I don't understand why a definite integral of a function in the top left quadrant of a graph is positive and one defined in the bottom left is negative.
there must be an error on my reasoning, this is what i think:
Using the Riemann sum to define integrals: $$ \int _{a}^b f(x) dx = \lim _{x \to \infty} \sum_{i=0}^\infty f(x_i) \frac{(b-a)}{n}$$ where $$b>a$$
if a function were to be on the top left quadrant, where $f(x)$ is somewhere between $0$ and $\infty$ and $x$ is between $-\infty$ and $0$, meaning that:
$f(x_i)$ is positive
$\frac{(b-a)}{n}$ is negative, because $b>a$ and $n>0$
therefore the multiplication will result in a negative number
$$ f(x_i) \frac{(b-a)}{n} < 0$$
similarly with the bottom left quadrant using the same logic both terms are negative $ f(x_i)<0$ and $ \frac{(b-a)}{n} <0$, therefore, multiplying both would make a positive result.
why is this wrong?
here is a picture in case my explanation was poor.
You state that $\frac{b-a}n$ is negative since $b>a$, but, as you will notice, this is false:
$$b>a\iff b-a>0\iff\frac{b-a}n>\frac0n=0$$
Thus, $\frac{b-a}n$ is always positive, and the sign becomes only affected by $f(x)$.