jak rozwiązać całke
\(\displaystyle{ \int - e^{\frac{-x^{2}}{2}}dx}\)
całka z e
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octahedron
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całka z e
Np. tak:
\(\displaystyle{ \mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}\mbox{d}t=\begin{vmatrix}t=\frac{z}{\sqrt{2}}\\\mbox{d}t=\frac{\mbox{d}z}{\sqrt{2}} \end{vmatrix} =\sqrt{\frac{2}{\pi}}\int_0^{x\sqrt{2}}e^{-\frac{z^2}{2}}\mbox{d}z\\
\mbox{erf}\left( \frac{x}{\sqrt{2}}\right) =\sqrt{\frac{2}{\pi}}\int_0^{x}e^{-\frac{z^2}{2}}\mbox{d}z\\
\int - e^{\frac{-x^{2}}{2}}\mbox{d}x=-\sqrt{\frac{\pi}{2}}\mbox{erf}\left( \frac{x}{\sqrt{2}}\right)+C}\)
\(\displaystyle{ \mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}\mbox{d}t=\begin{vmatrix}t=\frac{z}{\sqrt{2}}\\\mbox{d}t=\frac{\mbox{d}z}{\sqrt{2}} \end{vmatrix} =\sqrt{\frac{2}{\pi}}\int_0^{x\sqrt{2}}e^{-\frac{z^2}{2}}\mbox{d}z\\
\mbox{erf}\left( \frac{x}{\sqrt{2}}\right) =\sqrt{\frac{2}{\pi}}\int_0^{x}e^{-\frac{z^2}{2}}\mbox{d}z\\
\int - e^{\frac{-x^{2}}{2}}\mbox{d}x=-\sqrt{\frac{\pi}{2}}\mbox{erf}\left( \frac{x}{\sqrt{2}}\right)+C}\)