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Kestrel

Indefinite Integral Help?

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Posted (edited)

Y'all are smart here. Maybe someone can help me.

I'm stuck in Mathematical Analysis II. For some reason. I still don't know what this has to do with my major, but I'm here, and I have a final next week. My teacher isn't the best, but I've been able to teach myself most of the material via our online homework. Everything but indefinite integrals. Integrals in general stump me, but definites I can do in my TI-84, so if all else fails on the final I can use that. But there's no way to do indefinites in the TI-84 (as far as I'm aware), and I have no idea how to do any of these by hand. 

I'm not sure what level of these will be on the final either. The final has 2 multiple choice problems and 4 short answer. I think there will be a few straight math problems, but this class in general is supposed to be business math, so I'm expecting more word problems. This isn't the only thing on the final either, but I'd be really upset with myself if I don't at least try to learn these and it ends up blowing my A.

Here are some example problems she gave us specifically to study. I have the solutions to them too. I understand the first one okay, the rest make no sense. Under a spoiler for convenience, it's a large image.

Spoiler

8c421cc6bad275cde4dd0bd646e88dd8d6d8c74a

Anyway, maybe someone can help me with this? I've googled, gone to youtube, tried everything I can think of, and I always think I understand it, but when I go to the homework to do a real problem I can't solve it.

edit: The homework is nasty stuff like this btw. She NEVER went over problems this challenging in class, and I honestly think she just assigned the homework without even looking at it. So I don't think the final will be this hard. But the homework is annoying.

Q7CGGDW.png

Edited by Kestrel
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@Kestrel

The trick to use here is called "u-substitution." The idea is to find an expression and its derivative within the integral. When the integral is of polynomials, a good indication that u-sub is the right method is when there are two terms with degrees that differ by 1. You then substitute "u" for the higher degree polynomial, and du as the lower degree polynomial times the "dx" term.

Using problem 1 as an example, we can see that there is a x^3 term and an x^2 term, so we probably want to set the x^3 polynomial as u.

First write u=x^3+4. Then the derivative of u, du=3x^2 dx. We now substitute these expressions into the original example, giving the integral of u^2*du. We can evaluate this as 1/3 u^3. Then we substitute x^3+4=u back into the expression and get 1/3(x^3+4)^3. Then add a constant "c" since it is an indefinite integral.

This same method should generalize to problems 2 and 3. 

For e^(something) integrals, its probably worth memorizing that integral(e(a*x) dx) = 1/a * e^(a*x). We can verify this by taking the derivative of 1/a * e^(a*x) = 1/a * a * e^(a*x) = e^(a * x).

I hope that was helpful. If not, I can try again, or you could try looking up "u-substitution integrals" or something similar.

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@Araris Valerian

thank you!

I vaguely understand u-substitution, to the point where I can get the first problem. However, I'm mostly confused on where the 1/3 came from. With the first problem I used the powers of x formula, but on the second and third problem the solution work has a 1/3 before the integral. I'm not sure where that comes from?

Would it be possible for you to write down the e equations by hand and take a picture? My brain can't really wrap around how it's written there.

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Okay, so the 1/3 comes from the u substitution. Since u is some form of x^3, du=3x^2 dx. Since the original equation only contains x^2dx, and not 3x^2 dx, you divide both sides by 3 to get 1/3du=x^2dx.

I don’t have a great way to take a picture right now, but you can think of the e problems with u substitution as well. Take the e^(8x)dx from the last problem. We can write u=8x, and du=8dx, or dx=1/8du. When you substitute, you get the integral of 1/8*e^u du, which is 1/8e^u, or 1/8e^(8x).

Possibly another way to think of it is the chain rule. When you take the derivative of e to some power, you get the e bit back, and multiply by the derivative of the exponent. Here, when you take the integral of e to some power, you get the e bit back, and divide by the derivative of the exponent.

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The chain rule thing made the light bulb turn on! I think! That makes a lot more sense.

Now I have to figure out how to do some of these application problems.

This is the one that's really stumping me. I admit I guessed my way through the first part of this, but I have no idea how to do the second, since there's three (?) variables?

XuMXbJj.png

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With the first part, you’ve almost got it. When the investment has doubled, then $P will become $2P. So you can write the equation: e^(13n) * P = 2P. (Basically saying that the money after n years equals double the initial money). Ps cancel and your calculator should be able to do the rest.

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Posted (edited)

I can't believe I missed that! This class is rotting my brain out I think.

For now that might be all my questions, but I'll post if I have more aha. Thank you!

edit: Well that lasted *checks watch* ten minutes

how would you go about this, @Araris Valerian? I have no idea which of these to make "u"

9hywSP2.png

Edited by Kestrel
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Since there is a 5th degree polynomial and a 4th degree polynomial, I'd make the 5th degree one "u".

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I figured it out kinda by guessing and checking. I'm not sure something like that will be one the final, but I'm still confused on how to solve it. I couldn't quite figure out how to get it all to work together, if that makes sense.

I moved on, kinda confused here now.

5OTwEgt.png

I know how to get the anti-derivative of MR, but I'm not sure how to get the anti-derivative of MC (does the 30 stay the same or does it become 30x?). I'm sure I can figure out the rest of it if I find the proper profit equation, but I can't figure out the cost equation to help me get that.

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Okay, so you want to do a u-sub again. Set u=x+4, and then du=dx. You can factor out the 30, since it is a constant, and you get 30 times the integral of u^(1/2) du, which you can solve with the regular power rule.

For the previous problem, you want to set x^5-5x as u, and then du= 5x^4-5 dx, so (x^4-1)dx = 1/5 du. I think it just takes some practice to recognize what to choose as u, and until you've done it for a while it can seem kind of arbitrary.

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On 6/29/2019 at 7:39 PM, Araris Valerian said:

Okay, so you want to do a u-sub again. Set u=x+4, and then du=dx. You can factor out the 30, since it is a constant, and you get 30 times the integral of u^(1/2) du, which you can solve with the regular power rule.

For the previous problem, you want to set x^5-5x as u, and then du= 5x^4-5 dx, so (x^4-1)dx = 1/5 du. I think it just takes some practice to recognize what to choose as u, and until you've done it for a while it can seem kind of arbitrary.

okay! That makes sense. I'm slowly working my way through these and I think I know enough to at least do somewhat well on the final.

This isn't indefinite integrals anymore, but it's stumping me so I might as well post it. The last question I have on this homework. I can't seem to get it right, and of course there's no example in the text book.

QwKbcnx.png

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Okay, so you may be better off googling this one. A quick search for "producers surplus" seemed promising. However, the first step is to find the market equilibrium, which is the value of x where supply = demand. Using the equations you are given, we would get 169 - x^2 = x^2 + 10x +141. You can reduce this to a quadratic equation: 2x^2 + 10x - 28 = 0 --> 2(x + 7)(x - 2) = 0. Since x can't be negative, the market equilibrium is at x = 2. Using either the supply or demand equations, we can find the price to be $165.

Producer surplus is the area between the supply curve (p = x^2 + 10x + 141) and the constant price line at equilibrium (p = 165).

download.png.7a87855d4850050b3df3d2a665de7ff3.png

To find the area between curves, we evaluate an integral of the difference between then two:

Capture.PNG.d77389b0ab519d9a6c9d43bae426ca64.PNG

The bounds of the integral are 0, since selling less than 0 units doesn't make sense, and 2, since that is where the market equilibrium happens.

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Posted (edited)

Yes! Awesome, thank you! I had things mixed around (I was trying to use the demand function, not supply)
so I'm assuming, when the problem says "consumer's surplus" you use the demand function as f(x), and when it says "producer's surplus" you use the supply function?
I might have also been completely wayy off originally, too.

One last question. Then hopefully this madness will be over, as the test is tomorrow.

Capture.thumb.PNG.616957d851fb5030c8773576d830ef56.PNG

I don't really know where to begin. I'm assuming I take the integral of the equation above, then plug the income y (0) and and the consumption S (7.8 billion) back into the found equation and solve for C?

This one is from a powerpoint, not the online homework, so I have no idea if I'm on the right track here because I can't check my answers but...

I thought the resulting equation would be:

0.2y - 0.1e^-2y + C = S

An online calculator I found is telling me the -.1 is actually a +.05, the rest is fine. Does the .05 come from dividing the -.1 by the -2 coefficient in front of the y? So if it was actually, say -3y, I'd get .1/3e^-3y?

If that makes sense. Am I on the right track here?

Edited by Kestrel
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I'm pretty sure you're on the right track. You are exactly right about where the .05 comes from.

The one thing that might be off is that the equation is the marginal propensity to save, and you probably want the marginal propensity to consume, since the question is about the consumption function. (I may be totally off here, since it's been a while since I did econ). Just check to make sure that you are solving the right problem, because your math checks out. Marginal propensity to consume + marginal propensity to save = 1, so the math won't change that much.

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Posted (edited)

Okay, that makes sense! So if it is the marginal propensity to consume, how should I find that equation? I'm gonna go back and see if I can't find a problem like this in previous powerpoints (couldn't find one in the homework).

edit: Nope, none like it. There were problems that were the same type, (ie: either savings or consumption) but not both in one problem/going between the two.

edit 2: wait I got it now!

Edited by Kestrel
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