$$$10 \sqrt{b} - b^{2}$$$의 적분

$$$\int \left(10 \sqrt{b} - b^{2}\right)\, db$$$을(를) 구하시오.

각 항별로 적분하십시오:

$${\color{red}{\int{\left(10 \sqrt{b} - b^{2}\right)d b}}} = {\color{red}{\left(\int{10 \sqrt{b} d b} - \int{b^{2} d b}\right)}}$$

멱법칙($$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=2$$$에 적용합니다:

$$\int{10 \sqrt{b} d b} - {\color{red}{\int{b^{2} d b}}}=\int{10 \sqrt{b} d b} - {\color{red}{\frac{b^{1 + 2}}{1 + 2}}}=\int{10 \sqrt{b} d b} - {\color{red}{\left(\frac{b^{3}}{3}\right)}}$$

상수배 법칙 $$$\int c f{\left(b \right)}\, db = c \int f{\left(b \right)}\, db$$$을 $$$c=10$$$와 $$$f{\left(b \right)} = \sqrt{b}$$$에 적용하세요:

$$- \frac{b^{3}}{3} + {\color{red}{\int{10 \sqrt{b} d b}}} = - \frac{b^{3}}{3} + {\color{red}{\left(10 \int{\sqrt{b} d b}\right)}}$$

멱법칙($$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=\frac{1}{2}$$$에 적용합니다:

$$- \frac{b^{3}}{3} + 10 {\color{red}{\int{\sqrt{b} d b}}}=- \frac{b^{3}}{3} + 10 {\color{red}{\int{b^{\frac{1}{2}} d b}}}=- \frac{b^{3}}{3} + 10 {\color{red}{\frac{b^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- \frac{b^{3}}{3} + 10 {\color{red}{\left(\frac{2 b^{\frac{3}{2}}}{3}\right)}}$$

따라서,

$$\int{\left(10 \sqrt{b} - b^{2}\right)d b} = \frac{20 b^{\frac{3}{2}}}{3} - \frac{b^{3}}{3}$$

적분 상수를 추가하세요:

$$\int{\left(10 \sqrt{b} - b^{2}\right)d b} = \frac{20 b^{\frac{3}{2}}}{3} - \frac{b^{3}}{3}+C$$

$$$\int \left(10 \sqrt{b} - b^{2}\right)\, db = \left(\frac{20 b^{\frac{3}{2}}}{3} - \frac{b^{3}}{3}\right) + C$$$A