TITLE: Find polynomials $p_n$ on $[0,1]$ with integral $=3$ and converges pointwisely to $0$
QUESTION [6 upvotes]: Prove that there  is a sequence of polynomials $\{p_{n}\}$ such that $p_{n} \to 0$ pointwise on $[0,1]$, but such that 
$$\int_{0}^{1}p_{n}(x)dx=3.$$

My thought: 
  I was thinking of working with a seuqence of functions that has pointwise limits to equal to $0$ but the integral to equal $3$. And then using the Weiestrass approximation theorem to say that there is a $p_n$ that would approximate the functions. I just can't think of the functions that would work.

REPLY [5 votes]: Take $p_n(x) = 3x^n(1-x)(n^2+3n+2)$. Each $\displaystyle\int_0^1p_n  = 3$ and $\displaystyle\lim_{n \to \infty} p_n(x) = 0$ for $x \in [0,1]$.