Integral von $$$2^{- x}$$$

Bestimme $$$\int 2^{- x}\, dx$$$.

Sei $$$u=- x$$$.

Dann $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = - du$$$.

Somit,

$${\color{red}{\int{2^{- x} d x}}} = {\color{red}{\int{\left(- 2^{u}\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = 2^{u}$$$ an:

$${\color{red}{\int{\left(- 2^{u}\right)d u}}} = {\color{red}{\left(- \int{2^{u} d u}\right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=2$$$:

$$- {\color{red}{\int{2^{u} d u}}} = - {\color{red}{\frac{2^{u}}{\ln{\left(2 \right)}}}}$$

Zur Erinnerung: $$$u=- x$$$:

$$- \frac{2^{{\color{red}{u}}}}{\ln{\left(2 \right)}} = - \frac{2^{{\color{red}{\left(- x\right)}}}}{\ln{\left(2 \right)}}$$

Daher,

$$\int{2^{- x} d x} = - \frac{2^{- x}}{\ln{\left(2 \right)}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{2^{- x} d x} = - \frac{2^{- x}}{\ln{\left(2 \right)}}+C$$

$$$\int 2^{- x}\, dx = - \frac{2^{- x}}{\ln\left(2\right)} + C$$$A