Integral dari $$$\left(2 t - 1\right)^{2}$$$

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

Misalkan $$$u=2 t - 1$$$.

Kemudian $$$du=\left(2 t - 1\right)^{\prime }dt = 2 dt$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dt = \frac{du}{2}$$$.

Oleh karena itu,

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

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

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

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=2$$$:

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

Ingat bahwa $$$u=2 t - 1$$$:

$$\frac{{\color{red}{u}}^{3}}{6} = \frac{{\color{red}{\left(2 t - 1\right)}}^{3}}{6}$$

Oleh karena itu,

$$\int{\left(2 t - 1\right)^{2} d t} = \frac{\left(2 t - 1\right)^{3}}{6}$$

Tambahkan konstanta integrasi:

$$\int{\left(2 t - 1\right)^{2} d t} = \frac{\left(2 t - 1\right)^{3}}{6}+C$$

$$$\int \left(2 t - 1\right)^{2}\, dt = \frac{\left(2 t - 1\right)^{3}}{6} + C$$$A