1. Kanwar245

?

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

3. UnkleRhaukus

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

yes but when i do that it still seems ugly

5. UnkleRhaukus

do one half at a time

i shud make a u sub $u=1-x^7$

in the 1st one

8. UnkleRhaukus

i'd say so

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

the x^6 is a problem

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

12. UnkleRhaukus

hmm,

13. UnkleRhaukus

maybe that method isn't going to work

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'

15. UnkleRhaukus

where did you get this question?

*

19. shubhamsrg

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

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

21. shubhamsrg

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

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

23. SithsAndGiggles

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.

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

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

26. experimentX

this is a famous question on M.SE

27. experimentX