How do you find the left Riemann sum for #f(x) = e^x# on [0,In 2] with n = 40?

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Answer 1 · 2015-07-01T18:29:16

To find the left Riemann sum for #f(x) = e^x# on #[0,ln 2]# with n = 40, we need:

#f(x)=e^x#
#a=0# and #b=ln2#
#Delta x = (ln2-0)/40 = ln2/40#

The left endpoints are:

#0, ln2/40, 2ln2/40, 3ln2/40, 4ln2/40, . . . kln2/40, . . . 39ln2/40#

The Riemann sum is, therefore:

#L = sum_(k=0)^39 f(kln2/40)*Deltax#

#= sum_(k=0)^39 e^(kln2)/40* ln2/40#

# = ln2/1600 sum_(k=0)^39 e^(kln2)#

# = ln2/1600 sum_(k=0)^39 2^k#

But the sum is a simple geometric series with #a_0 =1# and #r=2#, so the sum is:

#sum_(k=0)^39 2^k = (1-2^40)/(1-2) = 2^40-1#

So we get:

#L = ln2/1600 * (2^40-1)#

If you want or need to do the arithmetic to finish, be my guest.