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Jonask
integration- radicals
\[\color{blue}{\huge\int_0^1\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}dx}\]
\[\color{blue}{\int_0^1\sqrt[3]{1-x^7}dx-\int\sqrt[7]{1-x^3}dx}\]
yes but when i do that it still seems ugly
do one half at a time
i shud make a u sub \[u=1-x^7\]
\[u=1-x^7\] \[\int\frac{u^{1/3}}{7x^6}du\]
\[\int\frac{u^{1/3}}{7(1-u)^{1/7}}\]
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'
where did you get this question?
steward calculus advanved problem
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)
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
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
\[\int_0^1 f(x)-f^{-1}(x) \space dx\] \[f(1)=f^{-1}(1)=0\]
this is a famous question on M.SE