Integral dari $$$- b^{2} t + \frac{1}{a^{2}}$$$ terhadap $$$t$$$

Temukan $$$\int \left(- b^{2} t + \frac{1}{a^{2}}\right)\, dt$$$.

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dt = c t$$$ dengan $$$c=\frac{1}{a^{2}}$$$:

$$- \int{b^{2} t d t} + {\color{red}{\int{\frac{1}{a^{2}} d t}}} = - \int{b^{2} t d t} + {\color{red}{\frac{t}{a^{2}}}}$$

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

$$- {\color{red}{\int{b^{2} t d t}}} + \frac{t}{a^{2}} = - {\color{red}{b^{2} \int{t d t}}} + \frac{t}{a^{2}}$$

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

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

Oleh karena itu,

$$\int{\left(- b^{2} t + \frac{1}{a^{2}}\right)d t} = - \frac{b^{2} t^{2}}{2} + \frac{t}{a^{2}}$$

Tambahkan konstanta integrasi:

$$\int{\left(- b^{2} t + \frac{1}{a^{2}}\right)d t} = - \frac{b^{2} t^{2}}{2} + \frac{t}{a^{2}}+C$$

$$$\int \left(- b^{2} t + \frac{1}{a^{2}}\right)\, dt = \left(- \frac{b^{2} t^{2}}{2} + \frac{t}{a^{2}}\right) + C$$$A