Integralen av $$$\frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}$$$
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Din inmatning
Bestäm $$$\int \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}\, dx$$$.
Lösning
Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{\sqrt{2}}{2 \sqrt{\pi}}$$$ och $$$f{\left(x \right)} = e^{- \frac{x^{2}}{2}}$$$:
$${\color{red}{\int{\frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}} d x}}} = {\color{red}{\left(\frac{\sqrt{2} \int{e^{- \frac{x^{2}}{2}} d x}}{2 \sqrt{\pi}}\right)}}$$
Låt $$$u=\frac{\sqrt{2} x}{2}$$$ vara.
Då $$$du=\left(\frac{\sqrt{2} x}{2}\right)^{\prime }dx = \frac{\sqrt{2}}{2} dx$$$ (stegen kan ses »), och vi har att $$$dx = \sqrt{2} du$$$.
Alltså,
$$\frac{\sqrt{2} {\color{red}{\int{e^{- \frac{x^{2}}{2}} d x}}}}{2 \sqrt{\pi}} = \frac{\sqrt{2} {\color{red}{\int{\sqrt{2} e^{- u^{2}} d u}}}}{2 \sqrt{\pi}}$$
Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\sqrt{2}$$$ och $$$f{\left(u \right)} = e^{- u^{2}}$$$:
$$\frac{\sqrt{2} {\color{red}{\int{\sqrt{2} e^{- u^{2}} d u}}}}{2 \sqrt{\pi}} = \frac{\sqrt{2} {\color{red}{\sqrt{2} \int{e^{- u^{2}} d u}}}}{2 \sqrt{\pi}}$$
Denna integral (Felintegral) har ingen sluten form:
$$\frac{{\color{red}{\int{e^{- u^{2}} d u}}}}{\sqrt{\pi}} = \frac{{\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erf}{\left(u \right)}}{2}\right)}}}{\sqrt{\pi}}$$
Kom ihåg att $$$u=\frac{\sqrt{2} x}{2}$$$:
$$\frac{\operatorname{erf}{\left({\color{red}{u}} \right)}}{2} = \frac{\operatorname{erf}{\left({\color{red}{\left(\frac{\sqrt{2} x}{2}\right)}} \right)}}{2}$$
Alltså,
$$\int{\frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}} d x} = \frac{\operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}$$
Lägg till integrationskonstanten:
$$\int{\frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}} d x} = \frac{\operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}+C$$
Svar
$$$\int \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}\, dx = \frac{\operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2} + C$$$A