Nov 29, 2013 · Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". This MIT page says, "The more common name for the antiderivative is the indefinite integral." One is free …

Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

@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 …

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

Feb 4, 2018 · The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function f …

Jun 8, 2015 · The integral we generally teach in a first calculus course actually depends on a parameterization of the interval we are integrating over, and perhaps most naturally generalizes to …

I was reading on Wikipedia in this article about the n-dimensional and functional generalization of the Gaussian integral. In particular, I would like to understand how the following equations are

Aug 26, 2015 · 2 I am trying to calculate the integral of this form: $\int_ {-\infty}^ {+\infty}e^ {-x^2}\cdot x^2dx$ I am stuck. I know the result, but I'd like to know the solution step-by-step, because, as some …

3 days ago · 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 …

One possible interpretation: a "normal" integral is simply a line integral where the path is straight and oriented along a particular axis. Thus, as soon as you perform a transformation to the integrand to …