- FTCI:

Let be continuous on and for in the interval , define a function by the definite integral: - FTCII:

Let be continuous on .

Let be any function such that , for any in .

Then we have: - FTCIII:

Let have a continuous derivative on the interval .

Then we have the integral: - FTCIV:

For a differentiable function of introduce its differential .

Then we have the integral, for a differentiable function with continuous derivative:

FTCIII has the intuitive phrasing: the integral of the instantaneous rate of change of a quantity is the total change.

Partial proof:

We defer for a moment the proof of FTCI.

We assume FTCI and prove the other parts of FTC from it.

- Proof of FTCII.

For FTCII,assume given a function , such that , for any in .

Then we want to show that:

So , for any in .

So the function is constant.

So .

So we get: - Proof of FTCIII and FTCIV.

Since FTCIV is just a rewrite of FTCIII, it suffices to prove FTCIII.

By FTCII, we have:

So will do.

With this choice of , we have:

- Find a formula for the function defined for any real by the integral:

But a function whose derivative is is the function .

Put .

Then .

So is constant.

But .

So for all .

So , for any real .

We have shown that, for any real , we have: - Evaluate the definite integral:

But one such function is , since .

So we get: - The velocity of a particle moving along the -axis is
.

- If the particle is at the origin at time , where is the particle at time ?

By FTCIII, the change in position from to , is given by:

But also we have , so we get:

To do the remaining integral, we observe that obeys the equation: , so FTCII applies to give: - How much distance does the particle cover in the time interval ?

The total distance covered is given by the integral of the absolute value of the velocity:

We see that in the interval , we have and , whereas in the interval , we have and , so we get:

- If the particle is at the origin at time , where is the particle at time ?