Calc: Integration and U Substitution (Def Integ)

starmanbrand

So, I am in the last two weeks of my short term business calc class. I have understood pretty much everything. But subbing stuff in for integrals is just confusing me and I am hoping someone here can provide a basic explanation of what I am doing. I have read the previous thread on the topic, but it was mostly about specific problems and I am having trouble with the concept of it.

For integration- I am basically just adding one to the power of the variable and then dividing the variable by that power

For U/DU- I am finding a base function with an exponent and turning the base into U. Then everything that is left over in the problem becomes part of dx. I have to make the f'=dx=du, correct? This is where I seem to be having the most trouble. My book seemed to have glossed over this important step with just one example and not much direction.

tl;dr- Please explain integration and u substitution to me in common english rather than math mumbojumbo. Thank you.

PaleCommander

u substitutions are essentially working backwards with the chain rule. When taking derivatives, if you had a function f of another function g, then f(g(x))' = f'(g(x)) * g'(x) (in English, you take the derivative of f, leaving g alone, then multiply by the derivative of g).

When integrating, if you're given a function v of another function u, where you know how to integrate v(x) (in your example, a power function, although it could just as easily be a cosine or something), and that's multiplied by the derivative of u(x), then you can use a u substitution, essentially reversing what you would do when differentiating.

I realize that might have been brain-melting, so let me try to illustrate with an example. you're given the integral from 0 to (square root of pi) of cos(x^2) * 2x * dx. You could integrate cos(x) just fine. You can (hopefully) see that the derivative of x^2 is 2x, so you make the following substitution:

u = x^2 (this will let you integrate cos(u), which you know how to do, instead of cos(x^2), which you don't),

du = 2x * dx (just differentiate each side of that first equation),

and now we substitute. The x^2 changes to a u, and the 2x * dx changes to a du. After changing the bounds of integration (simple, but easy to forget), we have the integral from 0 to pi of cos(u) * du, which you can integrate pretty easily as 1.

I know that was long and possibly jargon-laden, but I hope you can get something out of it. I'd be happy to explain anything that was unclear or that I breezed over.

starmanbrand

So, basically. Just so I can get this straight. lets say the problem is 3x/(2x+1)^4 [This is just something i made up while typing]

I would let d= 2x+1
that would make du= 3x. I would need to get this to equal 2 since d'=2? Or would I be trying to just solve to get dx alone by dividing the 3 so I have 1/3du= dx?

Tarantio

It's generally not best to simply make up one's own problems when trying to learn a particular technique, as not every technique is suitable for every problem.

One thing you're missing (unless it's a typo?) is that you don't ever set d= to anything. Just u, and du (which is the derivative of u). After that, you're going to want to find your integral with respect to u, and possibly plug the original term back in, rather than solving for anything algebraically.

I'll see if I can find some good examples.

Edit: Good Tutorial here:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/USubstitution.html

enlightenedbum

The simplest kind of example is something like:

int(2x/(x^2 + 1)dx)

In which cause you set u = x^2 + 1
Then you take the derivative of that: du/dx = 2x, so du = 2x dx

Regrouping the initial problem:

int(2x dx / (x^2 + 1))
we can substitute:
int(du/u)
ln u + C (drop the absolute value because x^2 + 1 is always positive)
ln (x^2 + 1) + C

is the answer.

What if that were say, 3x on top instead? Well, if you multiply by 2/3 inside the integral and 3/2 outside (multiplying by one overall) you get the problem we just did, and the answer is 3/2 ln (x^2 + 1) + C.

Savant

starmanbrand wrote: »

So, basically. Just so I can get this straight. lets say the problem is 3x/(2x+1)^4 [This is just something i made up while typing]

I would let d= 2x+1
that would make du= 3x. I would need to get this to equal 2 since d'=2? Or would I be trying to just solve to get dx alone by dividing the 3 so I have 1/3du= dx?

The "d" isn't a variable like "u" or "x" are, it's more of an operator. If you did substitution here you could set u = 2x + 1, and then du/dx = 2, or .5 du = dx. This is a little messier, because then you would have to substitute with x = (u - 1)/2, which would give the integral Int( (u - 1)/(4u^4) du ) = .25 * Int ( u^(-3) - u(-4) du), which is relatively simple. You would substitute back in x at the end and add C if it is indefinite, or change the boundaries of integration based on u = 2x + 1 if it is definite.

Substitution tends to be most useful in certain situations, while others integration by parts is more useful. It's something you just have to get a feel for.

starmanbrand

Thanks for the explanations guys. Don't know why I am having so much trouble with this. Between class yesterday, reading this thread, and especially that link with examples I am getting a better feel for it. Just going to have to practice more.

Thank ya, i will leave it unlocked incase anyone has any other tips though.