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 …

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 …

The indefinite integral cannot be expressed in terms of elementary functions. The integral is, quite unsatisfactorily, expressed in terms of the exponential integral $\mathrm {Ei} (x)$.

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 function $\\mathrm{f}$ is

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. With 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 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 and use …

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 …

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 integrals do not seek to …