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Am I right to assume that

$$\bar{y} = \frac{{\int_{x_0}^{x_1} f(x) dx - \int_{x_0}^{x_1} g(x) dx}} {x_1 - x_0}$$ where f(x) and g(x) are two polinomials, not crossing each other in this range, and y bar is the average height of the area between the two curves? If not, why? What would be the average height then?

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    Assuming $f(x) \geq g(x)$ for $x\in[x_0,x_1]$ then yes. – 2017-01-31
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    @Zubzub can $\bar{y}$ be negative.? – 2017-01-31
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    Well the goal is to ensure that either $f$ is above $g$ over the entire interval or it is entirely below in which case $\bar{y}$ will be negative but its absolute value will still represents the average height. You can more or less ignore the sign of $\bar{y}$ as long as you ensure that $f$ and $g$ don't cross in $[x_0,x_1]$. – 2017-01-31

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We assume that $f \ge g$ on $[x_0,x_1]$. Then $\int_{x_0}^{x_1} f(x) dx - \int_{x_0}^{x_1} g(x) dx=\int_{x_0}^{x_1} (f(x)-g(x)) dx $ is the area of

$A=\{(x,y) \in \mathbb R^2:x \in [x_0,x_1], g(x) \le y \le f(x)\}$

By the Mean value theorem, there is $ \xi \in [x_0,x_1]$ such that

$$\bar{y}=f(\xi)-g(\xi).$$

Therefore the area of $A$ is the area of the rectangle with the side lengths $x_1-x_0 $ and $f(\xi)-g(\xi).$