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. enter image description here

  • 3
    $\begingroup$ How does $(b-a)/n$ end up negative? Since $b>a$, it follows that $(b-a)/n>0$ for $n>0$. $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 20:59
  • $\begingroup$ yup you are absolutely right!, silly mistake on my side. $\endgroup$ – Joaquin Brandan Feb 20 '17 at 21:09

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 $f(x)$.

  • $\begingroup$ you are absolutely right, I made a silly mistake in my mind for some reason thinking that since i'm on the left side of the graph then $\delta x$ would be negative, thanks!. $\endgroup$ – Joaquin Brandan Feb 20 '17 at 21:06
  • $\begingroup$ Well, its certainly good to see yourself trying to prove these things. $\ddot{\stackrel{~~>}\smile}$ $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 21:07

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