Base9se I'm calling you out!

[madbouncy]

Yeah but it's still not the integral even with C. Integration by parts sucked, and so did lots of calculus. I might try it tonight if I get bored enough. Bought Guilty Gear Isuka so I'll be playing that a bunch.

[madbouncy]

Well integration by parts didn't work. I just ended up with Int(e^x^2) = x * e^x^2 - Int(2x^2 * e^x^2). Which is in no way easier to solve than the original equation so I have no idea. I suck too much at math to do it.

[szh]

Meaning that for a given perimiter, the maximum area occurs when you make a square. Somebody asked up in the start of the thread, so I figured I'd answer it.

A side bar: if the shape does not need to be "rectangular", i.e., with vertical and horizontal sides, then the maximum area for a given perimeter is ... a circle!

Z

[madbouncy]

From what I remember as long as you keep all the angles even, the more sides you give it, the more area you'll get out of the perimeter. Hence the circle pretty much being the most sides you can give it. Though the square is usually the most used because it's a lot more practical.

[yelnatsch517]

If you think integration by parts is hard, engineering is not your major. O man, 6 quarters of Calc is hell. Personally I think integration by trig substitution is way harder, just because I hate trig functions. BTW, all that is only considered 2nd quarter calc. Imagine what the next 4 quarters are like.

[madbouncy]

It's not hard, I just don't pay attention. Anyways, I won't be done with college for another...5 years about, and that's just for my BS. Nothing like taking a 3 year break to go into the army to afford college. I'm basicaly going to try my hardest to get stationed in germany.

[yelnatsch517]

It's not hard, I just don't pay attention. Anyways, I won't be done with college for another...5 years about, and that's just for my BS. Nothing like taking a 3 year break to go into the army to afford college. I'm basicaly going to try my hardest to get stationed in germany.

LOL, I never pay attention in class either. I always, and I mean always fall asleep in class. Funny thing is, my earliest class starts at 11:00.

[madbouncy]

lol, yeah I know what you mean, I just listen to music if I have to. My earliest is at 8 but I haven't been in weeks. Well I went one day because we had a test, that's the only time I go there. Then this class I do the same thing, so really, my earliest class is at 1pm. Except for like one day a week when homework is due in the 11am class. Either way, it beats the hell out of high school.

[AZhitman]

Damn.

Whoever said we're all a bunch of idiots has obviously never seen this thread!

Well done, guys - Y'allz some smart cookies!!!

<--- Failed college algebra twice and still graduated with Honors.

[base9se]

Well integration by parts didn't work. I just ended up with Int(e^x^2) = x * e^x^2 - Int(2x^2 * e^x^2). Which is in no way easier to solve than the original equation so I have no idea. I suck too much at math to do it.

There is no need for integration by parts...this integration formula you probably aren't too familiar with is what I used to get my solution.

[PantherRacer]

ok.....so is this what I have to look forward to when I leave high-school and go to college? tell me that's not required for computer programming and engineering! or automotive engineering!well I know it's not needed for paiting and bodywork and engine enhancement...cuz DAMN! P = fixed perimiterAsq = L * WP = L + L + W + W2W = 2L - PW = (2L - P)/2Asq = L * ((2L - P)/2)Asq(L) = (2L^2 - LP) / 2Asq'(L) = 1/2 * (4L - P)Asq'(L) = (4L - P) / 2(4L - P) / 2 = 0L = P/4Asq'(L) = 0 @ L = P/4Asq''(L) = 2, so L = P/4 is a max value



Int(e^x^2) = x * e^x^2 - Int(2x^2 * e^x^2)

[base9se]

Most engineering majors will take Calculus & Differential Equations. It might not mean anything to you right now, but wait till you understand. It'll all make sense. If you listen & have a good professor.

[madbouncy]

It's requried for engineering, it's only calc 2. By the way, that rule you showed for e doesn't work for this. Because you have a second x to take care of. When you take the derivative of e^x^2 you get another x in there, so you need to have your original equation have one to get rid of, except if you have x * e^x^2 then the chain rule goes in and you end up with too much stuff. I don't know how to do it, but I don't think I've learned anything that would work with that. I could probably look it up tonight when I'm doing my calc stuff if I have enough time, I'll see.

[base9se]

when you have u=x^2, then you can have x=u^(1/2), which will give you the value of that 2nd x term you're talking about.

[madbouncy]

Don't quite follow what you're saying. I tried having u in there like that but I didn't have any luck, sorry I'm pretty bad at math so I don't follow it easily without a lot of explanation.

[hannibal]

when you have u=x^2, then you can have x=u^(1/2), which will give you the value of that 2nd x term you're talking about.

Good call, man! I used that formula, but when someone said I was wrong I looked at it and realized I didnt have a 'dx' term. But the simple equations you wrote above solved that problem.So I was right, and didnt even know it. That's a shame

I actually got a F the first time I took Calc 2. On my second try, I got an A+ baby! (better teacher)

[madbouncy]

What'd you do to put it in, I'm curious how to solve it. I tried working out what you had before but I couldn't get it, I'm probably just messing it up, I usually always miss something when I do that stuff. Too lazy to be careful.

[base9se]

What'd you do to put it in, I'm curious how to solve it. I tried working out what you had before but I couldn't get it, I'm probably just messing it up, I usually always miss something when I do that stuff. Too lazy to be careful.

U substitution, which is u=x^2Then, take the derivative, which is du=2xdxTherefore dx=(du)/(2x)Then the equation looks like: (1/(2x)) integral e^u du(1/(2x)) is a constant now & can be set outside the integralThen all you have left to integrate is e^u duInt e^u du = e^uSolution: [[(1)/(u')]*(e^u)]+C => [[(1)/(2x)]*(e^(x^2))]+C

[Doomed2Walk]

I forgot how to derivatives, damn, I'm so stupid :[

[rbsileighty]

ok.....so is this what I have to look forward to when I leave high-school and go to college? tell me that's not required for computer programming and engineering! or automotive engineering! well I know it's not needed for paiting and bodywork and engine enhancement... cuz DAMN! P = fixed perimiter Asq = L * W P = L + L + W + W 2W = 2L - P W = (2L - P)/2 Asq = L * ((2L - P)/2) Asq(L) = (2L^2 - LP) / 2 Asq'(L) = 1/2 * (4L - P) Asq'(L) = (4L - P) / 2 (4L - P) / 2 = 0 L = P/4 Asq'(L) = 0 @ L = P/4 Asq''(L) = 2, so L = P/4 is a max value



Int(e^x^2) = x * e^x^2 - Int(2x^2 * e^x^2)

Don't let this stuff discourage you. Like most classes you will learn a lot of stuff and only use some of what you learned. I had someone ask me the other day how to do Trig sub and it took a while for it to come back and it finally did... but it's been a very long time since I've taken that class. Diff EQ wouldn't have been that bad if it weren't for a teacher I couldn't understand due to broken english... so I taught that one to myself and don't remember to much of it either.

My Calc classes didn't allow for calculators, but the engineering ones do... I highly recommend picking up a TI-89. I started back in the day with a 82 and finally picked up an 89 Titanium this year... should have started with this bad boy calculator and saved some $$ on the 83Plus I had after the 82 bit the dust!

My advice... do your best and don't worry about it. Engineering isn't so much of what you learn, but a way of learning. I'd say it's worth it...

[base9se]

It depends on your classes. Some let you use calculators, some not. Just learn the theory & learn how to apply it. Don't think too deep into it.

[madbouncy]

Most of the stuff we've posted you wont' have to remember. The biggest thing about engineering is just knowing what you need to solve the problem. My dad didn't memorize every formula, he had an entire book shelving unit full of stuff in his office. He just told me to make sure you can indentify what you need to know, that's also what my physics teacher told me. You'll very rarely ever be in a real life situation where you won't be allowed to look up an exact formula, you just need to know enough about it to look it up. For college, you just memorize it, test on it, forget it, it's all in a weeks work.

I still don't see how that works, I've never heard of being able to take an X out of an equation and just making it a constant and then throwing it back into the integral at the end so you get around the chain rule. Though it might just be something we haven't learned yet. I never did u substitution. I always just looked at it and did it, so when people use u it can get me a little messed up at times till I see what they're doing.

[base9se]

so does that mean you understand what i did? if not, read up on u substitution..check out this place http://www.hyper-ad.com/tutoring/math/usub.htm Didn't really read through it all the way..just kinda scrolled through it. Maybe that helps you understand something about u-substitution.

[madbouncy]

Yeah I understand all that. Those are all easy stuff where it works out nicely. I always ignored the whole u thing and just did it. It just all depends what you get use to, I got use to it without it, so that's the way I do it. I still dno't think that's the asnwer for it, just take the derivative of (e^x^2)/2x + C and you won't get e^x^2. It's all ****ed up, I left the sheet at my dorm of what it worked out too, but you have to remember the chain rules messes it up.

( g(x) * f'(x) - g'(x) * f(x) ) / g(x)^2

g(x) = 2xg'(x) = 2f(x) = e^x^2f'(x) = e^x^2 * 2x

( (2x * e^x^2 * 2x) - (2 * e^x^2) ) / (2x)^2( (4x^2 * e^x^2) - (2 * e^x^2) ) / 4x^2( 2 * e^x^2 * ((2x^2) - (1)) ) / 4x^2(e^x^2 * (2x^2 - 1) ) / 2x(e^x^2 * 2x^2 / 2x) - (1 / 2x)(e^x^2 / x) - (1 / 2x)

Alright, that doesn't look right but I can't think of anything else. I'm sure I messd something up in there, not sure what though.

[base9se]

you're making it too complicated for yourself..in math you always go on the tangent path (fastest & more simple). The derivative of u was taken because you need a value for dx (U substitution)When dx is determined, then its equal value is put back in the original equationTherefore giving you an equation of (1/(2x)) integral e^u du Making (1/(2x)) a constant because you're integrating with respect to uIntegrating will give you [[(1)/(2x)]*(e^(x^2))]+C

[skylndrftr]

hey base what program were those drawings done in...did i see someone else on earth who uses IDEAS drafting...::shudder::

anyways I just finished calc 3, got my c and thanked the lord I will probably never see another talylor or maclaurin series again... no onto double inegrals and the dreaded addition of z to calculus

[madbouncy]

I just don't think that works, because for it to be the integral the derivative of it has to become that. That's the whole point of the the integral. That was how we always checked our answers, you did the anti-derivative on the integral and then to check yourself, you took the derivative of your answer, and if it gave you the original problem, then you're good. u substitution is terrible for integration in my opinion. It's only good if you have u and du in the integral, and you really don't. All you have is u in e^x^2. You can't make up du in the integral and throw it in. It has to be in there in the first place. So you'd have to split e^x^2 into a u and a du, and I can't see how you'd do that. I'll ask my brother to try and solve it when he comes by, he usually teaches me anything my teachers can't.

[base9se]

you're just not seeing it. it's all good, just let me know what your brother gets.

[madbouncy]

Well I talked to my brother finally. He agrees with me that (e^x^2)/2x won't work because of the chain rule. However, he wasn't sure exactly how to take the integral besides that it would be really hard and he had an easier way to find it.

http://integrals.wolfram.com/

You can put in the integral, they use E for e, so just do like E^(x^2) and it'll give you a really ****ed up equation that it equals. I think e^(-x^2) is like sqrt(pi). But that doesn't help in this equation. You can see that it's definitly not an easy equation, and you're u substitution stuff definitly won't work. At least we have it cleared up though, too bad neither of us could actually do out the function instead of having to have a program do it.

[base9se]

this is the derivative formula for e Here's the website ... http://www.pen.k12.va.us/Div/W....htmlWhich makes sense. I don't rely on computer programs to do Calculus. I mean they can do simple Integration & Derivatives, but they just get too complex for complex integrals & derivatives. Don't get me wrong I use MathCAD for some of my classes, but I don't use them to do all my Calculus. More for graphing. I use formulas which give a more simple solution, never to complicate my answer.