Integral dari $$$\frac{2^{- t}}{5}$$$

Temukan $$$\int \frac{2^{- t}}{5}\, dt$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ dengan $$$c=\frac{1}{5}$$$ dan $$$f{\left(t \right)} = 2^{- t}$$$:

$${\color{red}{\int{\frac{2^{- t}}{5} d t}}} = {\color{red}{\left(\frac{\int{2^{- t} d t}}{5}\right)}}$$

Misalkan $$$u=- t$$$.

Kemudian $$$du=\left(- t\right)^{\prime }dt = - dt$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dt = - du$$$.

Integral tersebut dapat ditulis ulang sebagai

$$\frac{{\color{red}{\int{2^{- t} d t}}}}{5} = \frac{{\color{red}{\int{\left(- 2^{u}\right)d u}}}}{5}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = 2^{u}$$$:

$$\frac{{\color{red}{\int{\left(- 2^{u}\right)d u}}}}{5} = \frac{{\color{red}{\left(- \int{2^{u} d u}\right)}}}{5}$$

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

$$- \frac{{\color{red}{\int{2^{u} d u}}}}{5} = - \frac{{\color{red}{\frac{2^{u}}{\ln{\left(2 \right)}}}}}{5}$$

Ingat bahwa $$$u=- t$$$:

$$- \frac{2^{{\color{red}{u}}}}{5 \ln{\left(2 \right)}} = - \frac{2^{{\color{red}{\left(- t\right)}}}}{5 \ln{\left(2 \right)}}$$

Oleh karena itu,

$$\int{\frac{2^{- t}}{5} d t} = - \frac{2^{- t}}{5 \ln{\left(2 \right)}}$$

Tambahkan konstanta integrasi:

$$\int{\frac{2^{- t}}{5} d t} = - \frac{2^{- t}}{5 \ln{\left(2 \right)}}+C$$

$$$\int \frac{2^{- t}}{5}\, dt = - \frac{2^{- t}}{5 \ln\left(2\right)} + C$$$A