TITLE: Why is $\int_0^{x}\eta \cdot g(\eta) d \eta > 0 \;\;\forall \;\;x\in \mathbb{R} \setminus \{ 0 \} $ assuming that $ g(x) > 0$ for all x?
QUESTION [1 upvotes]: $g(x_1)$ is a nonlinear function with $ g(x_1) > 0 \;\;\forall \;\;x_1\in \mathbb{R} \setminus \{ 0 \} $ and $ g(x_1 = 0) = 0 $.
I have read in "The Variable Gradient Method of Generating Liapunov Functions with Application to Automatic Control Systems by John E. Gibson" that
$\int_0^{x_1}\eta \cdot g(\eta) d \eta > 0 \;\;\forall \;\;x_1\in \mathbb{R} \setminus \{ 0 \} $
but I don't really understand why.
The argumentation in the book was:
"... Since the
nonlinearity was specified as $y = x\cdot g(x)$, $g(x)$ is always
positive if the nonlinearity lies in the first and third
quadrant, and, under these conditions, the integral involving $g(x_1)$ is always positive ..."
In my case the nonlinearity lies in the first and secound quadrant, but I just can't understand why hence the integral $\int_0^{x_1}\eta \cdot g(\eta) d \eta$ is always positive.
Any tip or answer is helpful, I am trying to understand why it is the case.

REPLY [1 votes]: A simple derivation is the following:
Make the change of variable $\eta=\theta x$ in the integral. Then the integral can be written as
$$\int_0^x{\eta g(\eta)d\eta}=x^2\int_0^1\theta g(\theta x)d\theta>0 \qquad \forall x\in\mathbb{R}\setminus\{0\}
$$
since $\int_0^1\theta g(\theta x)d\theta>0 $ due to $\theta g(\theta x)>0$ for all $x\in\mathbb{R}\setminus\{0\}$  and $\theta\in(0,1]$.
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