Antiderivatives
One of the consequences of the mean value theorem we worked out was that two differentiable functions 
and 
on an interval 
differ by a constant if and only if their derivatives are the same: 
for all 
. Now let’s turn this around the other way.
We start with a function on an interval and define an “antiderivative” of to be a function 
on the same interval such that 
for . What the above conclusion from the mean value theorem shows us is that there’s only one way any two solutions could differ. That is if is some particular antiderivative of then any other antiderivative 
satisfies 
for some real constant 
. So the hard bit about antiderivatives is all in finding a particular one, since the general solution to an antidifferentiation problem just involves adding an arbitrary constant corresponding to the constant we lose when we differentiate.
Some antiderivatives we can pull out right away. We know that if 
then 
. Thus, turning this around, we find an antiderivative of 
, except if 
, because then we’ll have to divide by zero. We’ll figure out what to do with this exception later.
We can also turn around some differentiation rules. For instance, since ![\frac{d}{dx}\left[f(x)+g(x)\right]=f'(x)+g'(x)](https://s0.wp.com/latex.php?latex=%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Bf%28x%29%2Bg%28x%29%5Cright%5D%3Df%27%28x%29%2Bg%27%28x%29&bg=e6e6e6&fg=333333&s=0&c=20201002)
then if is an antiderivative of a function and an antiderivative of then 
is an antiderivative of 
. Similarly, the differentiation rule for a constant multiple tells us that 
is an antiderivative of 
for any real constant 
.
Between these we can handle antidifferentiation of any polynomial 
. Each term of the polynomial is some constant times a power of 
, so the constant multiple rule and the rule for powers of gives us an antiderivative for each term. Then we can just add these antiderivatives all together. We also only have one arbitrary constant to add since we can just add together the constants for each term to get one overall constant for the whole polynomial.
January 28, 2008 -
Posted by John Armstrong |
Analysis, Calculus
The Unapologetic Mathematician 
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