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Wardub
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Calc 3 help
« on: Jan 8th, 2008, 11:27am » |
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Hi guys it's me again. I'd like to thank everyone that helped me in Discrete Math. Now I'm taking Calc 3 or multi variable Calc. And i will probably need help in the future. So I'll have problems. My main problem now is I haven't taken Calc in like 8 months and I'm a bit rusty. I remember all the basic derivatives stuff I think  Anyways i need a refresher and I have a few things to ask. Hopefully my parents can find my notes and mail them to me.
I remember there is something special with the exponential function and I forgot how to take the derivative and anti derivative of that.
So could someone explain both of those for f(x) = e^-3x ?
Then I believe the anti derivative of 1/x is ln(x) correct? Just making sure.
Also I want to say the derivative of sin is cosine but i can't remember if its the other way around.
And lastly I can't remember how to do anti derivatives very well. I can do easy things like 6x^2 = 2x^3 but i Can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully.
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towr
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Re: Calc 3 help
« Reply #1 on: Jan 8th, 2008, 3:11pm » |
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on Jan 8th, 2008, 11:27am, Wardub wrote:I remember there is something special with the exponential function and I forgot how to take the derivative and anti derivative of that.
So could someone explain both of those for f(x) = e^-3x ? |
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Two things to know
d/dx ex = ex
d/dx ef(x) = f'(x) ef(x)
[The latter is just a case of the rule d/dx g(f(x)) = f'(x)g'(f(x)) ]
Quote:| Then I believe the anti derivative of 1/x is ln(x) correct? |
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Yes
Quote:| Also I want to say the derivative of sin is cosine but i can't remember if its the other way around. |
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sin -> cos -> -sin -> -cos -> sin
Remember the graphs, sin starts at zero and goes up, so its derivative must be positive at the start, so it must be cos.
Quote:| And lastly I can't remember how to do anti derivatives very well. I can do easy things like 6x^2 = 2x^3 but i Can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully. |
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Do you have a specific example?
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Wikipedia, Google, Mathworld, Integer sequence DB
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Hippo
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Re: Calc 3 help
« Reply #2 on: Jan 8th, 2008, 3:56pm » |
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on Jan 8th, 2008, 11:27am, Wardub wrote:| I can't do more complicated ones. also I forgot integration by parts. So if any of you could help, i would highly appreciate it. I think after I see a few It will all come back hopefully. |
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It is not so simple in general 
Per partes uses formula for derivative of product:
(f.g)'= f'.g+f.g'.
Uf you are looking for "antiderivatives" of f.g', it is f.g - antiderivative of f'.g.
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| « Last Edit: Jan 8th, 2008, 3:57pm by Hippo » |
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Icarus
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Re: Calc 3 help
« Reply #3 on: Jan 9th, 2008, 7:43am » |
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The big problem with integration by parts is trying to pick your two functions in such a way that you can integrate the one differentiate the other to produce an easier expression. The rule is
 udv = uv - vdu
or (in a more directly applicable form):
f(x)g'(x) dx = f(x)g(x) - f'(x)g(x) dx
Sometimes, there are obvious ways of writing the expression you are trying to integrate as a product, which allows you to easily figure out how to do something useful:
xex dx = xex - 1ex dx = xex - ex + C (f(x) = x, g'(x) = ex, g(x) = ex)
Often, though, you have to be creative to figure out what division of the integrand would be useful:
ln(x) dx = xln(x) - x/x dx = xln(x) - x + C. (f(x) = ln(x), g'(x) = 1, g(x) = x)
It is very easy to choose poorly, and therefore get something worse than you started with:
xcos(x) dx = x2cos(x)/2 + x2sin(x)/2 dx  (f(x)=cos(x), g'(x) = x, g(x) = x2/2)
Sometimes, you can seem to get nowhere, but in fact, if you do additional integrations by parts, the answer slips out in an unexpected direction:
excos(x) dx = excos(x) + exsin(x) dx = excos(x) + exsin(x) - excos(x) dx. Solving for the integral gives excos(x) dx = ex(cos(x) + sin(x))/2 + C.
(first int by parts: f(x)=cos(x), g(x)=g'(x)=ex; 2nd int by parts: f(x)=sin(x), g(x)=g'(x)=ex)
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Wardub
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Re: Calc 3 help
« Reply #4 on: Jan 9th, 2008, 8:32am » |
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Thanks everyone I am starting to remember  . Ok I knew there was something special about e^x so when taking the anti derivative is it e^(w/e)/ derivative of (w/e) for example e^2x = e^2x/2? Im pretty sure thats how you take them with the expo function.
And I get the cycle for sin to cos to -sin ... so thanks for that.
Lastly Icarus thanks for trying to help me understand integration by parts i still am not fully sure of how to do it. I remember it can often take doing the integration by parts multiple times. Anyway how i was taught was with the U V DU and DV but i cant remember exactly how to do it. For your example of xe^x which is the U? and which is the DV. I remember there was like a rule to which was which. Hmm I'm not explaining myself well. Could you for your example explain which is the U and why? also for the other V and DU/DV Because i think that would help me get it. Thanks again everyone. I'm starting my first chapter of calc 3 tonight, So i might post question but hopefully i get it Good thing is homework isn't graded and we have quizzes that are extracredit so I think I should do pretty well I just need to make sure I understand it for the tests. Thanks again everyone.
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Grimbal
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Re: Calc 3 help
« Reply #5 on: Jan 9th, 2008, 9:02am » |
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on Jan 9th, 2008, 8:32am, Wardub wrote:| ... so when taking the anti derivative is it e^(w/e)/ derivative of (w/e) for example e^2x = e^2x/2? Im pretty sure thats how you take them with the expo function. |
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No. Not if "w/e" means "whatever".
(ef(x))' = ef(x)·f'(x)
doesn't imply
(ef(x)) = ef(x)/f'(x)
It works in your example because f'(x) is a constant, i.e. f(x) = a·x.
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| « Last Edit: Jan 9th, 2008, 9:05am by Grimbal » |
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pex
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Re: Calc 3 help
« Reply #6 on: Jan 9th, 2008, 9:06am » |
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on Jan 9th, 2008, 8:32am, Wardub wrote:| Thanks everyone I am starting to remember . Ok I knew there was something special about e^x so when taking the anti derivative is it e^(w/e)/ derivative of (w/e) for example e^2x = e^2x/2? Im pretty sure thats how you take them with the expo function. |
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This only works if (w/e) is linear - indeed, an antiderivative of eax+b is (1/a)eax+b, if a is nonzero.
However, it does not work for nonlinear exponents. Clearly, eln x = x, so an antiderivative is (1/2)x2. Your "rule" would give an antiderivative eln x / (1/x) = x2, which is incorrect.
Taking it even further, consider ex^2. Taking antiderivatives as you propose leads to ex^2 / (2x), but it has been proven that no "elementary" antiderivative of ex^2 exists at all!
Edit: I'm slow...
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| « Last Edit: Jan 9th, 2008, 9:07am by pex » |
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Icarus
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Re: Calc 3 help
« Reply #7 on: Jan 10th, 2008, 11:29am » |
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xex dx = xex - 1ex dx = xex - ex + C
u = x; dv = exdx (so v = ex).
ln(x) dx = xln(x) - x/x dx = xln(x) - x + C.
u = ln(x); dv = dx (so v = x)
xcos(x) dx = x2cos(x)/2 + x2sin(x)/2 dx
u = cos(x); dv = xdx (so v = x2/2). A much better choice would have been to make u = x; dv = cos(x)dx, but that would spoil my example!
excos(x) dx = excos(x) + exsin(x) dx = excos(x) + exsin(x) - excos(x) dx. Solving for the integral gives excos(x) dx = ex(cos(x) + sin(x))/2 + C.
First integration by parts: u = cos(x); dv = exdx (so v = ex)
second integration by parts: u = sin(x); dv = exdx (so v = ex)
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
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