$$$x$$$에 대한 $$$f x \left(x - 1\right)$$$의 적분

$$$\int f x \left(x - 1\right)\, dx$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=f$$$와 $$$f{\left(x \right)} = x \left(x - 1\right)$$$에 적용하세요:

$${\color{red}{\int{f x \left(x - 1\right) d x}}} = {\color{red}{f \int{x \left(x - 1\right) d x}}}$$

Expand the expression:

$$f {\color{red}{\int{x \left(x - 1\right) d x}}} = f {\color{red}{\int{\left(x^{2} - x\right)d x}}}$$

각 항별로 적분하십시오:

$$f {\color{red}{\int{\left(x^{2} - x\right)d x}}} = f {\color{red}{\left(- \int{x d x} + \int{x^{2} d x}\right)}}$$

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

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

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

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

따라서,

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

간단히 하시오:

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

적분 상수를 추가하세요:

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

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