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Mathematics-Online course: Prepcourse Mathematics - Analysis - Integral Calculus

Fundamental Theorem of Integral Calculus

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If $ F$ is a primitive of a continuous function $ f$ $ \left(f=F^\prime\right)$ , then it is

$\displaystyle \int_a^b f(x)\,dx = F(b) - F(a) $

which is commonly abbreviated as

$\displaystyle \int_a^b f = \left[ F \right]_a^b \,. $

A definite integral can thus be calculated as the difference of the primitive's values at the interval's boundary points.
(Authors: Höllig/Hörner/Abele)
The exponential function's derivative again is the exponential function, and thus it is

$\displaystyle \int_a^b e^x\,dx = e^b-e^a\,. $

So, the area below the graph within the interval $ [a,b]$ is equal to the area of a rectangle with breadth $ 1$ and the values' difference as its height.

\includegraphics[width=0.7\linewidth]{hs_bild}

The derivative of the logarithmis function $ F(x)=\ln(x)$ is given by $ F'(x) = 1/x\,,\, x\in \mathbb{R}^+$ and thus

$\displaystyle \int_a^b 1/x\,dx = \ln(b)-\ln(a) \,,\quad a,b \in \mathbb{R}^+\,. $

(Authors: Höllig/Hörner/Abele)

A planet of mass $ M$ induces a gravitation field in which a body of mass $ m$ experiences a force $ F(x) = \gamma \frac{mM}{x^2}$. Here, $ \gamma$ is the gravitational constant and $ x$ is the barycenters' distance.

A primitive for $ 1/x^2$ is $ -1/x$. In order to move a body from the distance $ a$ to the distance $ b$, the work

$\displaystyle \int_a^b F(x)\,dx = \gamma mM \int_a^b \frac{1}{x^2}\,dx= -\left[ \gamma mM/x \right]_a^b = \gamma mM(1/a-1/b) $

is thus required.

With $ a$ as the planet's radius and $ b \to \infty$, equating with the kinetic energy yields the so-called escape velocity $ v$, that is the speed necessary to leave a planet's gravitation field:

$\displaystyle \frac{m}{2}v^2 = \gamma \frac{mM}{a} \quad \Rightarrow \quad v = \sqrt{\gamma \frac{2M}{a}}\,. $

Along the equator this is for the earth $ v = 11.2\,$km/s.
(Authors: Höllig/Hörner/Abele)
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  automatically generated 9/18/2007