Respuesta :
Answer:
[tex] \int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C[/tex]
Step-by-step explanation:
For this case we need to find the following integral:
[tex] \int x(8-x)^{3/2}dx[/tex]
And for this case we can use the substitution [tex] u = 8-x [/tex] from here we see that [tex] du = -dx[/tex], and if we solve for x we got [tex] x = 8-u[/tex], so then we can rewrite the integral like this:
[tex] \int x(8-x)^{3/2}dx= \int (8-u) u^{3/2} (-du)[/tex]
And if we distribute the exponents we have this:
[tex] \int x(8-x)^{3/2}dx= - \int 8 u^{3/2} + \int u^{5/2} du[/tex]
Now we can do the integrals one by one:
[tex] \int x(8-x)^{3/2}dx= -8 \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{7/2}}{\frac{7}{2}} +C[/tex]
And reordering the terms we have"
[tex] \int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C[/tex]
And rewriting in terms of x we got:
[tex] \int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C[/tex]
And that would be our final answer.
