# If f(x) = e^(x)− 3, 0 ≤ x ≤ 2, find the Riemann sum with n = 4 correct to six decimal places, taking the sample points to be midpoints.If f(x) = e^(x)− 3, 0 ≤ x ≤ 2, find the Riemann...

If f(x) = e^(x)− 3, 0 ≤ x ≤ 2,

find the Riemann sum with *n* = 4 correct to six decimal places, taking the sample points to be midpoints.

If f(x) = e^(x)− 3, 0 ≤ x ≤ 2,

find the Riemann sum with *n* = 4 correct to six decimal places, taking the sample points to be midpoints.

M4=__________

### 1 Answer | Add Yours

The Riemann sum is computed by `sum_(i=1)^nf(c_i)Deltax_i`

On the interval `0<=x<=2` with `n=4` we can let `Deltax=(2-0)/4=1/2` . We are told to use the midpoints of the intervals, so `c_1=1/4,c_2=3/4,c_3=5/4,c_4=7/4`

Then the Riemann sum is

`1/2[e^(1/4)-3+e^(3/4)-3+e^(5/4)-3+e^(7/4)-3]`

`~~1/2[-1.715975-.883000+.490343+2.745603]`

`~~1/2[.636971]~~.318486`

**So the Riemann sum is .318486**

This compares to the actual value of `~~.389056`