Question

someone know how to do this integral?

Answer 1

\[\int\limits_{0}^{9}\frac{ 1 }{ \sqrt[3]{x-1} }dx\]

Answer 2

did you try using a u substution,
u = x-1
?

Answer 3

^^ that will do it

Answer 4

would then be \[\int\limits_{0}^{9}\frac{ 1 }{\sqrt[3]{u}}du\]

Answer 5

yep then change it to u^(-1/3)

Answer 6

don't forget to change the limits
when x= 0, u =...
when x = 9, u =... ?

Answer 7

\[u ^{-\frac{ 4 }{3 }}\]

Answer 8

@felavin, that is derivative ... for integrating you add 1 to exponent

Answer 9

when x=0 u=1 and when x=9 does u=8^-1/3

Answer 10

u = x-1

when x = 0
u = 0-1 = -1

when x =9
u = 9-1 =8

Answer 11

does it become \[\int\limits_{-1}^{8}\frac{ 3 }{ 2}u ^{\frac{ 2 }{ 3}}du\]

Answer 12

since you have already done the integration of u^(-1/2) , no need to put integral sign again

\(\large [\frac{ 3 }{ 2}u ^{\frac{ 2 }{ 3}}]^8_{-1}\)

plug in the limits

Answer 13

of u^(-1/3) ****

Answer 14

for my answer i get 61/2 is that the answer?

Answer 15

nopes, how u get that ?

Answer 16

is it 67/2?

Answer 17

\(\large [\frac{ 3 }{ 2}u ^{\frac{ 2 }{ 3}}]^8_{-1} \\ \large =\large [\frac{ 3 }{ 2}8 ^{\frac{ 2 }{ 3}}] - \large [\frac{ 3 }{ 2}(-1) ^{\frac{ 2 }{ 3}}] \\ \large = \dfrac{3}{2}[4-1] =...\)

Answer 18

@hartnn 's work looks correct