Wsrtjtyk

If I think of an integral, $int_a^b f(x) dx$ as roughly $sum f(x)delta x$ where $delta x$ is very very small, then can't I write this sum as

$$
f(x)x - f(x)(x-delta x) + f(x-delta x)(x-delta x) +dots -f(a)a
$$

Then, since $delta x$ is so small, assuming $f(x)$ is "smooth" in some sense, shouldn't $f(x) approx f(x-delta x)$ so $$f(x)(x-delta x) approx f(x-delta x)(x-delta x)$$

and then all terms would cancel out except for $f(x)x$ and $f(a)a$?

I.e. wouldn't this give $int_a^b f(x)dx approx f(x) x - f(a)a$?

Why does this not give a decent approximation?

For example, I think this is always exact for a constant function. I would then think it should be pretty good for any function that doesn't change too quickly at any point.

but what is "too quickly"

I thought it would also be pretty good for a linear function, but this is not the case...

Sure, $f(x) approx f(x-delta x)$, so you can pair off most of the terms in pairs that are approximately $0$. But they are only approximately zero, and there are a lot of them ($(b-a)/delta x$ of them, and $delta x$ is very small!). So, the total sum of them may still be non-negligible because there are so many of them.

Indeed, using the same reasoning, you could just say that in $sum f(x)delta x$ each term is very small, since $f(x)$ is bounded and $delta x$ is small. Your approach would then say that $0$ is a good approximation to the integral. Of course this is wrong, because although each term is small (roughly proportional to $delta x$), the number of terms is large in a way that cancels it out (proportional to $1/delta x$).

We can be more precise and say that $f(x)-f(x-delta x)approx f'(x) delta x$ if $f$ is differentiable. So, the errors in your terms are also roughly proportional to $delta x$, at least as long as $f'(x)$ is bounded and stays away from $0$. We can thus expect a non-negligible total error when we add up $(b-a)/delta x$ such errors.

To be even more precise, we can expect the error to be $int_a^b xf'(x), dx$, since each term of the error has the form $(x-delta x)(f(x)-f(delta x))approx xf'(x)delta x$. We can verify this rigorously: if you integrate $int_a^b f(x), dx$ by parts with $u=f(x)$ and $v=x$ (assuming $f$ is continuously differentiable) you get $$int_a^b f(x),dx=xf(x)bigg|_a^b-int_a^b xf'(x),dx=(bf(b)-af(a))-int_a^b xf'(x),dx.$$ The first term is exactly what you find with your approximation, so the error is indeed $int_a^b xf'(x),dx$. In this sense, then, the error you are making is akin to saying that the derivative of $xf(x)$ should be $f(x)$, which is what you get by differentiating just the $x$ factor but ignoring the fact that $f(x)$ may also be changing.