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

$g(x_1)$ 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.

Answer 1

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 $\theta\in(0,1]$.