Question

integration- radicals

Answer 1

?

Answer 2

\[\color{blue}{\huge\int_0^1\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}dx}\]

Answer 3

\[\color{blue}{\int_0^1\sqrt[3]{1-x^7}dx-\int\sqrt[7]{1-x^3}dx}\]

Answer 4

yes but when i do that it still seems ugly

Answer 5

do one half at a time

Answer 6

i shud make a u sub
\[u=1-x^7\]

Answer 7

in the 1st one

Answer 8

i'd say so

Answer 9

\[u=1-x^7\]
\[\int\frac{u^{1/3}}{7x^6}du\]

Answer 10

the x^6 is a problem

Answer 11

\[\int\frac{u^{1/3}}{7(1-u)^{1/7}}\]

Answer 12

hmm,

Answer 13

maybe that method isn't going to work

Answer 14

http://www.wolframalpha.com/input/?i=integral+%281-x%5E7%29%5E%7B1%2F3%7D-%281-x%5E3%29%5E%7B1%2F7%7D ugly as well
but i think we are supposed to use 'by parts'

Answer 15

where did you get this question?

Answer 16

http://books.google.com.tr/books?id=v4-1Rk7uaEcC&pg=PA618&hl=tr&source=gbs_toc_r&cad=4#v=onepage&q&f=false

Answer 17

steward calculus advanved problem

Answer 18

*

Answer 19

what if we use the property
integral ( f(x) from x=a to b = f(a+b-x) from x= a to b )?

Answer 20

in this case a+b=0+1
int f(1-x)=f(x)

Answer 21

yep, am not too sure if that'll work though, but probably should work ?

Answer 22

ok i have to go i;ll try it thanks to everyone for the effort

Answer 23

According to WA, the definite integral is 0: http://www.wolframalpha.com/input/?i=integrate+%28%281-x%5E7%29%5E%281%2F3%29+-+%281-x%5E3%29%5E%281%2F7%29%29+over+%5B0%2C1%5D

Maybe there's some detail we're not getting about the function itself.

Answer 24

interesting result maybe this property is because of the function is actually an inverse of itself i mean
\[f(x)=\sqrt[3]{1-x^7}\]
then
\[f^{-1}(x)=\sqrt[7]{1-x^3}\]
should this mean anything

Answer 25

\[\int_0^1 f(x)-f^{-1}(x) \space dx\]
\[f(1)=f^{-1}(1)=0\]

Answer 26

this is a famous question on M.SE

Answer 27

http://math.stackexchange.com/questions/139393/how-to-show-that-int-01-left-sqrt31-x7-sqrt71-x3-right-dx/139398#139398