La presente nota de clases aborda el estudio de algunos conceptos fundamentales del cálculo integral de una variable real. En ella se exponen las ideas básicas de la diferencial y la antiderivada de ...
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...