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- Fu(U) = P[U<= u) = integral from -inf to +inf of 1 du = u120 B (29 words) - 16:51, 20 October 2008
- *i.e. the first integral will look something like this: <math> f_z(z)= \int \limits_{0}^{\infty} \la2 KB (344 words) - 17:00, 21 October 2008
- Another integral to convolute is <math> f_z(z)= \int \limits_{z}^{\infty} \lambda e^{-\lambd196 B (37 words) - 18:58, 21 October 2008
- *E[1/x] = integral(<math>\lambda e^{-\lambda x}</math> * (1/x)) dx * this integral is undefined182 B (28 words) - 14:48, 10 November 2008
- I would suggest splitting the double integral up. (Think of a double integral as a nested for loop -- integrating "slowly" over the outside loop and "qui1 KB (167 words) - 18:33, 9 December 2008
- E[x-q(x))^2] = Integral from -inf to inf (x-q(x))^2*fx(x)dx =integral from 0 to 1 (x-q(x))^2dx253 B (48 words) - 08:44, 10 December 2008
- Riemann Sum for the integral719 B (133 words) - 10:49, 14 October 2008
- Evaluate the Integral:1 KB (259 words) - 08:19, 1 October 2008
- Evaluate the integral: Good work. That last integral is easier to look at if you write <math>e^{-x}</math> in place of <math>\fr1 KB (260 words) - 07:50, 3 October 2008
- ...a <math> \frac{t}{p} </math> so I would have a ''dt''. That led me to the integral below. Does it make sense and does anyone know how to integrate the proble I don't know how to use this integral, but I did some manipulation and got this:1 KB (270 words) - 09:43, 7 October 2008
- A(t) = the integral of e^(-x) dx from 0 to t V(t) = the integral of Pi*[e^(-x)]^2 dx from 0 to t1 KB (245 words) - 18:31, 6 October 2008
- ...with the limits of integration when you take the derivative of a definite integral?645 B (120 words) - 18:05, 6 October 2008
- ...pi/2 instead of pi/4 because you have to bring out a 2 before you take the integral meaning that you have to multiply the first part of the answer from above b Now in the first integral substitute <math>v=2x</math> Therefore <math>dv=2dx</math> and when x=0, v=2 KB (315 words) - 14:23, 8 October 2008
- <math>\int\frac{6*2du}{1+u^2}</math> an easily-integrated integral. :) [[User:Jhunsber|Jhunsber]]794 B (147 words) - 14:30, 8 October 2008
- ...heir powers are equal), you can use this trick to drastically simplify the integral. It's a case that I don't think we covered in reading or lecture, but it d3 KB (584 words) - 10:12, 21 October 2008
- ...it eventually, but the inverse sin just gets worse and worse. Actually the integral of the inverse sin is just the inverse sin minus some radical. So it just c This looks better... but then I can't figure out how to solve that integral. Anyone? I've tried using the bottom as dv and going back to the inverse si2 KB (289 words) - 12:27, 14 October 2008
- ...es to <math>-\frac{3}{2}</math>, And I use partial fractions on the second integral: I solve for the first integral, leaving:1 KB (224 words) - 08:12, 14 October 2008
- Again we want to estimate the error for this integral on the interval x is between 0 and 13 KB (599 words) - 08:47, 13 November 2008
- That works wonder if the first part of the integral is x to the third power, but in this case, you end up with an uneliminatabl858 B (146 words) - 11:37, 1 November 2008
- ...ide of the equation the closer we get to <math>\frac{\pi}{4}</math>. This integral can therefore be called the error function.10 KB (1,816 words) - 15:32, 8 December 2008
