Integral dari $$$2 \sqrt{2} t$$$

Temukan $$$\int 2 \sqrt{2} t\, dt$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ dengan $$$c=2 \sqrt{2}$$$ dan $$$f{\left(t \right)} = t$$$:

$${\color{red}{\int{2 \sqrt{2} t d t}}} = {\color{red}{\left(2 \sqrt{2} \int{t d t}\right)}}$$

Terapkan aturan pangkat $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=1$$$:

$$2 \sqrt{2} {\color{red}{\int{t d t}}}=2 \sqrt{2} {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=2 \sqrt{2} {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$

Oleh karena itu,

$$\int{2 \sqrt{2} t d t} = \sqrt{2} t^{2}$$

Tambahkan konstanta integrasi:

$$\int{2 \sqrt{2} t d t} = \sqrt{2} t^{2}+C$$

$$$\int 2 \sqrt{2} t\, dt = \sqrt{2} t^{2} + C$$$A