} f \approx f(x)h

by continuity, so A'

(x)=f(x)

.

Any antiderivative F differs from A by a constant, giving

\int

_a^

b f = A(b) = F(b)-F

(a).

Limiting cases

b = a

⟶

A = 0

No interval, no area.

f > 0 everywhere⟶ A increases with bPositive height always adds area — A is monotonic.

f changes sign⟶ A can decreaseBelow-axis area is subtracted; the integral is signed.

What if…

What if the upper limit is itself a function g(x)?

Chain rule: $d/dx \int$ _a^ ${g(x)} f dt = f(g(x))\cdot g'(x)$ .

What if f has a jump discontinuity?

The area function stays continuous but develops a kink — its derivative jumps with f.

1

Distance from constant acceleration

Given ·

v:: at
a lim:: 0
b lim:: T

Find · displacement

Steps

Antiderivative of at $is \tfrac{1}{2}at^{2}$
Evaluate at T minus at 0
$x = \tfrac{1}{2}aT^{2}$

Result ·

x = \int _0^T

at

dt = \tfrac{1}{2}aT^{2}

2

Derivative of an area function

Given ·

G:: $\int _0^x \cos t dt$

Find · G'(x)

Steps

Part 1 says $d/dx \int _0^x f = f(x)$
Here $f = \cos$
So $G'(x) = \cos x$

Result ·

G'(x) = \cos x

Mean Value Theorem→gradient theoremGreen's Theorem→Taylor Expansion→