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## The Fundamental Theorem of Calculus (FTC)

There are four somewhat different but equivalent versions of the Fundamental Theorem of Calculus.
• FTCI:
Let be continuous on and for in the interval , define a function by the definite integral:

Then is differentiable on and , for any in .
• 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:

Note that FTCIII and FTCIV are just rewrites of each other.
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:

As in FTCI, define for in by the formula:

Then by FTCI, we have for any in , so , for any in .
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:

where is any function such that .
So will do.
With this choice of , we have:

giving us the required result and we are done.
Sample uses of FTC.
• Find a formula for the function defined for any real by the integral:

By FTCI, we have: .
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:

Then for example we have:

• Evaluate the definite integral:

By FTCII, we have:

where is any function such that .
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:

where is the position at time .
But also we have , so we get:

Here we used the given fact that .
To do the remaining integral, we observe that obeys the equation: , so FTCII applies to give:

So the particle is at position at time .
• 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:

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

Next: Using the mean value Up: Internet Calculus II Previous: Solutions
George A. J. Sparling 2002-01-27