Integralen av $$$y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)$$$ med avseende på $$$x$$$

Bestäm $$$\int y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)\, dx$$$.

Förenkla integranden:

$${\color{red}{\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x}}} = {\color{red}{\int{2 x y^{\frac{9}{2}} \left(2 x^{2} - y\right) d x}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=2 y^{\frac{9}{2}}$$$ och $$$f{\left(x \right)} = x \left(2 x^{2} - y\right)$$$:

$${\color{red}{\int{2 x y^{\frac{9}{2}} \left(2 x^{2} - y\right) d x}}} = {\color{red}{\left(2 y^{\frac{9}{2}} \int{x \left(2 x^{2} - y\right) d x}\right)}}$$

Låt $$$u=2 x^{2} - y$$$ vara.

Då $$$du=\left(2 x^{2} - y\right)^{\prime }dx = 4 x dx$$$ (stegen kan ses »), och vi har att $$$x dx = \frac{du}{4}$$$.

Alltså,

$$2 y^{\frac{9}{2}} {\color{red}{\int{x \left(2 x^{2} - y\right) d x}}} = 2 y^{\frac{9}{2}} {\color{red}{\int{\frac{u}{4} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{4}$$$ och $$$f{\left(u \right)} = u$$$:

$$2 y^{\frac{9}{2}} {\color{red}{\int{\frac{u}{4} d u}}} = 2 y^{\frac{9}{2}} {\color{red}{\left(\frac{\int{u d u}}{4}\right)}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

$$\frac{y^{\frac{9}{2}} {\color{red}{\int{u d u}}}}{2}=\frac{y^{\frac{9}{2}} {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=\frac{y^{\frac{9}{2}} {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$

Kom ihåg att $$$u=2 x^{2} - y$$$:

$$\frac{y^{\frac{9}{2}} {\color{red}{u}}^{2}}{4} = \frac{y^{\frac{9}{2}} {\color{red}{\left(2 x^{2} - y\right)}}^{2}}{4}$$

Alltså,

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(2 x^{2} - y\right)^{2}}{4}$$

Förenkla:

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4}$$

Lägg till integrationskonstanten:

$$\int{y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right) d x} = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4}+C$$

$$$\int y^{\frac{7}{2}} \left(4 x^{3} y - 2 x y^{2}\right)\, dx = \frac{y^{\frac{9}{2}} \left(- 2 x^{2} + y\right)^{2}}{4} + C$$$A