$FF$ ist Stammfunktion von$ff$, wenn gilt $F^{\mathrm{\prime }}\left(x\right)=f\left(x\right)F\text{'}\backslash left(x\backslash right)=f\backslash left(x\backslash right)$.

Also: $F(x)=-{\displaystyle \frac{2}{\sqrt{x}}}=-2\cdot x^{-\frac{1}{2}}\Rightarrow F^{\mathrm{\prime }}(x)=-2\cdot (-\frac{1}{2}\cdot x^{-\frac{3}{2}})=\frac{1}{\sqrt{x^3}}=f(x)F(x)=-\backslash dfrac\{2\}\{\backslash sqrt\{x\}\}=-2\backslash cdot\; x^\{-\backslash frac\{1\}\{2\}\}\backslash Rightarrow\; F\text{'}(x)=-2\backslash cdot(-\backslash frac\{1\}\{2\}\backslash cdot\; x^\{-\backslash frac\{3\}\{2\}\})=\backslash frac\{1\}\{\backslash sqrt\{x^3\}\}=f(x)$ .

Die Fläche unter dem Graphen von $ff$ in den Grenzen von $11$ bis $b>0b>0$ berechnet sich als

$A={\int }_1^b\frac{1}{\sqrt{x^3}}=-\frac{2}{\sqrt{b}}+2A=\backslash int^b\_1\backslash frac\{1\}\{\backslash sqrt\{x^3\}\}=-\backslash frac\{2\}\{\backslash sqrt\{b\}\}+2$.

Man soll $bb$ so bestimmen, dass $A=1A=1$ ist:

$2-\frac{2}{\sqrt{b}}=1\iff \frac{2}{\sqrt{b}}=1\iff b=42-\backslash frac\{2\}\{\backslash sqrt\{b\}\}=1\backslash Leftrightarrow\backslash frac\{2\}\{\backslash sqrt\{b\}\}=1\backslash Leftrightarrow\; b=4$.