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

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


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Claude Leibovici, Aqua, Shailesh, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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)


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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @user167920 I calculated an integral. It's calculus, I think. $\endgroup$ – Michael Rozenberg Nov 23 '17 at 4:49
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.