
What is an integral? - Mathematics Stack Exchange
Dec 15, 2017 · A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path …
What is the integral of 0? - Mathematics Stack Exchange
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 …
What is the integral of 1/x? - Mathematics Stack Exchange
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.
integral from zero to zero - Mathematics Stack Exchange
Feb 4, 2014 · it seems obvious that this integral is zero and so is the limit but what theorem we are using here? I see it's connected to Riemann sums with an interval=zero Right ? The …
What is the difference between an indefinite integral and an ...
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 …
How do I integrate $\\sec(x)$? - Mathematics Stack Exchange
Sep 27, 2013 · My HW asks me to integrate $\sin (x)$, $\cos (x)$, $\tan (x)$, but when I get to $\sec (x)$, I'm stuck.
What does it mean for an "integral" to be convergent?
Feb 17, 2025 · The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression …
solving the integral of $e^ {x^2}$ - Mathematics Stack Exchange
The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary …
How to calculate the integral in normal distribution?
If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.
calculus - Is there really no way to integrate $e^ {-x^2 ...
@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, …