Question

resolver la integral ∫ limite superior de 5 inferior 4 de x^2 √(x-4) dx

Answer 1

$\int\limits^5_4 { x^{2} \sqrt{x-4} } \, dx$

$u = x - 4$

$du = dx$

$x = u + 4$

si x = 4 → u = 0

si x = 5 → u = 1

$\int\limits^1_0 { (u+4)^{2} \sqrt{u} } \, du$

$\int\limits^1_0 { \sqrt{u} ( u^{2} + 8u + 16)} \, du$

$\int\limits^1_0 { (u^{ \frac{5}{2} }+8 u^{ \frac{3}{2} } + 16 u^{ \frac{1}{2} } ) } \, du$

$= \frac{2}{7} u^{ \frac{7}{2} } + \frac{16}{5} u^{ \frac{5}{2} } + \frac{32}{3} u^{ \frac{3}{2} }$

evaluado entre 0 y 1

$= \frac{2}{7} + \frac{16}{5} + \frac{32}{3} = \frac{1486}{105}$