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Suppose we have a curved graph such as $y=x^2$ for $a< b< c$ and $a,b,c\neq 0$. It is not possible that the area from $0$ to $a$ can be equal to the area under the graph from $b$ to $c$. This is true for any graph of the form $x^n$ where $n>1$ and $n$ is a natural number. This is my conjecture. Please can anyone prove it. Which basically means that

$\int_0^a x^2\, dx\neq\int_b^c x^2\, dx$

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closed as off-topic by user99914, Claude Leibovici, Aqua, Shailesh, Nosrati Nov 23 '17 at 15:55

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  • $\begingroup$ You need to be using MathJax by now. For that reasons I am down-voting this question. Please reply when you have made this appropriately legible so I can undo my vote. $\endgroup$ – gen-z ready to perish Nov 23 '17 at 4:43
  • $\begingroup$ i am in 10th standard. need to study social science as well. i dont know mathjax $\endgroup$ – user167920 Nov 23 '17 at 4:44
  • $\begingroup$ I as well started using this site in the tenth grade. That didn’t stop me. $\endgroup$ – gen-z ready to perish Nov 23 '17 at 4:45
  • $\begingroup$ can you suggest a site to learn mathjax. it would be much appreceated $\endgroup$ – user167920 Nov 23 '17 at 4:45
  • $\begingroup$ Click here and it will take you to a Math Meta tutorial. It’s also linked in my first comment. $\endgroup$ – gen-z ready to perish Nov 23 '17 at 4:46
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If so $$\int\limits_{0}^ax^2dx=\int\limits_{b}^cx^2dx,$$ or $$\frac{a^3}{3}=\frac{c^3}{3}-\frac{b^3}{3},$$ or $$a^3+b^3=c^3,$$ which is impossible by the Fermat's Last Theorem (if you mean that $a$, $b$ and $c$ are naturals)

https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

If you can not mean that $a$, $b$ and $c$ are naturals then it's possible:

take $c=\sqrt[3]{a^3+b^3}.$

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  • $\begingroup$ well yes precisely. i know eventually it takes the form of fermats last theorem. i want a proof related to calculus not elliptic curves. $\endgroup$ – user167920 Nov 23 '17 at 4:42
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    $\begingroup$ @user167920 I calculated an integral. It's calculus, I think. $\endgroup$ – Michael Rozenberg Nov 23 '17 at 4:49
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    $\begingroup$ Why impossible? How about $a=1$, $b=2$, and $c=\sqrt[3]{9}$? Nowhere in the OP does it say that $a,b,c$ are natural numbers. $\endgroup$ – zipirovich Nov 23 '17 at 5:09
  • $\begingroup$ @zipirovich It means that $a$, $b$ and $c$ they are naturals. See the previous comment of user167920 about elliptic curves. $\endgroup$ – Michael Rozenberg Nov 23 '17 at 5:15
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    $\begingroup$ If $a,b,c$ are natural, then you're right. And I can see that you mentioned that in the parentheses. But the OP does NOT say that. (It only says that $n$ is a natural number.) So the OP's conjecture, as currently stated, is false. If they mean that $a,b,c$ are natural, then they should say so. $\endgroup$ – zipirovich Nov 23 '17 at 5:22

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