Integral von $$$\frac{1}{t^{102}}$$$

Bestimme $$$\int \frac{1}{t^{102}}\, dt$$$.

Wenden Sie die Potenzregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-102$$$ an:

$${\color{red}{\int{\frac{1}{t^{102}} d t}}}={\color{red}{\int{t^{-102} d t}}}={\color{red}{\frac{t^{-102 + 1}}{-102 + 1}}}={\color{red}{\left(- \frac{t^{-101}}{101}\right)}}={\color{red}{\left(- \frac{1}{101 t^{101}}\right)}}$$

Daher,

$$\int{\frac{1}{t^{102}} d t} = - \frac{1}{101 t^{101}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{t^{102}} d t} = - \frac{1}{101 t^{101}}+C$$

$$$\int \frac{1}{t^{102}}\, dt = - \frac{1}{101 t^{101}} + C$$$A