Has any1 got tips for coping under exam conditions. Anxiety is bad when ever it happens, but when you have several 2 hour exams coming up that count for a large part of your degree, you really dont want to be dealing with anxiety as well as trying to get a good grade! If Anybody has any tips on controlling anxiety and coping within an exam, I would love to hear about it!
I hate to use this thread in the way it was intended, but could anyone suggest how to find the first crossover point between the integral of e^(-x^2)+sin(x/12pi) and the integral of cos(x-x^2)^(xsin(lnx))?
Joined: Apr 18, 2005 Posts: 299 Location: United States of America
Posted: Thu Feb 15, 2007 4:10 pm Post subject:
I'm not exactly sure what you mean by crossover point, but if it is just where the two curves intersect, the would it be possible to plot them to find where they intersect? Or set them equal to each other and solve for x...I'm guessing this isn't as simple as that. By the way, that second integral is possibly the nastiest one I have ever seen.
_________________ "No, try not. Do, or do not. There is no try" - Yoda
"People usually fail when they are on the verge of success. So give as much care to the end as to the beginning" - Lao Tsu
I'm not exactly sure what you mean by crossover point, but if it is just where the two curves intersect, the would it be possible to plot them to find where they intersect? Or set them equal to each other and solve for x...I'm guessing this isn't as simple as that. By the way, that second integral is possibly the nastiest one I have ever seen.
The crossover point is where the weight of a function with a lower initial value (f(0)) surpasses the weight of a function with a higher initial value (g(0)). So, in this case, the intersection of the second integrals.
I concur on the nasty bit. My teacher is somewhat sadistic. I asked her about it and she said the e^(-x^2) part is called the Hearth Function or something like that and that it can't be integrated (or at least, that is what I gleaned), but past a certain point, its integral can be approximated by integrating 1/x^2. I suppose I could try the graphing approach, provided my calculator or winplot can handle it, but the problem is not from a solve-by-graphing section, so I'm sure it's supposed to be solved exactly by non-graphic means. As for the second integral, I don't even know where to begin.
Joined: Apr 18, 2005 Posts: 299 Location: United States of America
Posted: Mon Feb 19, 2007 5:28 pm Post subject:
mienaino wrote:
The crossover point is where the weight of a function with a lower initial value (f(0)) surpasses the weight of a function with a higher initial value (g(0)). So, in this case, the intersection of the second integrals.
I concur on the nasty bit. My teacher is somewhat sadistic. I asked her about it and she said the e^(-x^2) part is called the Hearth Function or something like that and that it can't be integrated (or at least, that is what I gleaned), but past a certain point, its integral can be approximated by integrating 1/x^2. I suppose I could try the graphing approach, provided my calculator or winplot can handle it, but the problem is not from a solve-by-graphing section, so I'm sure it's supposed to be solved exactly by non-graphic means. As for the second integral, I don't even know where to begin.
Probably doesn't matter anymore, but the only way i can think to solve that would be through graphing it...it's probably possible to write a program to iterate until the equations are approximately equal as well, and get a point from that. I went ahead and tried to get the double integral of those functions in derive...the first one could be integrated, but had an error function in it. Derive couldn't integrate the second one at all...so i have no idea, that problem just sucks.
_________________ "No, try not. Do, or do not. There is no try" - Yoda
"People usually fail when they are on the verge of success. So give as much care to the end as to the beginning" - Lao Tsu
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