Math Explained in English: Indefinite Integration by Substitution

This is my first attempt to put a symbol-bound bit of math into English, for ease of my own comprehension and amusement of others. It took me something like an hour to parse out the explanation in the textbook because it relied on a dense bunch of nested functions, variables flying about left and right, and so forth. So I’m rewriting it to cut the comprehension time down to about ten minutes or so.
First, you should see the symbol-bound version. This guy’s web page is incredibly good on the subject.

Here’s my summary.

So, you’ve got this function you need to integrate. Suppose it’s kind of messy, like say [imagine an integration symbol here] 20x(x2+5)2 dx. Well, if you can express that as a function (f) of a second function (g), multiplied by the derivative of that second function (or a constant multiple of that derivative), then you can just substitute the variable “u” for that second function, drop the multiplication by the derivative out of your consciousness (changing dx to du to represent it, and keeping any constant multiple), and integrate the resulting much simpler function.

Sooo… in the example I gave, the g function (“u”) would be x2+5. The f function is then u2 (not, shockingly, the band), because that meets the condition described above (the second function being buried within the first function). The derivative of the g function is 2x, which is a tenth of the 20x we conveniently have sitting around collecting dust. So we can express the new function to be integrated as: [integral symbol] u2 (10)du, which, by the constant multiple rule, becomes 10 [integral symbol] u2 du, for which the integral is easy: 10u3/3 + C. Abracadabra!

Take that, textbook!

Math people: did I screw this up? Time to chime in…
Also, who wants to find me a text-sized gif or jpg integral symbol clipart?

4 Comments to Math Explained in English: Indefinite Integration by Substitution

  1. Matt says:

    To be honest, I had a very hard time parsing your explanation. Granted, I may be biased on the side of [being used to reading] pristine mathematical exposition, but your explanation doesn’t really seem to get us anything but superfluity, in my opinion.

    The whole idea behind u-substitutions is the chain rule: d/dx [f(u)] = d/du [f] * d/dx [u].

    I think the reason that your explanation is difficult to follow is kind of clear: look at all the clauses in your second sentence! =)

    The idea should be presented something like this:

    Given a complex integrand, exploit the chain rule and look for possible choices for a new function u(x) so that the integrand can be recast as f(u)du. Defining u=x2+5, in your example, we note then that du=2x. With this information we can recast the integrand

    20x(x2+5)2 = 10*2x*((x2+5)2

    as

    10*du*u2 (*).

    [It's always important to define u and then immediately look at du before trying to recast the integrand so you don't waste any time -- sometime the expression for du just doesn't help you. Try u = 20x in this example.]

    Now (*) is an integral we can handle. Integrating with respect to u, we have

    (10/3)u3+C.

    But u=x2+5, so …

    I have to run to lunch with some friends, but hopefully you get my idea?

  2. Yes, your explanation is much clearer!

    I wrote this post as an attempt to accurately replicate the thought process that takes my not-used-to-mathematical-symbology brain through the steps necessary to get one set of mathematical symbols to another. My brain is going to be less efficient at performing this task than the brain of someone who is fluent in the symbols. As we’ve seen, this is reflected in the difference between our explanations. :-)

    (I need to start writing shorter sentences in general.)

  3. Eric says:

    Matt’s criticism is offbase since he resorts to the “symbol-bound” approach that you were attempting to avoid. And his exposition is not much clearer than your own. Actually, it’s a bit worse because it obfuscates one of the mysterious parts of the procedure (ie computing the differential then ‘dropping it’), whereas your approach exposes it, even if it doesn’t deal with it.

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