# Thread: Substitution Rule Confusion

1. ## Substitution Rule Confusion

Hey There,

My question involves the u-substitution rule for calculus. I do not understand it intuitively, and am relying purely on a mechanical recital, so I do not understand what happens to the "du"?

here is an example of my confusion:

$\displaystyle \int $$\displaystyle (x^2)$$\displaystyle (2x)$dx
= $\displaystyle \int $$\displaystyle udu = (x^2)/2 + C so I set \displaystyle u = x^2 and \displaystyle du = 2xdx. But where does the "du" go? It always seems to disappear, but I am not good enough at mathematics to understand why. Thanks for your input. -Warren 2. Originally Posted by Warrenx Hey There, My question involves the u-substitution rule for calculus. I do not understand it intuitively, and am relying purely on a mechanical recital, so I do not understand what happens to the "du"? here is an example of my confusion: \displaystyle \int$$\displaystyle (x^2)$$\displaystyle (2x)dx = \displaystyle \int$$\displaystyle u$du
= (x^2)/2 + C

so I set $\displaystyle u = x^2$ and $\displaystyle du = 2xdx$.

But where does the "du" go? It always seems to disappear, but I am not good enough at mathematics to understand why.

Thanks for your input.

-Warren
The same thing that happens to the "$\displaystyle \int$". The anti-derivative or integral of f(x) is $\displaystyle \int f(x) dx= F(x)+ C$ where F is any function such that F'(x)= f(x). The pair, $\displaystyle \int dx$ or $\displaystyle \int du$, or $\displaystyle \int dy$, is a symbol telling you to do a specific operation- after you have done the operation they are no longer needed. It the same thing that "happens" to $\displaystyle \sqrt$ when you write $\displaystyle \sqrt{4}= 2$.

3. Originally Posted by Warrenx
Hey There,

My question involves the u-substitution rule for calculus. I do not understand it intuitively, and am relying purely on a mechanical recital, so I do not understand what happens to the "du"?

here is an example of my confusion:

$\displaystyle \int $$\displaystyle (x^2)$$\displaystyle (2x)$dx
= $\displaystyle \int $$\displaystyle udu = (x^2)/2 + C so I set \displaystyle u = x^2 and \displaystyle du = 2xdx. But where does the "du" go? It always seems to disappear, but I am not good enough at mathematics to understand why. Thanks for your input. -Warren Am I right in guessing that you're worried not only about the du or the dx disappearing but the (2x) also? This seems likely, as your working is muddled enough to say \displaystyle \int$$\displaystyle (x^2)$$\displaystyle (2x)dx = \displaystyle \int$$\displaystyle u$du
= (x^2)/2 + C

when you mean

$\displaystyle \int $$\displaystyle (x^2)$$\displaystyle (2x)$dx
= $\displaystyle \int$$\displaystyle u$du
= (u^2)/2 + C
= (x^4)/2 + C

If so, then the answer to your question is: it disappears because you're going backwards through the chain rule. Look again at what the chain rule does going forwards: it produces a product, one 'half' of which is a kind of by-product - I'm referring to the derivative of the inner function of the composite function you start with. The chain rule instructs you to differentiate with respect to the inner function (descend the straight dashed line) and then multiply the result by the derivative of the inner with respect to x (having descended the straight continuous line). So you get a product. But with integration you're going backwards, so it makes sense that one part of the product you start with disappears. Compare going downwards through this shape...

... in order to differentiate the top level, with going up from bottom to top - finding the anti-derivative of the bottom level.

That's what happens to the 2x. I.e...

(Try to see it in both directions. By the way, it's a funny example because you could do it via the power rule, but perhaps that makes it a good one.)

Quite apart from that problem, you may now or later on want to ask what happens to the du just for the same reason you might ask what happens to the dx. I.e. nothing to do with the 2x, but just in its own right. This is a question about Leibniz notation. In fact this notation is (as you probably suspect) meant to help people (among other things) work backwards through the chain rule, by means of u-substitution. But it doesn't always seem to. So you might like to try the diagrams, or else, as many do, use Lagrange notation (see Notation for differentiation - Wikipedia, the free encyclopedia) where possible. And notice that Halls of Ivy is telling you to treat the whole enveloping pair $\displaystyle \int\ dx$ as one symbol that says "the anti-derivative with respect to x of..." (of whatever function you put in between)... and not necessarily (at this stage in your aquaintance with anti-differentiation) use u-substitution.

Hope that helps, or doesn't further confuse.

_________________________________________

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4. Haha I realize that I had it muddled. It was very late and I was sleepy so I apologize. Thank you very much for your detailed replies Tom@ballooncalculus & HallsofIvy, cleared things up for me.

### In u substitution what happens to the du

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