The Unapologetic Mathematician

Mathematics for the interested outsider

Antiderivatives

One of the consequences of the mean value theorem we worked out was that two differentiable functions f and g on an interval (a,b) differ by a constant if and only if their derivatives are the same: f'(x)=g'(x) for all x\in(a,b). Now let’s turn this around the other way.

We start with a function f on an interval (a,b) and define an “antiderivative” of f to be a function F on the same interval such that F'(x)=f(x) for x\in(a,b). 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 F is some particular antiderivative of f then any other antiderivative G satisfies G(x)=F(x)+C for some real constant C. 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 F(x)=x^n then F'(x)=nx^{n-1}. Thus, turning this around, we find an antiderivative of f(x)=x^n=\frac{x^{n+1}}{n+1}, except if n=-1, 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) then if F is an antiderivative of a function f and G an antiderivative of g then F+G is an antiderivative of f+g. Similarly, the differentiation rule for a constant multiple tells us that cF is an antiderivative of cf for any real constant c.

Between these we can handle antidifferentiation of any polynomial P(x). Each term of the polynomial is some constant times a power of x, so the constant multiple rule and the rule for powers of x 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.

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January 28, 2008 - Posted by | Analysis, Calculus

5 Comments »

  1. [...] Before continuing with methods of antidifferentiation, let’s consider another geometric problem: integration. Here’s an [...]

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  2. [...] again start with a continuous function , but now we take any antiderivative , so that . Then the FToC asserts [...]

    Pingback by The Fundamental Theorem of Calculus II « The Unapologetic Mathematician | February 14, 2008 | Reply

  3. [...] function this is will have as its derivative. We can see that has this derivative, and we know that any two antiderivatives differ by a constant. That is, for some real constant . But we can also [...]

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  4. [...] . Its derivative must be , and we can check that also has this derivative, so these two functions can only differ by a constant. Clearly we want , since at that point we’re “integrating” over a degenerate [...]

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  5. [...] of the theorem tells us that we know a function whose derivative is : the function defined by . And we know that any two functions with the same derivative must differ by a constant! That is, there is some [...]

    Pingback by The Fundamental Theorem of Calculus (all together now) « The Unapologetic Mathematician | September 30, 2008 | Reply


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