$$$14 - 4 r^{2}$$$ 的積分
您的輸入
求$$$\int \left(14 - 4 r^{2}\right)\, dr$$$。
解答
逐項積分:
$${\color{red}{\int{\left(14 - 4 r^{2}\right)d r}}} = {\color{red}{\left(\int{14 d r} - \int{4 r^{2} d r}\right)}}$$
配合 $$$c=14$$$,應用常數法則 $$$\int c\, dr = c r$$$:
$$- \int{4 r^{2} d r} + {\color{red}{\int{14 d r}}} = - \int{4 r^{2} d r} + {\color{red}{\left(14 r\right)}}$$
套用常數倍法則 $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$,使用 $$$c=4$$$ 與 $$$f{\left(r \right)} = r^{2}$$$:
$$14 r - {\color{red}{\int{4 r^{2} d r}}} = 14 r - {\color{red}{\left(4 \int{r^{2} d r}\right)}}$$
套用冪次法則 $$$\int r^{n}\, dr = \frac{r^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,以 $$$n=2$$$:
$$14 r - 4 {\color{red}{\int{r^{2} d r}}}=14 r - 4 {\color{red}{\frac{r^{1 + 2}}{1 + 2}}}=14 r - 4 {\color{red}{\left(\frac{r^{3}}{3}\right)}}$$
因此,
$$\int{\left(14 - 4 r^{2}\right)d r} = - \frac{4 r^{3}}{3} + 14 r$$
加上積分常數:
$$\int{\left(14 - 4 r^{2}\right)d r} = - \frac{4 r^{3}}{3} + 14 r+C$$
答案
$$$\int \left(14 - 4 r^{2}\right)\, dr = \left(- \frac{4 r^{3}}{3} + 14 r\right) + C$$$A