Question

show that the integral is 0

Answer 1

\[\large \color{green}{\int_0^1\sqrt[m]{1-x^n}-\sqrt[n]{1-x^m}dx} \]

Answer 2

\[m<n\]

Answer 3

the original question is \[m=3,n=7\]

Answer 4

i think you could start like this\[\int\limits_{0}^{1}(1-x ^{7})^{1/3}dx - \int\limits_{0}^{1}(1-x ^{3})^{1/7}dx\]
then you'll get
\[ \left[ \frac{ (1-x ^{7})^{4/3} }{ 4/3}\frac{ 1 }{ -7x ^{6} }\right]_{0}^{1}\] - \[ \left[ \frac{ (1-x ^{3})^{8/7} }{ 8/7}\frac{ 1 }{ -3x ^{2} }\right]_{0}^{1}\]

i dont know if this is right.... but i think you should try changing the limits before integrating because if you don't, you'll get a zero in the denominator.... which is like a mathematical catastrophe :O

Answer 5

i know there was a way to change the limits before integrating it... i 'll try to remember it...

Answer 6

oh you can do this,
change the limits according to the property\[\int\limits_{-a}^{a} f(x) dx = 2 \int\limits_{0}^{a} f(x) dx \]
therefore,

\[\int\limits_{0}^{a} f(x) dx = \frac{ 1 }{ 2 } \int\limits_{-a}^{a} f(x) dx \]

then you can integrate it the way i did and substitute -1 and 1 as the limits

Answer 7

doing that will get you zero.... probably..

Answer 8

make a substitution
\[y=\sqrt[m]{1-x^n},y'=-\frac{n}{m}(1-x^n)^{\frac{1}{m}-1}=-\frac{n}{m}y^{1-m}dx\]

\[\int _0^1\sqrt[n]{1-x^m}dx=-\frac{m}{n}\int_1^0 y(y^{m-1})dy=-\frac{m}{n}\int_1^0 y^mdy =\frac{m}{n}\int_0^1 y^mdy \]
by symmetry

\[I=\frac{n}{m} \int_0^1 y^m-\frac{m}{n}\int_0^1 y^n dy\]
\[=\frac{n}{m(m+1)}y^{m+1}-\frac{n}{m(n+1)} y^{n+1}\Large |_0^1\\=\frac{n}{m(m+1)}-\frac{m}{n(n+1)}\]

Answer 9

we also notice that \[y^n=1-x^m,y^m=1-x^n\\\implies \sqrt[n]{1-x^m}=\sqrt[m]{1-x^n}\]

Answer 10

whoa..... okayyyy. i just discovered im a real n00b when it comes to integration.. :)

Answer 11

i cus say the above is enough to prove the statement since if \[f(x)=\sqrt[m]{1-x^n},f^{-1}(x)=\sqrt[n]{1-x^m}\]
so maybe we should consider\[\int f(x)-\int f^{-1}(x)\]

Answer 12

but wierd thing is it doesnt give zero but some