calculus question?

Question

in an effort to better understand how calculus works (instead of just solving problems), i have a maybe simple-maybe difficult question to ask regarding u substitution.

In integration, when you set something equal to u, why must you take du/dx. How exactly does that work? To make my Q clearer, here is an example.

definite integral of [(3x^2 + 2x)/(x^3 + x^2)]dx

u = x^3 + x^2
du = (3x^2 + 2x) dx <-----------what exactly is going on here??

integral du/u is ln|u| so answer is ln|x^3 + x^2|...

Sorry if this question sucked, but I really would like to know more in depth...

Answer 1

When you are integrating the rate of change of the variables involved in the integration must be present with the variables. It means if you are to integrate x then "dx" must also be present there to integrate it.

Hence, when you substitute "u" for "x", "du" must be present in the integral. That is why you evaluate du/dx and substitute dx in terms of du in the original integral in order to solve the integral when the variable is represented in terms of "u".

I hope I kept things simple and clarified to you the point well.

Answer 2

I'm not sure if you wanted an explanation in first principles, but here's how I like to look at it:

u = x^3+x^2, so du/dx = 3x^2+2x.

The equation you're given can be rewritten as the integral of (du/dx)dx/u, which equals the integral of du/u because the dx's cancel out.

As for the bit where you write du = (3x^2+2x) dx, that is simply a rearranged form of du/dx = 3x^2+2x. It's a separable differential equation.

Answer 3

Your question is an excellent one. Your confusion is probably brought about by the very efficient short-hand notation that is used in calculus. Notation in mathematics is extremely important and calculus notation is among the best of them all. The point here is that integration by substitution is really the chain rule for differentiation in disguise. It is hard to be clear without proper symbolic typing, but with your example the full procedure is as follows:

Let y=Indefinite integral[(3x^2 + 2x)/(x^3 + x^2)]dx, then by definition, dy/dx=(3x^2 + 2x)/(x^3 + x^2). Now let u=x^3 + x^2 , then du/dx=3x^2 + 2x. But dy/dx=dy/du times du/dx (chain rule), therefore (3x^2 + 2x)/(x^3 + x^2)=dy/du times (3x^2 + 2x). Cancellation makes this dy/du = 1/u etc.

The dx du stuff is just short-hand for all of this. Calculus is intrinsically subtle and the more one thinks about it the more subtle it is!

Answer 4

It is a very good question so you shouldn't berate yourself.

∫f(x)dx is a convenient notation for the anti-derivative of f. One way to think about it is to look at f(x) as dF/dx. The notation mirrors the idea of finding out the anti-derivative by 'apparently' canceling out the dx;

∫f(x)dx = ∫[dF/dx]dx = dF.

U-substitution is just the observation that if F is a function of u and u is differentiable with respect to x. That is dF/dx = [dF/du](du/dx)

Answer 5

d stands for derivative.

u substitution is basically finding the derivative of an expression, instead of a term.

All in all, dx is just a way to get rid of variables.

Answer 6

You are relating a differential (small change) of one function,x, to another function,u. The confusing part is that students are not initially taught to take differentials, but rather derivatives with respect to an argument which is

= differential/(differential of an argument)

Are you familiar with implicit differentiation? The differential acts on all arguments equally. For example, in:

y^2 =5x
d(y^2 =5x)
d(y^2) = d(5x)
2y(dy) = 5dx

This equation relates changes in y to changes in x.

As mentioned, often people want dy/dx to begin with, which you get when you divide the above by dx and solve for dy/dx. Because dx/dx =1, and conveniently dissappears, you've probably not noticed that this is how differentiation works.

You can solve for the rate of change of x with y, or y with x, or simply relate the changes between the two as in the above.
Again, you are relating changes in one variable/function to another. For the example above, an infinitesmal change in y is (solving for dy)

(5/(2y))dx = (5/(2(5x))dx (which is (dy/dx)dx = dy)

This is in terms of x and an infinitesmal change in x, but in terms of y, an infinitesmal change in y is just that, simply dy.

Viewing derivatives in terms of differentials is ultimately more useful. Say both x and y vary with respect to another argument, t. To relate these rates to each other simply divide the expresion above by dt, you'll get dy/dt in terms of dx/dt and x and y.

To drive all of this home, say I have f= x^2 and I want to find the derivative of f with respect not to x, but to x^2. We know the answer should be 1, but how is this done?

Remember we wanted the derivative of f with respect to x^2, that's

d/d(x^2)[ x^2]

but d(x^2) = 2xdx

(Remember the chain rule, the derivative of a function is the derivative of the function first, then the derivative of what's inside that function. What's "inside" a function of x, is "x", it doesn't just dissapear unless you divide by dx (are interested in rates of change with respect only to x) so d(f(x)) = f'(x)x' =f'(x)dx = df.)

This means d/(d(x^2)) = (1/(2x))(d/dx) which is likely more familiar (d/dx is the derivative with respect to x).

But d/dx(x^2) =2x,

so d/(d(x^2))[ f ] = 1 as expected.

One correction to the above answers I'd like to add is that [dF/dx]dx = dF not ∫[dF/dx]dx = dF

∫[dF/dx]dx = ∫dF = delta(F) a FINITE change in F, e.g. F2 - F1. This is called the "Fundamental theorem of calculus". You can find more about it on Wikipedia.