Question

I remember this question that has been bugging me for a long time.
Evaluate \(\Large \int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}\ dx\). The answer's supposed to be 0, but I don't know how it ever came to this. Any ideas?

Answer 1

i wish they were bigger :( i cant read the indeces boohoo

Answer 2

oh they're scary :/ im sure the trick is in the u

Answer 3

looks like this can only be evaluated using hypergeometric functions
http://mathworld.wolfram.com/HypergeometricFunction.html
thus it can't be integrated by normal means
of course since you are looking at definite integral , there are ways to approximate area under curve to verify result is zero

Answer 4

Yeah, well, that wasn't discussed in the book where I found the problem, so I doubt I'd think of that.
I think some sort of symmetry is at play here. I just don't know what.

Answer 5

\[\int_0^1 \sqrt[3]{1- (1-x)^7} - \sqrt[7]{1-(1-x)^3} \]1 cancels out.

Answer 6

But you'd have to expand (1-x)^7 and (1-x)^3. :(

Answer 7

At 0 both the functions must be Zero, right?

Answer 8

Yeah, then...?

Answer 9

http://www.wolframalpha.com/input/?i=%281-x^7%29^%281%2F3%29-%281-x^3%29^%281%2F7%29+from+0+to+1

Answer 10

@blockcolder , You will cry(how easy the problem is) after seeing the solution of the problem. Do you want it ?? :P

Answer 11

Just give me a hint. I'll try to work it out on my own.

Answer 12

Ok.

\[\int\limits_{}^{}f(p)dp = \int\limits_{}^{}f(x)dx\]

That's all :D

Answer 13

You mean I gotta prove that
\[\Large \int_0^1 \sqrt[3]{1-x^7}\ dx=\int_0^1 \sqrt[7]{1-x^3}\ dx\]
??

Answer 14

I think he meant change of variable.

Answer 15

@Ishaan94 , yes

Answer 16

x=-u? does it helps?

Answer 17

Nope

Answer 18

Ok. You guys have 5 min :P After that I will give the solution :)

Answer 19

I ran out of tricks. I give up.

Answer 20

\[\large \sqrt[7]{1-x^3}= p\]
\[\large {1-x^3}= p^7\]
\[\large {x^3}= 1-p^7\]
\[\large {x}= \sqrt[3]{1-p^7}\]
Now using the property
\[\int\limits\limits\limits_{}^{}f(p)dp = \int\limits\limits\limits_{}^{}f(x)dx\]
we get
\[\large \int\limits\limits\limits_{0}^{1}\sqrt[7]{1-x^3}= \int\limits\limits\limits_{0}^{1}\sqrt[3]{1-x^7}\]

Substitute this in your question and get your answer :D

Answer 21

Well, do tell me if there are any flaws in my solution. I always tend to make mistakes :)

Answer 22

You'd have to be assuming that x=p for the penultimate line, but how'd you prove that?

Answer 23

Ok. Let me say it this way
\[\large \int\limits_{0}^{1}\sqrt[3]{1-p^7} \space dp= \int\limits_{0}^{1}\sqrt[3]{1-x^7}\space dx\]

Answer 24

Then? After this?

Answer 25

Trust me bro. My brain is having some blockage now :P. I am not able to find the right reasoning but this seems to me the way to go. What do you guys think?

Answer 26

\[\sqrt[3]{1-x^7} = p\]This is how you started, \(1 - x^7 = p^3 \implies x = \sqrt[7]{1-p^3}\)
And it stresses on some kind of symmetry I can not see.
\[\int f(p)dp = \int f(x) dx\]It's fine up till now but after this I really don't understand what you did. I am having a headache now, can't even concentrate on the problem.

Answer 27

And also your solution kinda proves it for nearly every interval, while I am pretty sure the answer is Zero only the in restricted domain from 0 to 1. Correct me If I'm wrong but I don't think you considered the intervals while attempting the problem?

Answer 28

Yep. So it seems I am wrong alltogether. :(

Answer 29

I have posted the question on math.se. I had to. The question is really twisted.

http://math.stackexchange.com/questions/139393/problem-integration

Answer 30

@Ishaan94 LOL, I have never seen this kind of proof before :D

Answer 31

Me neither shivam. I will have to understand it and I am quite sure understanding it is only gonna severe my headache.

Answer 32

But it is right. Only problem with the last step. Rest all I got it :)

Answer 33

My above statement was in reference to the solution link given by @Ishaan94 . Just for clarification :D

Answer 34

Hmm the inequality approach is really nice.

Answer 35

I could have tried Integration by parts :/ hmm

Answer 36

Well, sorry for offtopic but trust me I was not able to answer even a single question asked on the site which Ishaan94 linked me to :/

Answer 37

Hmm don't worry shivam, even FoolForMath doesn't answer much. Most of the guys on math.stackexchange are academicians, and those of our age (around 16-17-18) are IMO kids.