# Indefinite integration of general polynomial. Is this correct?

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.

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

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