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.

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 …

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.

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 …

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 …

Mar 17, 2015 · A different approach, building up from first principles, without using cos or sin to get the identity, $$\arcsin (x) = \int\frac1 {\sqrt {1-x^2}}dx$$ where the integrals is from 0 to z. …

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

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 …

@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, …

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 …