Integration of e^x using riemann sums?

So you have:

(1 + e^(1/n) + e^(2/n) + ... + e^((n-1)/n))/n

The geometric series 1 + r + r^2 .. + r^(n-1) has value (r^n-1)/(r-1)

So in this case, r=e^(1/n), and you get

(e-1)/(e^(1/n)-1)

Dividing by n, then, we get:

(e-1)/(n*(e^(1/n)-1))

So you need to determine what:

lim n->oo of n*(e^(1/n)-1) is.

You can let h=1/n, then this is:

lim h->0 (e^h-1)/h

So, that limit is the derivative of e^x at x=0, which is 1.

So the original expression has limit: e-1.

To get a draw close on how integrals artwork, you do the Riemann sum. additionally it incredibly is prepare in algebra. in case you notice what a soreness it incredibly is to do it as a shrink of sums, then you get to have a good time with how magic & effectual the FTC is. additionally it incredibly is sturdy prepare to do the algebra besides, just to benefit your artwork and be responsive to the way the bounds fall out. it incredibly is conceivable the class is only too sluggish on your great recommendations of expertise, too. yet understanding info via hand could nicely be a sturdy self-discipline. If pissed off with sluggish %. of attending to three bigger math, possibly see in case you are able to detect a duplicate of something called Princeton companion to arithmetic - that could desire to maintain you busy. Oh, yea! i forgot to characteristic this ... yet one greater reason is that regularly you desire to combine something and don't have an antiderivative obtainable for it. And yet another proper reason is which you do a matching operation to a Riemann sum (with a shrink) once you evaluate an fundamental numerically. (yet if so there's a commerce-off with a value of delta that gets too close to to 0 and starts inflicting too many errors from numerical computational aspects, as adversarial to the approximation errors of representing the fundamental as a finite sum.)