Answer:
Since 0 < x < 2 and n = 4 we see that each sub interval is of lenght (b - a)/n = (2 - 0)/4 = 1/2
[0, 1/2] [1/2, 1] [1, 3/2] [3/2, 2]
The midpoints are t0 = 1/4, t1 = 3/4, t2 = 5/4, and t3 = 7/4
Riemann sum is
Sum from i = 0 to 3 {f(ti) * (b-a)/n}
Sum from i = 0 to 3 {(e^(ti) - 5) * 1/2}
(1/2) Sum from i = 0 to 3 {e^(ti) - 5}
(1/2) [e^(t0) - 5 + e^(t1) - 5 + e^(t2) - 5 + e^(t3) - 5]
(1/2) [e^(1/4) + e^(3/4) + e^(5/4) + e^(7/4) - 20]
