Answer
True
Work Step by Step
All the polynomials of degree n can be represented as:
$p_n(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ , $a_n\ne0$
Therefore, by reverse power rule, the antiderivative of $p_n(x)$ is given by:
$q(x)= \frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^{n}}{n}+...+\frac{a_1x^2}{2}+a_0x+C$ , C being the constant of integration.
$a_n\ne0$ , therefore $\frac{a_n}{n+1}\ne0$ and hence q(x) is a polynomial of degree n+1 as asserted.
