How Do You Calculate the Integral of e^x from 0 to 1 Using Riemann Sums?

[angel_eyez]

Reimann's Law..maajoor Helpp!

1. Use Riemann sum ( do NOT use fundamental theorum of Calculus) to calculate
(integral) b= 1 and a =0 e^x dx

Attempt:

(delta x)=b-a/n 1-0/n =1/n
xi = 1/n

(integral) b=1 a=0 sigmaf(xi)deltax
limn->infin sigma f(i/n)1/n
use eqn to get...limn->infin 1/nsigma{e^(i/n)}
limn->infin 1/n e^1/ni
limn->infi 1/ne^1/n n(n+1)/2
limn->infin 1/n e^1/2(n+1)

then sub for n (table n | Rn where n =40,100,500,1000,5000 and Rn should be close to 1.72(got that using FTC2)...so the eqn must be wrong :D

3.

 

[HallsofIvy]

1. Use Riemann sum ( do NOT use fundamental theorum of Calculus) to calculate
(integral) b= 1 and a =0 e^x dx

Attempt:

(delta x)=b-a/n 1-0/n =1/n
xi = 1/n

(integral) b=1 a=0 sigmaf(xi)deltax
limn->infin sigma f(i/n)1/n
use eqn to get...limn->infin 1/nsigma{e^(i/n)}

Use what equation? This is the crucial part!
You can rewrite this sum as
\frac{1}{n} \sum_{i=0}^n (e^{1/n})^i
That's a geometric series with "common ratio" e^1/n: its sum is
\frac{1}{n}\frac{1- e^{(n+1)/n}}{1- e^{1/n}}

limn->infin 1/n e^1/ni
limn->infi 1/ne^1/n n(n+1)/2
limn->infin 1/n e^1/2(n+1)

then sub for n (table n | Rn where n =40,100,500,1000,5000 and Rn should be close to 1.72(got that using FTC2)...so the eqn must be wrong :D

3.