Integration is usually taught as a set of rules for turning one expression into another. The idea underneath is far simpler, and it can be seen in a single picture: the area piling up under a function is the height of its integral function.

Note Two graphs, one number

Take the constant function $f(x) = 1$ . Its integral function is $F(x) = x + C$ , and for now we can set the constant aside and use $F(x) = x$ .

Stop at

x = 2

. The region under

f

from the start to that point is a rectangle of width

2

and height

1

, so its area is

2

Now read the lower graph at the same place: $F(2) = 2$ . The area above and the height below are the same number, and they stay the same number wherever you stop.

Example Moving the stopping point

Same two graphs, three different stopping points.

Stop at

2

: area

= 2 \times 1 = 2

, and

F(2) = 2

Stop at

4

: area

= 4 \times 1 = 4

, and

F(4) = 4

Stop at

3.9

: area

= 3.9

, and

F(3.9) = 3.9

⟹ area under $f$ up to a point $=$ value of $F$ at that point

Nothing here depends on the numbers being easy. The rule is what defines the integral function in the first place.

Note A rising function forces a steepening integral

The constant case hides something, because every step to the right adds the same slab of area. Take $f(x) = x$ instead, where each step adds more area than the last.

The strip from

1

2

has area

1.5

. The strip from

3

4

is the same width but has area

3.5

More area arriving per step means $F$ has to climb faster to keep matching it. That is why the lower curve bends upward: its slope at each point is exactly the height of $f$ there.

Definition The area between two points

If $F$ records the area accumulated up to a point, then the area between $a$ and $b$ is what has accumulated by $b$ , minus what had already accumulated by $a$ .

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

The upper limit $b$ is evaluated first and the lower limit $a$ is subtracted from it. The constant $C$ never survives this subtraction, which is why it does not appear in a definite integral.

Example The area under

f(x) = x

from 1 to 4

The integral function is $F(x) = \dfrac{x^{2}}{2}$ .

F(4) = \dfrac{16}{2} = 8

F(1) = \dfrac{1}{2} = 0.5

Subtract:

8 - 0.5

⟹ $\displaystyle\int_{1}^{4} x\,dx = 7.5$

The region is a trapezium with parallel sides $1$ and $4$ and width $3$ , so its area is $\tfrac{1}{2}(1 + 4)(3) = 7.5$ . The geometry and the integral agree, as they must.

Summary

The area under $f$ up to a point equals the value of $F$ at that point.
Move the stopping point and both numbers move together.
A taller $f$ delivers more area per step, so $F$ climbs more steeply.
The slope of $F$ at any point is the height of $f$ there.
For the area between $a$ and $b$ , evaluate $F$ at both and subtract.
The constant $C$ cancels in that subtraction, so definite integrals do not carry it.

A Function and Its Integral Function