integral
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Kaller integralet ditt for I:
[tex]I\,=\,sqrt{3}\int_0^1{1\over \cos^2(\frac{\pi}{6}(x+1))}{\rm dx}[/tex]
[tex]I\,=\, sqrt3 \cdot\,\frac{6}{\pi}[ \tan(\frac {\pi}{6}(x+1))]_0^1\,=\,\frac{6sqrt3}{\pi}(sqrt3\,-\,\frac{1}{sqrt3})\,=\,\frac{12}{\pi}[/tex]
[tex]I\,=\,sqrt{3}\int_0^1{1\over \cos^2(\frac{\pi}{6}(x+1))}{\rm dx}[/tex]
[tex]I\,=\, sqrt3 \cdot\,\frac{6}{\pi}[ \tan(\frac {\pi}{6}(x+1))]_0^1\,=\,\frac{6sqrt3}{\pi}(sqrt3\,-\,\frac{1}{sqrt3})\,=\,\frac{12}{\pi}[/tex]
La verken mennesker eller hendelser ta livsmotet fra deg.
Marie Curie, kjemiker og fysiker.
[tex]\large\dot \rho = -\frac{i}{\hbar}[H,\rho][/tex]
Marie Curie, kjemiker og fysiker.
[tex]\large\dot \rho = -\frac{i}{\hbar}[H,\rho][/tex]