1
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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 $f(x)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.