# 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=__________ 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...

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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`

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