Calc help

[RavingManiac]

I seem to be having a brain-fart and can't seem to see something that's probably very obvious.

I'm taking multi-variable calc and we just finished up with the differentiation part of the course and are moving on to integration. I'm going through the homework and can't seem to see the very first step of the integration for this particular problem. The teacher posts solutions to all the problems online so I have most of it worked out...

The problem is (2x+3y)^2 dxdy

The solution to the problem gives the first step as [(2x+3y)^3]/6

There was a year in between Calc 2 and Calc 3 so i ended up forgetting a lot of the integration stuff unfortunately. Simply put, why is the denominator 6?

[Null]

First post in over a year is for math?

Sorry useless post on my part since I can't help.

[Woozie]

I'm not even sure what's being asked. Do they want you to integrate that with respect to x and then with respect to y?

If so, then simply expand the binomial and do normal integration (integrate with respect to x and then y, or y and then x. Either way you'll get the same answer).

If not, can you post the exact wording of the problem?

[Bugpop]

check out the first one...

List of integrals of rational functions - Wikipedia, the free encyclopedia

[Kaylia]

Assuming it's just multiple integral and not a partial derivative equation (you wouldnt be able to write it like this anyway), you need to do it like Woozie said.

Basically, do a first integral like you would if everything else was constant, and after that, you integrate a 2nd time using the 2nd variable.



i tried doing it once, and didn't find the same equation, so idk, there might be an error in your solution (might be me too)

[RavingManiac]

Well, the question is to integrate over a region, the choice of dxdy or dydx is left up to you. The region is a triangle with vertices at points (-1,0) (0,1) and (1,0). The function is the (2x+3y)^2.

In the posted solution, the professor chose to go dxdy. The first step in the solution is [(2x+3y)^3]/6. Do I have to use U substitution where my U is the 2x+3y and then I'm looking at U^2? Or should I just expand the binomial and go from there? Six of one, half a dozen of the other?

I'm sorry if I'm still being unclear. I'm trying to do too many things at once and not focusing as much as I should be.

[Kaylia]

Oh gdi...the "answer" was the result of the first integral.


[(2x+3y)^3]/6 should be integral f(y) = [(2x+3y)^3]/6 dy

[Kaylia]

Well, the question is to integrate over a region, the choice of dxdy or dydx is left up to you. The region is a triangle with vertices at points (-1,0) (0,1) and (1,0). The function is the (2x+3y)^2.

In the posted solution, the professor chose to go dxdy. The first step in the solution is [(2x+3y)^3]/6. Do I have to use U substitution where my U is the 2x+3y and then I'm looking at U^2? Or should I just expand the binomial and go from there? Six of one, half a dozen of the other?

I'm sorry if I'm still being unclear. I'm trying to do too many things at once and not focusing as much as I should be.

There is never a single method to do an integral. You just have to pick your poison. In your case, I would say it's safer to do them using "u" or expanding the ² term. Using tables works, but you won't have it during your exam, so start practicing now.



I've done the first integral using 3 different method here

1st: expanded the ² term
2nd: used the table
3rd: integral on u instead of x


Spoilered cause it's bigger than I though

Spoiler: show

http://img689.imageshack.us/img689/7181/integral.jpg

[Woozie]

Well, the question is to integrate over a region, the choice of dxdy or dydx is left up to you. The region is a triangle with vertices at points (-1,0) (0,1) and (1,0). The function is the (2x+3y)^2.

In the posted solution, the professor chose to go dxdy. The first step in the solution is [(2x+3y)^3]/6. Do I have to use U substitution where my U is the 2x+3y and then I'm looking at U^2? Or should I just expand the binomial and go from there? Six of one, half a dozen of the other?

I'm sorry if I'm still being unclear. I'm trying to do too many things at once and not focusing as much as I should be.

Oh gdi...the "answer" was the result of the first integral.


[(2x+3y)^3]/6 should be integral f(y) = [(2x+3y)^3]/6 dy

Between these two posts, I think I'm starting to understand what they're asking for.

My first instinct to integrate any equation when something is squared is to simply factor it out (because it makes the solution incredibly obvious).

Your question didn't make sense when you neglected to mention you were integrating over a region. I was thinking the same thing Kaylia said (that you were somehow trying to write a partial differential equation). But the way it was written didn't make sense like that, so my second instinct was that you meant to integrate with respect to one variable, and then to the other.

So if you're integrating over a region, then just integrate the function with respect to one variable and treat the other as a constant. Apply the appropriate boundaries. Then do the second.

Now I see where your equation [(2x+3y)^3]/6 comes from (thanks to Kaylia). "The first step" was kinda vague because you can integrate in either order since the bounds don't depend on either variable.

So yeah, do what Kaylia said and remember to apply the appropriate end points after each integration.

[Kaylia]

So yeah, do what Kaylia said and remember to apply the appropriate end points after each integration.

It's not a bad idea to evaluate the function at the very end in case you have to reuse the general form later in the problem.



I'm not sure I understand the whole triangle thing. It just mean the function is evaluated between x [-1,1] and y [0,1]?

[Woozie]

Wait, the region is a triangle? I misread, I thought it was a square. So the order in which he integrate is going to be important because he's going to have a boundary that depends on either x or y, depending on how he defines his boundaries.

I'll get to this tomorrow but I really need to work on my own homework for now.

[Woozie]

Let me explain this a little more. It's hard to do without a graph in front of me. I would draw a picture and scan it for you, but my scanner doesn't work until I install the drivers for my printer, which I don't feel like doing right now.

Here's one way to think of the boundaries:

You can say x goes from -1 to 1. Now, y has a minimum value of 0, and a max of 1. But if you think of x as going [-1,1], then you can't say y is [0,1] because then you'd have a square region. So if you decide to say x goes [-1,1] then you have to say y is bound by the lines y = 0, the line going from (-1,0) to (0,1) (the equation of this line is y = x+1), and the line going from (0,1) to (1,0) (the equation of this line is y = -x+1).

So in this case, we'd break the region down into two segments. The first would be the left half of the tringle, where x is bound by [-1,0] and y is bound by [0,x+1] (because the bottom bounding line was y=0 and the top bound is bound by x+1. The region is bound on the right by x=0, but the x part of the boundaries already takes care of this third bound).

Try integrating this region in either order and applying the appropriate bounds (so when you integrate with respect to y, the upper limit of integration will be the equation x+1. You're probably more used to the upper and lower limits being numbers [like integrating from 2 to 5, for example]. In this case, one of the limits is a function).

You'll notice here that integrating doesn't work in both orders. When the boundaries are functions, the order in which you integrate your answers will affect your final answer. In one case, your answer will be a number (which is what you want) and in the other case, your answer wont make sense (it should come out to be a function). Obviously, the answer that was a number is the correct way. If you've done problems like this before, then you know which order to integrate and why. If not, then I'm not going to tell you. You're going to have to see for yourself.

For the second half of the region (notice that the way I defined the bounds above, we only integrated over half of the region), you use similar reasoning as I did above. What three lines bound our triangle? Two of those lines are constant, and one is a function (y=-x+1, as I've already given you). Integrate over this region and apply the appropriate boundaries. You should be able to tell which order to integrate now if you did the first part.

After integrating both regions, add the two answers together to get your final answer.




Note that this could have been done by looking at the region in a slightly different way (and this second way is actually a bit easier). This time we're going to say that y is bound by [0,1]. Now we can't say x is bound by [-1,1] or else we'll have a square region. We need to say that x is bounded on the right by x=-y+1 and on the left by x=y-1. This time it's a bit simpler. Instead of breaking this into two regions, we can just say that the boundaries of integration are [0,1] for the dy, and [y-1,-y+1]. So in this case you have a single integral that looks like this

Spoiler: show

Set up your integral first, and then check your answer against mine

Spoiler: show

No seriously, try it yourself first, or else it's cheating. This is homework, afterall.

Spoiler: show

You cheating bastard, set up the freakin integral

Spoiler: show

The answer is either

http://latex.codecogs.com/gif.latex?...}(2x+3y)^2dydx

or

http://latex.codecogs.com/gif.latex?...}(2x+3y)^2dxdy

I'll leave it up to you to figure out which of these two gives the correct answer. If you can't tell just by looking, then evaluate both expressions and see which one makes sense and which doesn't.

Hopefully this helps. If it doesn't or if you're confused about anything, let us know.

[Bugpop]

Well, the question is to integrate over a region, the choice of dxdy or dydx is left up to you. The region is a triangle with vertices at points (-1,0) (0,1) and (1,0). The function is the (2x+3y)^2.

In the posted solution, the professor chose to go dxdy. The first step in the solution is [(2x+3y)^3]/6. Do I have to use U substitution where my U is the 2x+3y and then I'm looking at U^2? Or should I just expand the binomial and go from there? Six of one, half a dozen of the other?

I'm sorry if I'm still being unclear. I'm trying to do too many things at once and not focusing as much as I should be.

Did you look at the first entry in that link...

the integral of (ax+b)^n is [(ax+b)^(n-1)]/[a(n+1)]
in this case the integral of (2x+3y)^2 dx is [(2x+3y)^(2+1)]/[2(2+1)] or [(2x+3y)^3]/6

edit:The substitution Kaylia shows is a much easier way to solve and your prof would likely prefer it.

Expanding, integrating, then factoring back is time consuming for cubic functions. The added steps add potential for mistakes.

[Kaylia]

Expanding, integrating, then factoring back is time consuming for cubic functions. The added steps add potential for mistakes.

Using the easiest method isn't always the best way to learn though. He won't have integral table during his exam. I agree it's the fastest way if all you want is an answer asap.

[Woozie]

Did you look at the first entry in that link...

the integral of (ax+b)^n is [(ax+b)^(n-1)]/[a(n+1)]
in this case the integral of (2x+3y)^2 dx is [(2x+3y)^(2+1)]/[2(2+1)] or [(2x+3y)^3]/6

edit:The substitution Kaylia shows is a much easier way to solve and your prof would likely prefer it.

Expanding, integrating, then factoring back is time consuming for cubic functions. The added steps add potential for mistakes.

The original function is quadratic. Expanding a quadratic function is not time consuming at all and it's very unlikely that he'd make a mistake. For me, it would be the least likely way to make a mistake for a quadratic function. With u substitution I have a tendency to forget constants.

I've also noticed that people have a tendency to be really good at recognizing and integrating stuff that looks like polynomials. When I'm tutoring, if I give someone 9x^4-24x^2+16, they're like "that's easy", but if I give them (3x^2-4)^2, they're usually like "wtf???"

[Kaylia]

Woozie, what method do you uses to factor quadratic function (serious question)? It's still giving me a lot of trouble when I'm trying to find an answer with rational number.

I agree u substitution is more confusing than expanding a quadratic (minus the factorization for me!), but I have a hunch he will have to use this method sooner or later to solve different type of integral (trigo). It's a good idea to use it a couple of time on simple problem, and get used to the method.

[Woozie]

for x^2 + bx + c, I just ask myself "what two numbers can I multiply together to get c, and add together to get b?" That alone is enough to do the most of the quadratic equations I typically run into (because most textbooks set up the equations to be easily factorable). If the answer is complicated enough so that I can't figure it out that way, then the quadratic equation is probably quicker than all the other shortcut methods I learned in high school. So those are my only two approaches.

Actually, if asking myself that question doesn't give me the answer, I just type "factor(<expression>)" into my calculator.

If it's ax^2 + bx + c, I can usually look at it for like 30 seconds and sort of figure out what it has to be. Again, if I can't figure it out just by looking at it, I go straight to the quadratic equation (or calculator). I don't even remember half the stuff we did in high school to factor because I never use those methods.

[Kaylia]

I guess I meant something along the line of ax² + bxy + cy² more than a quadratic. There is time where you can figure it out looking at it, but more often than not, I get lost trying to solve it mentally.

[Trajan]

There is never a single method to do an integral. You just have to pick your poison. In your case, I would say it's safer to do them using "u" or expanding the ² term. Using tables works, but you won't have it during your exam, so start practicing now.



I've done the first integral using 3 different method here

1st: expanded the ² term
2nd: used the table
3rd: integral on u instead of x


Spoilered cause it's bigger than I though

Spoiler: show

http://img689.imageshack.us/img689/7181/integral.jpg

It might be because it's late, but I can't get (1) to equal (2) or (3).

If you expand (2x + 3y)^2 you get 4x^2 + 12xy + 9y^2. So the problem is Double Integral(4x^2 + 12xy + 9y^2)dxdy.

Integrate with respect to x you should end up with Integral( (4/3)x^3 + 6yx^2 + 9xy^2 )dy.

Integrate with respect to y and get (4/3)yx^3 + 3y^2x^2 + 3xy^3.

Expanding (3) gives you..(1/6)(2x + 3y)^3 = (1/6)(8x^3 + 36yx^2 + 36xy^2 + 27y^3) = (4/3)x^3 + 6yx^2+ 6xy^2 + (9/2)y^3.

[Kaylia]

It might be because it's late, but I can't get (1) to equal (2) or (3).

If you expand (2x + 3y)^2 you get 4x^2 + 12xy + 9y^2. So the problem is Double Integral(4x^2 + 12xy + 9y^2)dxdy.

Integrate with respect to x you should end up with Integral( (4/3)x^3 + 6yx^2 + 9xy^2 )dy.

Integrate with respect to y and get (4/3)yx^3 + 3y^2x^2 + 3xy^3.

Expanding (3) gives you..(1/6)(2x + 3y)^3 = (1/6)(8x^3 + 12yx^2 + 18xy^2 + 27y^3) = (4/3)x^3 + 2yx^2+ 3xy^2 + (9/2)y^3.

Well...shit I don't know why it's not working. I dont see why not, but i get the same thing.