Question: Suppose that the interval 0 x 1 is divided into 100 subintervals with a width of x = .01. Show that the sum

Suppose that the interval 0 ≤ x ≤ 1 is divided into 100 subintervals with a width of Δx = .01. Show that the sum[3e-0.01] Ax + [3e-0.02] Ax + [3e-0.03] Ax +. + [3e-] Ax

[3e-0.01] Ax + [3e-0.02] Ax + [3e-0.03] Ax +. + [3e-] Ax is close to 3(1-e).

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