find the integral of e^x as the limit of riemann sums, obviously answer is e^x but geting there is a pain?

Answer 1

We divide the interval from a to b into N strips of equal width w = (b - a)/N.

The (i+1)-th strip runs from x0 = a + wi to x1 = a + w(i+1), with minimum and maximum values
y0 = e^x0 = A * W^i
y1 = e^x1 = A * W^(i+1) = y0 * W
where A = e^a and W = e^w

The lower and upper sum are (i = 0, ..., N-1)
S0 = SUM w * y0 = w * A * SUM W^i
... = w * A * (W^N - 1) / (W - 1)
S1 = S0 * W

Now W^N - 1 = e^(Nw) - 1 = e^(b-a) - 1 = B / A - 1, where B = e^b, so

S0 = w * A * (B/A - 1) / (W - 1)
... = (B - A) * w / (W - 1)

In the limit N --> oo, w --> 0 and
w / (W - 1) = w / (e^w - 1) --> 1.

Therefore, lim S0 = B - A.

Likewise, lim S1 = lim (W * S0) = B - A because W = e^w --> 1.

So the Riemann sums converge to
B - A = e^b - e^a, QED