Approximating an integral with rectangles? Legal? I've come across this while navigating on $\text{WiKiPeDiA}$:
$\sum_{i=1}^{n}f(t_i)\Delta_i$ Basically we can illustrate that as: 

This means that we approximate curved functions with rectangles, same approach done by Leibiniz to introduce the concept of integral. But I'm asking, is it legal? I mean we are just approximating it, if we draw rectangles till infinity then we wouldn't have the exact answer because the graph is curved, it is not smooth. This approach reminds me of the famous "Proof Pi=4" 

Will we know that it is wrong and that Pi=3.14... Is the same for calculus, and why?
In response to some comments, let $C$ be a circle of radius $r$, we can divide (as in the picture) into 4. You get it if you see the picture 

 A: Yes. This is "legal" for continuous curves. 
The heart of the issue is whether your approximation is actually converging to the object in question or not.
For area under a curve, one can show that the Riemann sums (sums of areas of rectangles) better and better approximate the area under the curve as the number of rectangles increases. Thus -- Voila! -- the limit gives us the area under the curve (in fact we may define area in this way). 
On the other hand, the $\pi=4$ false proof tricks you into thinking the approximation is approaching the circle when it really isn't -- or at least its arc length isn't. 
The curve you are using to approximate the circle is getting more and more jagged - not smoother and smoother. This kind of indicates that it's a poor approximation of a circle.
Or from a another viewpoint, the square containing the circle gives an overestimate for the circumference. Similarly one could inscribe a square (or diamond) and get an underestimate of $2\sqrt{2}$. As you continue to "flip corners over" the estimates remain constant revealing that your "approximations" aren't going anywhere. 
You must be careful anytime you send things off to infinity. Finitely many steps don't necessarily predict the ultimate outcome. 
Edit: More on why Riemann sums converge to the area...
This can be better seen using Darboux integrals. Darboux uses a set of rectangles that over-estimate the area and then a second set of rectangles that under-estimate the area. As we use more and more rectangles, the upper bound keeps on creeping down and the lower bound keeps on creeping up. For nice enough functions (such as continuous functions) these two numbers approach each other. So these approximations are getting better and better as the number of rectangles increases.
In addition one can prove that any choice of rectangles (as long as their widths are heading to zero) will tend to this same number (mostly because it's squeezed between these bounds).
For the squaring the circle $\pi=4$ false proof the estimate never gets better.
A: The question is getting rather broad, but let's see if I can answer a part of it.
One answer to the question of why the area under a continuous function is a limit of Riemann sums is that we define it that way.  But for that to make sense, we need the see that the limit exists, and also that this definition corresponds with our physical intuitions about area.
To see this informally, notice that any lower Riemann sum (e.g. the area of the red region) gives an underestimate of the area under the curve, and similarly an upper Riemann sum (e.g. the area of the green region) gives an overestimate.  This is just using the intuitive fact that the area of a subset is at most the area of the entire set.
(Here we are using a signed notion of "area" in which areas below the $x$-axis count as negative.)
Therefore the true area, if such a thing exists, must be at least the value of any lower Riemann sum and at most the value of any upper Riemann sum. 
That is, an upper sum gives an upper bound and a lower sum gives a lower bound.
Considering any pair consisting of a lower sum and upper sum, e.g. red and green above, does not pin down the true area entirely, but only gives us some "approximation interval" in which it must lie if it exists.
By using thinner rectangles in our upper and lower sums, we can get a smaller upper bound and a larger lower bound, therefore giving us a smaller approximation interval.  In fact, by using sufficiently thin rectangles we can make this interval as small as desired.  (This is not entirely obvious, and you shoudn't take my word for it, but perhaps consult an analysis textbook instead.)
This is the step where we use continuity of the function, and indeed if it were not continuous then all of the upper sums could be much larger than all of the lower sums, so that our "approximations" are not approximating anything in particular.
Now the notion of limit comes in.  Because the length of the approximation intervals is approaching zero as the width of the rectangles approaches zero, the completeness property of the real numbers implies that there is a (unique) real number that is contained in all of the approximation intervals—that is, it is above all the lower bounds given by lower Riemann sums and below all the upper bounds given by upper Riemann sums.  The true area under the curve is defined to be (and intuitively must be) this number.

Why doesn't a similar argument work for arc length, you might ask?
Well, the reason is just that there isn't a similar argument for arc length.  If you try to make a similar argument, then I will try to refute it. In the meantime all I can say is that the particular argument of the comic doesn't work, because there is no reason to think that the "approximations" are good approximations to anything at any step.  Indeed, if we accept the notion of true arc length in this case as being given by the geometric formula for circumference of a circle, then they are all equally bad approximations.
Taking the limit of equally bad approximations, we still get an equally bad approximation in the limit.
