Funktion $$$\sqrt{2} x^{3} \left(x^{2} - 4\right)$$$ integraali

Määritä $$$\int \sqrt{2} x^{3} \left(x^{2} - 4\right)\, dx$$$.

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\sqrt{2}$$$ ja $$$f{\left(x \right)} = x^{3} \left(x^{2} - 4\right)$$$:

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

Expand the expression:

$$\sqrt{2} {\color{red}{\int{x^{3} \left(x^{2} - 4\right) d x}}} = \sqrt{2} {\color{red}{\int{\left(x^{5} - 4 x^{3}\right)d x}}}$$

Integroi termi kerrallaan:

$$\sqrt{2} {\color{red}{\int{\left(x^{5} - 4 x^{3}\right)d x}}} = \sqrt{2} {\color{red}{\left(- \int{4 x^{3} d x} + \int{x^{5} d x}\right)}}$$

Sovella potenssisääntöä $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=5$$$:

$$\sqrt{2} \left(- \int{4 x^{3} d x} + {\color{red}{\int{x^{5} d x}}}\right)=\sqrt{2} \left(- \int{4 x^{3} d x} + {\color{red}{\frac{x^{1 + 5}}{1 + 5}}}\right)=\sqrt{2} \left(- \int{4 x^{3} d x} + {\color{red}{\left(\frac{x^{6}}{6}\right)}}\right)$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=4$$$ ja $$$f{\left(x \right)} = x^{3}$$$:

$$\sqrt{2} \left(\frac{x^{6}}{6} - {\color{red}{\int{4 x^{3} d x}}}\right) = \sqrt{2} \left(\frac{x^{6}}{6} - {\color{red}{\left(4 \int{x^{3} d x}\right)}}\right)$$

Sovella potenssisääntöä $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=3$$$:

$$\sqrt{2} \left(\frac{x^{6}}{6} - 4 {\color{red}{\int{x^{3} d x}}}\right)=\sqrt{2} \left(\frac{x^{6}}{6} - 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}\right)=\sqrt{2} \left(\frac{x^{6}}{6} - 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}\right)$$

Näin ollen,

$$\int{\sqrt{2} x^{3} \left(x^{2} - 4\right) d x} = \sqrt{2} \left(\frac{x^{6}}{6} - x^{4}\right)$$

Sievennä:

$$\int{\sqrt{2} x^{3} \left(x^{2} - 4\right) d x} = \frac{\sqrt{2} x^{4} \left(x^{2} - 6\right)}{6}$$

Lisää integrointivakio:

$$\int{\sqrt{2} x^{3} \left(x^{2} - 4\right) d x} = \frac{\sqrt{2} x^{4} \left(x^{2} - 6\right)}{6}+C$$

$$$\int \sqrt{2} x^{3} \left(x^{2} - 4\right)\, dx = \frac{\sqrt{2} x^{4} \left(x^{2} - 6\right)}{6} + C$$$A