The FTC (e^x^2)by IsomaDan Tags: None 

#1
Nov1312, 03:15 AM

P: 10

The function, p(x;y), of two variables is defined for x>y>0, and satisfies
We furthermore know that dp(x,y)/dx = (e^x^2) and that p(y; y) = 0 I now need to write p(x,y) as a definite integral of the form int (f(t)dt, with lower bound t=H and upper bound x. I suppose I need the info p(y; y) = 0 to get the bounds, but not quite sure how. Anyone can give me a hint :) Dan 



#2
Nov1312, 10:11 AM

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For a fixed y, you know p(y, y) = 0. Based on that, how do you get p(x; y) = 0 with dp(x,y)/dx = (e^x^2)?
If that does not help, here is a slightly easier problem: For a fixed y, you can reduce it to a 1dimensional problem: f(y)=0, df/dx=(e^x^2). Can you write down f(x) with an integral? 



#3
Nov1312, 10:57 AM

P: 10

Dear MFB. Thanks so much for your answers. That's much appreciated.
Well, with regards to your answer, I suppose (but do not know) I evaluate the definte integral of dp(x,y)/dx = (e^x^2), with the bounds being an arbitrary constant and set it equal to zero? Is that correct, or am I still far off? best Dan 



#4
Nov1312, 11:13 AM

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P: 10,824

The FTC (e^x^2)
I don't understand what you plan to do, or why.
If you know the value of a onedimensional function at one point and its derivative everywhere, how can you get the value of the function everywhere? You don't have to evaluate the integral. If g(0)=3 and g'(x)=x, how does g look like? 



#5
Nov1312, 11:23 AM

P: 10

Find the integral, and insert the "restriction", to find the constant!




#6
Nov1312, 11:39 AM

P: 10

I do know that, my problem however, is it impossible to integrate the function as it stands there.
best Dan 



#7
Nov1312, 12:20 PM

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#8
Nov1312, 01:02 PM

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P: 21,026

I'm locking this thread as it has been crossposted. The other thread is here: http://www.physicsforums.com/showthread.php?t=651819.



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