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 …

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 …

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.

Lowercase t-like symbol is a greek letter "tau". Here it represents an integration (dummy) variable, which "runs" from lower integration limit, "0", to upper integration limit, "t". So, the convolution …

@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, …

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 …

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 …

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.

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. …

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 …