Wiki-Quellcode von Lösung Horn von Torecelli
Version 19.1 von Niklas Wunder am 2023/10/24 16:20
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| author | version | line-number | content |
|---|---|---|---|
| 1 | a) Das Volumen V eines Rotationskörpers lässt sich durch {{formula}} V(x)=\pi \cdot \int_a^b (f(x))^2 \;dx {{/formula}} bestimmen. | ||
| 2 | Für die gegebene Funktion f erhält man demnach | ||
| 3 | {{formula}} | ||
| 4 | \begin{align*} | ||
| 5 | & g(x) & =\: & 0\\ | ||
| 6 | \Rightarrow\: & \frac{1}{2}(x^2-4x+3) & =\: & 0\\ | ||
| 7 | \Rightarrow\: & x^2-4x+3 & =\: & 0\\ | ||
| 8 | \end{align*} | ||
| 9 | |||
| 10 | \begin{align*} | ||
| 11 | &\Rightarrow x_{1,2}=\frac{4\pm\sqrt{4^2-4\cdot3}}{2}=\frac{4\pm2}{2}\\ | ||
| 12 | &\Rightarrow x_1=1;\: x_2=13 | ||
| 13 | \end{align*} | ||
| 14 | {{/formula}} | ||
| 15 | {{formula}}f(x) = 0 | ||
| 16 | |||
| 17 | \Rightarrow \frac12\left(x-1\right)\left(x-3\right)=0 | ||
| 18 | |||
| 19 | \Rightarrow x-1=0\vee x-3=0 | ||
| 20 | |||
| 21 | \Rightarrow x_1=1; x_2=3{{/formula}} | ||
| 22 | iodhiohdcio | ||
| 23 | {{formula}}dsd\begin{align*} | ||
| 24 | & g(x) & =\: & 0\\ | ||
| 25 | \Rightarrow\: & \frac{1}{2}(x^2-4x+3) & =\: & 0\\ | ||
| 26 | \Rightarrow\: & x^2-4x+3 & =\: & 0\\ | ||
| 27 | \end{align*}{{/formula}} | ||
| 28 | hksdhkfh |