# U-substitution, Is this allowed?

1. Mar 26, 2014

### MexterO

Hi Everyone, I have a question on u-substitution it's a conceptual one.

Let's say I have the integral ∫x-y dx. where y is strictly just some constant, very simple integral.

I know the integration for that is simply (x^2)/2 - yx.

However, my classmate told me that I could use u-substitution. I told him you could either
do this in your head or use the concept of linearity for integrals. I said it would be silly to use u substitution.

I find it peculiar though since if I use u substitution on this problem I get a whole different anti-derivative.

I let u = x-y and I get du = dx, I then substitute it in and I get of course the anti derivative to be equal to ((x-y)^2)/2.

What is going on here, am I forgetting something or am I breaking a calculus rule?

Last edited: Mar 26, 2014
2. Mar 26, 2014

### pwsnafu

It isn't. The correct answer is $\frac{x^2 }{ 2} - yx +C$ for some constant $C$.

Again it is $\frac{(x-y)^2}{2} + D$ for some constant $D$. Then expand, and you see $C = \frac{y^2}{2} + D$.

3. Mar 26, 2014

### SteamKing

Staff Emeritus
Before you lose a lot of sleep over this, expand your second result [(x-y)^2]/2 and see what you get: (x^2 - 2xy + y^2)/2

Now, since this was an indefinite integration in the first place, your first approach should have given you x^2/2 - xy + C, where C is that oft overlooked constant of integration. Similarly, using u-substitution, the integral will initially come out as (x^2 - 2xy + y^2)/2 + C. But, since y is strictly a constant, the second result can be expressed as (x^2 - 2xy)/2 + C1, which after a little more manipulation becomes x^2/2 - xy + C1. The different approaches give the same result to within the difference of some unknown constant, which is consistent with indefinite integration.

You can always check by differentiating the result of an indefinite integral and seeing if the original integrand is returned.

4. Mar 26, 2014

### MexterO

Constant of Integration

Oh, I knew this would come back and bite me and yes I have lost one nights sleep over this lol. Could you for the life of me show me how to manipulate it to become x^2/2 - xy + C1, also is it safe to then assume that if this were a definite integral would u-substitution still yield the same results or is this just for indefinite integrals that's the part.

That's where I'm pulling my hair out because of that y^2 it doesn't seem computationally the same. My teacher says it's "absorbed" into the constant of integration, but that doesn't change the fact that if this were a definite integral and y where still strictly a constant that the results should be different between both techniques.

Last edited: Mar 26, 2014
5. Mar 26, 2014

### D H

Staff Emeritus
This kind of thing happens all the time, where two ways of approaching an integral give apparently contradictory results. For example, consider $\int 2\sin(x)\cos(x)\,dx$. One approach is via the u substitution $u=\sin x$, which immediately leads to $\int 2\sin(x)\cos(x)\,dx = \sin^2 x +C_1$. Another approach is to realize that the integrand is $\sin(2x)$, which leads to $\int 2\sin(x)\cos(x)\,dx = -\frac 1 2 \cos(2x)+C_2$. These two solutions differ by a constant.

6. Mar 26, 2014

### MexterO

Still so weird, then what about the definite case...

Okay, yes I see that those are legitimate ways of solving that integral, but what irks and has always irk me since calculus 1 is the fact that there is no way that those two solutions are the same.

I know, you said that the two solutions differ by a constant but shouldn't there be a unique antiderivative for each function or is there no math rule that says that must be so. I mean it should be possible then to manipulate either of the two solutions you provide for that integral and get the other, but I don't see how that could happen.

Further, then how would I know which integral to use then in the definite case, if I'd get a different integral just because I decide to solve it in a different way. I got a headache just thinking about it.

Last edited: Mar 26, 2014
7. Mar 26, 2014

### arildno

"but shouldn't there be a unique antiderivative for each function or is there no math rule that says that must be so."

No, there are no unique antiderivatives to ANY function.

This shouldn't surprise you:

Can two different functions have IDENTICAL slopes everywhere, yet still remain strictly different?

Sure!

Functions whose graphs are strictly PARALLELL (i.e, only differing by a constant), DO have this property!

8. Mar 26, 2014

### MexterO

Okay, that made things I little more clearer and I can see the point you made, and it makes total sense. Thanks for that. :) However, as DH mentioned with his example won't those two different solutions yield different results if it were a definite integral. You have to admit that those are two different functions so there has to be a difference if I went and solve the integral in one way versus the other for the definite case.

9. Mar 26, 2014

### D H

Staff Emeritus
No, they won't yield different results because $\sin^2(x)$ and $-\frac 1 2 \cos(2x)$ differ by a constant (1/2). The constant of integration vanishes when an antiderivative is used to evaluate a definite integral.

To illustrate this, consider $\int_0^{\pi/2} 2\sin(x)\cos(x)\,dx$.

Using $\sin^2(x)$ as the antiderivative yields $\sin^2(\frac{\pi} 2) - \sin^2(0) = 1 - 0 = 1$.

Using $-\frac 1 2 \cos(2x)$ as the antiderivative yields $-\frac 1 2\left(\cos(2\frac{\pi} 2) - \cos(0)\right) = -\frac 1 2\left((-1) - 1\right) = 1$.

Last edited: Mar 26, 2014
10. Mar 26, 2014

### MexterO

Alright, yup it does check out that they do indeed yield the same result. It still weirds me out though it feels I'm missing something regarding my original question. What the heck happened to -y^2 after manipulating the expressions and letting it get absorbed into the constant of integration.

Last edited by a moderator: Mar 28, 2014
11. Mar 26, 2014

### SteamKing

Staff Emeritus
It's included in the constant of integration. Still, differentiate the result including -y^2 and see if you don't get the original integrand.

12. Mar 26, 2014

### DrewD

Instead of thinking of the antiderivative of a function as being a function, think of it as the set of ALL functions that, when differentiated, reproduce the original function. So $\int f(x)\,dx=F(x)+C$ does not mean that "the function $F(x)+C$ is the integral of $f(x)$", but instead it means "The set of functions defined as $F(x)+C$ for every real value of $C$, is the integral of $f(x)$". Then the two functions, despite being different functions, are both in the same set. When you write $\int(x-y)\,dx$ this notation describes the set, not one function. Of course we usually choose a particular function from the set to make life easier.

I think that we are often taught so much about functions leading up to calculus that we expect the operation of integration to behave similarly. But integration is not a function from the set of integrable functions to the set of integrable functions and there is no reason that it needs to be.

Maybe this will make you feel better or maybe not.

13. Mar 27, 2014

### pwsnafu

Technically speaking it's "The set of functions defined as $F(x) +N(x)$ where $N'(x) = 0$, is the antiderivative of $f(x)$". There are situations where $N(x)$ is non-constant and still have zero derivative.

14. Mar 27, 2014

### MexterO

So is it safe for me to assume that no matter what the anti-deriviative looks like as long as I'm able to differentiate it and get the integrand again then I was able successfully able to integrate? Constants are negligible like that -y^2 since it goes away with differentiation.

15. Mar 27, 2014

### DrewD

$N(x)$ has to be constant with respect to $x$ by the MVT. Under what circumstances could it vary with $x$? Wouldn't the MVT apply to any integrable function?

Yes, by definition, if you differentiate the supposed antiderivative and get back the original function, you have integrated correctly. The only situation in which this does not hold is when you have incorrectly differentiated

16. Mar 27, 2014

### MexterO

Thank you everyone, for helping me understand this topic. I'm fully confident now. Thank you all. :D If something related to this topic pops up again in my class do I mention it here again or do I start a new thread? What's the best way. I'm new to the forums.