Integration intuition help...

this question made me re-visit the basics of integration as the summation of increasingly thin rectangles. But I am getting stuck with it.

given
$f(x)=3x$
The area of each rectangle of width $\delta x$ between y=0 and the function is $3x\delta x$, but I can't figure out how to get these to sum to $\frac{3}{2}x^2$ (which is the indefinite integral of 3x.

As $\delta x \rightarrow 0$ why doesn't each rectangle reduce to a zero area? I realise that in the limit this is presumably counterbalanced by the fact that there is an infinite number of them, but how is this represented mathematically?

You might want to work out $\int_0^b3xdx$. Imagine you're chopping up the interval [0,b] into n equally sized segments, so you have n rectangles with their bottom-right corners at b/n, 2b/n, 3b/n, ..., nb/n=b. Each rectangle has width b/n and heights 3b/n, 3*2b/n, etc.

So to calculate the integral you would work out $\sum_{r=1}^n 3(br/n) (b/n)$ and let n tend to infinity. So here you have each rectangle tending to zero area counterbalanced by the number of rectangles tending to infinity, which doesn't tell you much about what the overall limit is. However with some algebra you can rewrite this summation in a way that makes it more obvious what the limit is:

$\sum_{r=1}^n 3(br/n) (b/n) =3\frac{b^2}{n^2} \sum_{r=1}^nr = 3\frac{b^2}{n^2} \frac{1}{2}n(n+1) = \frac{3}{2}b^2\times\frac{n+1}{n} = \frac{3}{2}b^2\times(1+\frac{1}{n})\rightarrow \frac{3}{2}b^2$ as n tends to infinity.

You might want to work out $\int_0^b3xdx$. Imagine you're chopping up the interval [0,b] into n equally sized segments, so you have n rectangles with their bottom-right corners at b/n, 2b/n, 3b/n, ..., nb/n=b. Each rectangle has width b/n and heights 3b/n, 3*2b/n, etc.

So to calculate the integral you would work out $\sum_{r=1}^n 3(br/n) (b/n)$ and let n tend to infinity. So here you have each rectangle tending to zero area counterbalanced by the number of rectangles tending to infinity, which doesn't tell you much about what the overall limit is. However with some algebra you can rewrite this summation in a way that makes it more obvious what the limit is:

$\sum_{r=1}^n 3(br/n) (b/n) =3\frac{b^2}{n^2} \sum_{r=1}^nr = 3\frac{b^2}{n^2} \frac{1}{2}n(n+1) = \frac{3}{2}b^2\times\frac{n+1}{n} = \frac{3}{2}b^2\times(1+\frac{1}{n})\rightarrow \frac{3}{2}b^2$ as n tends to infinity.

( This all concerns the last line of the explanation )

I understand everything till the part where the sigma sign disappears and the
(1/2*n)*(n+1) comes in. How did that come to be?

Also, when you multiplied (3br/n) by(b/n) , why did (3b^2/n^2) appear before the sigma sign and the r after the sigma sign?

Sorry for the questions
Either way, thank you for the detailed explanation

That's exactly what I was looking for. Thanks!