Is it valid to calculate an integral like a telescoping series? I was wondering how valid it is to consider computing integrals as though they are telescoping series. For example:
Using this as our model
$\sum_a^b [(n+1)^2-n^2]$ works, except the distance between points in an integral is dx not 1.
$\int_a^b [(x+dx)^2-x^2]=[(a+dx)^2-a^2]+[(a+2dx)^2-(a+dx)^2]+...+[b^2-(b-dx)^2]+[(b+dx)^2-b^2]$
Alright, so the above simplifies to:
$\int_a^b [(x+dx)^2-(x)^2]=(b+dx)^2-a^2$
Now carrying out algebra in the integral we get:
$\int_a^b [(x+dx)^2-x^2]=\int_a^b 2xdx+(dx)^2=2\int_a^b xdx+\int_a^b (dx)^2$
So now I kind of notice that an integral is an infinite sum of infinitely small things so that first integral I got will give us a number. But the second one is really an infinite sum of infinitely small things multiplied again by an infinitely small thing (dx) so the fact that it's squared overtakes it and in the "limit" this goes away. Similarly dx goes away on the other side of the "equality".
$2\int_a^b xdx+\int_a^b (dx)^2 = (b+dx)^2-a^2$
becomes $2\int_a^b xdx = b^2-a^2$
and it's fairly obvious that following this method will always end up giving me the right thing for polynomials since the coefficient out front of the integral will match the number on pascal's triangle and all the dx^2 and higher terms will disappear with my limiting scheme.
But I've never seen/heard of this before. How valid is this 'telescoping integral'?

The integration analogue of telescoping series is integrals of derivatives, which we can compute by the fundamental theorem of calculus:
$$\int_a^b f'(x)\,dx = f(b) - f(a).$$
