1
$\begingroup$

I was reading some notes of a guy I was tutoring the other day on basic calculus. He noted that if $$\int{x^n dx}=\frac{x^{n+1}}{n+1}+c,$$ then that can be extrapolated to all polynomials. He wrote, if $$p(x)=a_nx^n+a_{n-1}x ^{n-1}+\cdots+a_2x^2+a_1x+a_0,$$ then $$\int{p(x)dx}=\int{a_nx^n+a_{n-1}x ^{n-1}+\cdots+a_2x^2+a_1x+a_0}$$ $$=\int{\sum_{i=1}^n{a_ix^i}dx}=\sum_{i=0}^{n}{a_i\int{x^idx}}$$ $$=\sum_{i=0}^na_i \left(\frac{x^{i+1}}{i+1} \right)+c$$ $$=a_0 x+\frac{a_1x^2}{2}+\frac{a_2x^3}{3}+\cdots+\frac{a_nx^{n+1}}{n+1}+c, \space n\neq-1.$$

It looks to be correct, I was just checking if it is rigorously sound.

$\endgroup$
  • 1
    $\begingroup$ Yes, a sum of integrals equals an integral of the sum, so this is completely valid, just don't forget about the integration constant. $\endgroup$ – orion Jun 2 '14 at 8:23
  • $\begingroup$ Thank you very much. That was an accidental omission of $c$. $\endgroup$ – user124862 Jun 2 '14 at 8:23
  • 1
    $\begingroup$ Moreover you should have written $n\neq -1$... $\endgroup$ – sirfoga Jun 2 '14 at 8:51
  • $\begingroup$ You are the tutor? $\endgroup$ – Taladris Jun 3 '14 at 15:43
  • $\begingroup$ Yes, the lad I am tutoring is 10 years old ... $\endgroup$ – user124862 Jun 7 '14 at 2:12
2
$\begingroup$

This community wiki solution is intended to clear the question from the unanswered queue.


This proof looks very well written.

I think the key here is that $$\int \sum_{k = 0}^n a_kx^k = \sum_{k = 0}^n a_k \int x^k$$ and this follows from linearity of the integral.

If you had been dealing with infinite sums, i.e., series, you would need some added hypothesis to make that swap, but since you're only dealing with finite sums, you don't need to worry!

As @Foga mentioned, you should probably add $n \neq -1$ for the first statment, otherwise you'd be dividing by zero.

Also, kudos to the 10 year old for seeing this!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy