Integral von $$$9 e^{\frac{t}{2}}$$$

Bestimme $$$\int 9 e^{\frac{t}{2}}\, dt$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=9$$$ und $$$f{\left(t \right)} = e^{\frac{t}{2}}$$$ an:

$${\color{red}{\int{9 e^{\frac{t}{2}} d t}}} = {\color{red}{\left(9 \int{e^{\frac{t}{2}} d t}\right)}}$$

Sei $$$u=\frac{t}{2}$$$.

Dann $$$du=\left(\frac{t}{2}\right)^{\prime }dt = \frac{dt}{2}$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = 2 du$$$.

Somit,

$$9 {\color{red}{\int{e^{\frac{t}{2}} d t}}} = 9 {\color{red}{\int{2 e^{u} d u}}}$$

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

$$9 {\color{red}{\int{2 e^{u} d u}}} = 9 {\color{red}{\left(2 \int{e^{u} d u}\right)}}$$

Das Integral der Exponentialfunktion lautet $$$\int{e^{u} d u} = e^{u}$$$:

$$18 {\color{red}{\int{e^{u} d u}}} = 18 {\color{red}{e^{u}}}$$

Zur Erinnerung: $$$u=\frac{t}{2}$$$:

$$18 e^{{\color{red}{u}}} = 18 e^{{\color{red}{\left(\frac{t}{2}\right)}}}$$

Daher,

$$\int{9 e^{\frac{t}{2}} d t} = 18 e^{\frac{t}{2}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{9 e^{\frac{t}{2}} d t} = 18 e^{\frac{t}{2}}+C$$

$$$\int 9 e^{\frac{t}{2}}\, dt = 18 e^{\frac{t}{2}} + C$$$A