“La logística se convierte en la cara visible de la empresa: la primera imagen que se lleva el cliente”, dice Juan Manuel. Con esa mirada, profundiza en la importancia de diferenciarse en un mercado ...
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.
Surface Integral over a sphere Ask Question Asked 11 years, 8 months ago Modified 11 years, 8 months ago
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 functions such as $\frac {x^3} {3} +C$. However, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ...
What would you set the limits if you need to calculate the area of an infinitesimal ring in cartesian coordinates i.e. $\int dx \int dy $.. where you only want to integrate on the infinitesimal ring.. I know in polar that will be 2πrdr but how will you get it in caartesian using double integral
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 find the area under a curve. I think of them as finding a weighted, total displacement along a curve.
I have a question about the following property, which I didn't know so far: Why does the Itō integral have zero expectation? Is this true for every integrator and integrand? Or is this restricte...