Chapter 09

Integral: Area and Accumulation, a Bridge to Probability

Integration is the inverse of differentiation. It is used for area under a curve, cumulative quantities, and for probability and expectation.

Math diagram by chapter

Select a chapter to see its diagram below. View the flow of basic math at a glance.

The gap between the rectangles and the curve shrinks as you use more slices; in the limit you get the exact area (the integral).

The definite integral represents the area under the curve. Find an antiderivative and plug in the upper and lower limits.

What is the integral?

The integral is the inverse of differentiation. If differentiation is like slicing bread very thin (finding the rate of change), integration is putting those thin slices back together to get the original amount (area or total quantity). The symbol

\int

(integral) is a stretched letter S for Sum. The Fundamental Theorem of Calculus rigorously proves that differentiation and integration are inverse operations. Thanks to it, we can compute definite integrals by finding an antiderivative instead of taking limits directly.

It is the precise tool for area under a curve. For a wiggly curve, we add infinitely many very thin rectangles (width almost 0) to get the exact area. Mathematically it means the total of accumulated change.

A definite integral has a fixed start

a

and end

b

. To compute it we first find the antiderivative

F

(the function whose derivative is the integrand), then take $F(b) - F(a)$ — i.e. “later state minus initial state” — to get the accumulated result (area or total amount).

Essential for cumulative quantities in daily life. A car’s speed

v

changes every moment; integrating it over time

t

gives total distance. Integrating the flow from a tap gives total water in the bathtub. Whenever we “add up” a changing quantity, we use integration.

Key to continuous probability. For continuous data (height, weight), “probability that height is between 170–180 cm” is the area under the probability density curve over that interval — i.e. the integral. So the integral value is the probability of that event.

It underlies AI decision-making. When AI chooses under uncertainty, it computes expectation (each outcome times its probability, summed). That sum is mathematically an integral. Generative models (VAE, Diffusion) and reinforcement learning (expected cumulative reward) cannot move a single step without integration.

In physics, work and energy: integrating force over distance gives work; integrating acceleration gives velocity, integrating velocity gives position — used to predict trajectories and motion.

In economics, integration is used to find total demand or supply over time and to compute consumer and producer surplus for market efficiency.

AI performance and optimization: AUC (Area Under Curve) is literally an integral. Normalization (making total probability 1) also involves integration over the whole range. Every time we treat data flow probabilistically inside a network, integration is at work.

For a definite integral, use ① limits →

② antiderivative →

③

F(\text{upper})-F(\text{lower})

The antiderivative undoes the derivative of the expression inside

\int

Rules:

x^n

— raise the power and divide by the new exponent; constant

c

—

cx

;

e^x

— unchanged; sums — term by term. Differentiate your

F

to verify; subtract (upper) − (lower).

An indefinite integral is

F+C

only. For “value at

x=k

”, usually take

C=0

and substitute.

Simplest example:

\displaystyle\int_0^2 3\,dx

Antiderivative

3x

→

(3\cdot 2)-(3\cdot 0)=6

. Answer 6.

The table below goes from easy to varied integrands. Each row: antiderivative, then evaluate at both limits and subtract.

Problem $\int_0^2 3\,dx$
Solution $3x$

Problem $\int_1^3 2x\,dx$
Solution $x^2$

Problem $\int_0^2 (1+x)\,dx$
Solution $x+x^2/2$

Problem $\int 2x\,dx = x^2+C$
Solution $2^2=4$

Problem	Solution
$\int_0^2 3\,dx$	Antiderivative $3x$ → $(3\cdot 2)-(3\cdot 0)=6$
$\int_1^3 2x\,dx$	Antiderivative $x^2$ → $3^2-1^2=8$
$\int_0^2 (1+x)\,dx$	Antiderivative $x+x^2/2$ → $(2+2^2/2)-0=4$
$\int 2x\,dx = x^2+C$ , at $x=2$ ( $C=0$ )	$2^2=4$

Problem types and how to solve

Type $Constant definite$
Description $\int_a^b c\,dx$
How to get the answer $cx$

Type $Linear definite$
Description $\int_a^b (mx+k)\,dx$
How to get the answer $\frac{m}{2}x^2 + kx$

Type $Antiderivative value$
Description $x=k$
How to get the answer $k$

Type	Description	How to get the answer
Constant definite	$\int_a^b c\,dx$	Antiderivative $cx$ . (upper) − (lower) = $c(b-a)$ .
Linear definite	$\int_a^b (mx+k)\,dx$	Antiderivative $\frac{m}{2}x^2 + kx$ . Compute $F(b)-F(a)$ .
Antiderivative value	Indefinite integral given, value at $x=k$	Substitute $k$ into that expression. ( $C=0$ ⇒ $F(k)$ .)

Example (constant definite)

Find

\int_0^2 3\,dx

Solution

Antiderivative is

3x

(3\times 2)-(3\times 0)=6

. → Answer 6

Example (linear definite)

Find

\int_1^3 2x\,dx

Solution

Antiderivative is

x^2

3^2-1^2=9-1=8

. → Answer 8

Example (antiderivative value)

Given

\int 2x\,dx = x^2+C

, find the value at

x=2

. (

C=0

)

Solution

Substitute

2

into

x^2

→

4

. → Answer 4

What is the integral?

\int

A definite integral has a fixed start

a

and end

b

. To compute it we first find the antiderivative

F

(the function whose derivative is the integrand), then take $F(b) - F(a)$ — i.e. “later state minus initial state” — to get the accumulated result (area or total amount).

Essential for cumulative quantities in daily life. A car’s speed

v

changes every moment; integrating it over time

t

gives total distance. Integrating the flow from a tap gives total water in the bathtub. Whenever we “add up” a changing quantity, we use integration.

In economics, integration is used to find total demand or supply over time and to compute consumer and producer surplus for market efficiency.

For a definite integral, use ① limits →

② antiderivative →

③

F(\text{upper})-F(\text{lower})

The antiderivative undoes the derivative of the expression inside

\int

Rules:

x^n

— raise the power and divide by the new exponent; constant

c

—

cx

;

e^x

— unchanged; sums — term by term. Differentiate your

F

to verify; subtract (upper) − (lower).

An indefinite integral is

F+C

only. For “value at

x=k

”, usually take

C=0

and substitute.

Simplest example:

\displaystyle\int_0^2 3\,dx

Antiderivative

3x

→

(3\cdot 2)-(3\cdot 0)=6

. Answer 6.

The table below goes from easy to varied integrands. Each row: antiderivative, then evaluate at both limits and subtract.

Problem $\int_0^2 3\,dx$
Solution $3x$

Problem $\int_1^3 2x\,dx$
Solution $x^2$

Problem $\int_0^2 (1+x)\,dx$
Solution $x+x^2/2$

Problem $\int 2x\,dx = x^2+C$
Solution $2^2=4$

Problem	Solution
$\int_0^2 3\,dx$	Antiderivative $3x$ → $(3\cdot 2)-(3\cdot 0)=6$
$\int_1^3 2x\,dx$	Antiderivative $x^2$ → $3^2-1^2=8$
$\int_0^2 (1+x)\,dx$	Antiderivative $x+x^2/2$ → $(2+2^2/2)-0=4$
$\int 2x\,dx = x^2+C$ , at $x=2$ ( $C=0$ )	$2^2=4$

Problem types and how to solve

Type $Constant definite$
Description $\int_a^b c\,dx$
How to get the answer $cx$

Type $Linear definite$
Description $\int_a^b (mx+k)\,dx$
How to get the answer $\frac{m}{2}x^2 + kx$

Type $Antiderivative value$
Description $x=k$
How to get the answer $k$

Type	Description	How to get the answer
Constant definite	$\int_a^b c\,dx$	Antiderivative $cx$ . (upper) − (lower) = $c(b-a)$ .
Linear definite	$\int_a^b (mx+k)\,dx$	Antiderivative $\frac{m}{2}x^2 + kx$ . Compute $F(b)-F(a)$ .
Antiderivative value	Indefinite integral given, value at $x=k$	Substitute $k$ into that expression. ( $C=0$ ⇒ $F(k)$ .)

Example (constant definite)

Find

\int_0^2 3\,dx

Solution

Antiderivative is

3x

(3\times 2)-(3\times 0)=6

. → Answer 6

Example (linear definite)

Find

\int_1^3 2x\,dx

Solution

Antiderivative is

x^2

3^2-1^2=9-1=8

. → Answer 8

Example (antiderivative value)

Given

\int 2x\,dx = x^2+C

, find the value at

x=2

. (

C=0

)

Solution

Substitute

2

into

x^2

→

4

. → Answer 4