Lower bound for a specific integral Let $p(x)$ be a polynomial with real coefficients and $f:[a,b]\rightarrow\mathbb{R}$ be an integrable (possibly continuous) function. Is there a lower bound for $\int_{a}^{b}p(x)f(x)dx$?
I believe the answer is "yes" because $p(x)$ is continuous and assumes a max and a min. value on the compact set $[a,b]$. Call the minimum value $\alpha$ and knowing $\int_{a}^{b}f(x)dx=k$ and we could have: $\int_{a}^{b}p(x)f(x)dx\geq\alpha k$. I'm not entirely sure. Is the hypothesis "continuous" needed too?
edit: $p(x)$ is positive on $[a,b]$.
 A: I liked the question. I don't have an answer exactly how you want it, but I'll write my idea. 
I'll assume $f$ continuous.

Theorem 1. A function $g$ is Riemann integrable if only if the set of discontinuities of $g$ is a null set.

So, every continuous function is integrable. Thus, the function $g(x) = x^n$ is integrable.

Theorem 2. Let $f,g: [a,b] \to \mathbb{R}$ integrable functions, then $fg$ is integrable.

So, if $f$ is an integrable function and $g(x) = x^n$, $fx^n$ is integrable.
Now, the fuction $x^n: [a,b] \to \mathbb{R}$ is bounded, that is, $|x^n| \leq M_{n}$. Since $f$ is continuous in the compact $[a,b]$, $|f| \leq M$. Then $|fx^n| \leq MM_{n}$. Therefore
$$\left|\int fx^n\right| \leq \int|fx^n| \leq \int MM_{n} = MM_{n}(b - a)$$

Let $p$ be a polynomial of degree $n$, that is
$$p(x) = a_nx^n + \cdots a_1x^1 + a_0.$$
Then by linearity
$$\int fp = a_n\int fx^n + \cdots a_1\int fx + a_0\int f,$$
then
$$\left|\int fp\right| \leq M\underbrace{(a_{n}M_{n} + \cdots + a_{1}M_{1} + a_{0})}_{K}(b-a).$$
So,
$$\left|\int fp\right| \leq MK(b-a)$$
and so,
$$MK(a-b) \leq \int fp \leq MK(b-a)$$


EDIT. Consider $p(x) \geq 0$. For any $x \in [a,b]$, we have $\inf_{[a,b]}f \leq f(x) \leq \sup_{[a,b]}f$, then
$\inf_{[a,b]}f p(x) \leq f(x)p(x) \leq \sup_{[a,b]}f p(x)$. Thus
$$\inf_{[a,b]}f\int p(x) \leq \int f(x)p(x) \leq \sup_{[a,b]}f\int p(x).$$
Then, there is $d \in [\inf f, \sup f]$ such that
$$d\int p(x) = \int f(x)p(x).$$
Since $f$ is continuous, there is $c \in [a,b]$ such that $d = f(c)$. Therefore,
$$\int fp = f(c) \int p.$$
If $p(x) = x^{n}$, we have
$$\int fx^{n} = f(c) \int x^{n} = \frac{x^{n+1}}{n+1}.$$
Then,
$$\int fp = a_{n}f(c_{n})\frac{x^{n+1}}{n+1} + \cdots + a_{1}f(c_{1})\frac{x^{2}}{2} + a_{0}f(c_{0})x.$$
Take $K = \max\{|a_{n}f(c_{n})|\}$, then
$$\left|\int fp\right| \leq K\sum_{i=0}^{n}\left|\frac{x^{i+1}}{i+1}\right|$$
