Limites a partir da definição de Integração

This section should be pretty intuitive: just draw a picture and the result here will be "obvious".

Suppose $f(x)$ is bounded in the interval $[a,b]$; that is, for all $a\leq x\leq b$, $$m \leq f(x) \leq M$$ for some finite reals $m$ and $M$.

Then $m(b-a)\leq \int_{a}^{b}f(x)dx \leq M(b-a)$.

One way to "see" this is, as mentioned above, to draw a picture. Consider $f(x)=\sin(x)+3$ on $[0,\pi]$. Then clearly $m=3$ and $M=4$ will bound f on this interval (though of course the theorem holds for less tight bounds, too).

It should be quite apparent that $m(b-a)$, which measures the area of the green dotted rectangle, is less than the area under the curve from 0 to $\pi$.

Proving this isn't hard. In any of the sums by which we defined the integral, we can write that since $m\leq f(x_i)$ for all $i$, $$\sum m \Delta x \leq \sum f(x_i) \Delta x$$ where the $x_i$ are defined in whichever sum we chose to use.

The sum on the left, however, is just $m(b-a)$: the m can be factored out, as it is a constant, and then $\sum\Delta x$ is just the length of the interval, $b-a$. Then as we take the limit, the right hand side remains greater than or equal to $m(b-a)$; hence $$m(b-a) \leq \int_{a}^{b}f(x)dx.$$ A similar argument works for the claims about $M$.