$$$x^{2} \left(x - 1\right)$$$'nin integrali

Bulun: $$$\int x^{2} \left(x - 1\right)\, dx$$$.

Expand the expression:

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

Her terimin integralini alın:

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

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=3$$$ ile uygulayın:

$$- \int{x^{2} d x} + {\color{red}{\int{x^{3} d x}}}=- \int{x^{2} d x} + {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \int{x^{2} d x} + {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Kuvvet kuralını $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=2$$$ ile uygulayın:

$$\frac{x^{4}}{4} - {\color{red}{\int{x^{2} d x}}}=\frac{x^{4}}{4} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{4}}{4} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Dolayısıyla,

$$\int{x^{2} \left(x - 1\right) d x} = \frac{x^{4}}{4} - \frac{x^{3}}{3}$$

Sadeleştirin:

$$\int{x^{2} \left(x - 1\right) d x} = \frac{x^{3} \left(3 x - 4\right)}{12}$$

İntegrasyon sabitini ekleyin:

$$\int{x^{2} \left(x - 1\right) d x} = \frac{x^{3} \left(3 x - 4\right)}{12}+C$$

$$$\int x^{2} \left(x - 1\right)\, dx = \frac{x^{3} \left(3 x - 4\right)}{12} + C$$$A