Where the Constant C Goes — Moosa Academy

Every indefinite integral ends in $+ C$ . Put limits $a$ and $b$ on the sign and it is suddenly gone. Nothing was dropped and no rule was bent — $C$ simply stops mattering, and it is worth seeing exactly why.

Note What C actually stands for

$C$ records the starting state of the system — where the accumulation began before any of the integrating happened.

In the tank picture, $C$ is how much water was already sitting in the tank when you started the clock. An empty tank gives $C = 0$ . A tank that was half full gives some larger $C$ . The flow going in afterwards is identical either way.

Theorem C cancels between limits

$\big(F(b) + C\big) - \big(F(a) + C\big) = F(b) - F(a)$

A definite integral subtracts one value of the antiderivative from another.

The same

C

sits in both of them.

Subtraction removes it, whatever value it had.

This is not a special case or a convention. It happens for every constant, every function and every pair of limits, which is why a definite integral never carries a $+ C$ .

Note A difference, not a total

Changing

C

slides the whole line up or down without tilting it. Both ends move by the same amount.

The rise between $t = 1$ and $t = 3$ is therefore untouched. A definite integral asks for that rise, and the rise does not care where the line started.

Example The same tank, two starting points

Water flows in at $5$ litres per minute. How much collects between minute $1$ and minute $3$ ?

The antiderivative is

F(t) = 5t + C

Start empty, so

C = 0

F(3) - F(1) = 15 - 5

$= 10$ litres

Start with

20

litres, so

C = 20

F(3) - F(1) = 35 - 25

$= 10$ litres

Two different tanks, two different totals in them, one identical answer. The question was never how much water is in the tank — it was how much arrived during those two minutes.

Example Carrying C through the algebra

Evaluate $\int_{1}^{3} 5\,dt$ , keeping $C$ in place the whole way.

F(t) = 5t + C

F(3) = 15 + C

F(1) = 5 + C

F(3) - F(1) = (15 + C) - (5 + C)

= 15 + C - 5 - C

$= 10$

The $C$ terms subtract themselves away on the second-to-last line. Leaving $C$ out of a definite integral is not a shortcut — it is what the algebra does on its own.

Summary

$C$ stands for the starting state — what was already there before the accumulation began.
An indefinite integral cannot know that starting state, so it keeps $C$ .
A definite integral asks only for the change between two points.
Changing $C$ shifts both endpoints equally, so the change is unaffected.
Algebraically $(F(b) + C) - (F(a) + C) = F(b) - F(a)$ .
$C$ never disappears. It cancels.