What is the integral of 1 x? - Mathematics Stack Exchange $\begingroup$ "Answers to the question of the integral of 1x are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers" --- not completely correct: if they are both negative it also works This is an improper integral and does not converge in the remaining cases $\endgroup$ –
calculus - Is there really no way to integrate $e^{-x^2 . . . $\begingroup$ @user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral
Newest integration Questions - Mathematics Stack Exchange For questions about the properties of integrals Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag(s) that describe the type of integral being considered This tag often goes along with the (calculus) tag
Lesser-known integration tricks - Mathematics Stack Exchange Countless times, I've seen students make sign errors in this type of integral that could have been avoided by organizing the computations according to these simple rules Apparently this is being taught as a trick in some schools, judging from this clip from the 1988 movie Stand and Deliver
How to calculate the integral in normal distribution? It goes without saying that if you're trying to find a CDF, you need to add limits and evaluate the definite integral In the second equation you'll notice that I used "a" as the (upper) limit variable And the question is talking about the CDF, so the lower limit is negative infinity $\endgroup$ –