Integralen av $$$\sqrt{14} \sqrt{t}$$$

Bestäm $$$\int \sqrt{14} \sqrt{t}\, dt$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=\sqrt{14}$$$ och $$$f{\left(t \right)} = \sqrt{t}$$$:

$${\color{red}{\int{\sqrt{14} \sqrt{t} d t}}} = {\color{red}{\sqrt{14} \int{\sqrt{t} d t}}}$$

Tillämpa potensregeln $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=\frac{1}{2}$$$:

$$\sqrt{14} {\color{red}{\int{\sqrt{t} d t}}}=\sqrt{14} {\color{red}{\int{t^{\frac{1}{2}} d t}}}=\sqrt{14} {\color{red}{\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\sqrt{14} {\color{red}{\left(\frac{2 t^{\frac{3}{2}}}{3}\right)}}$$

Alltså,

$$\int{\sqrt{14} \sqrt{t} d t} = \frac{2 \sqrt{14} t^{\frac{3}{2}}}{3}$$

Lägg till integrationskonstanten:

$$\int{\sqrt{14} \sqrt{t} d t} = \frac{2 \sqrt{14} t^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{14} \sqrt{t}\, dt = \frac{2 \sqrt{14} t^{\frac{3}{2}}}{3} + C$$$A