Answer to Question #3458 in Calculus for louis

Question

Calculate the left Riemann sum for the given function over the given interval, using the given value of n. (When rounding, round answers to four decimal places.) f(x) = e^−x over [−6, 6], n = 6

Answer 1

Let f: D → R be a function defined on a subset D of the real line R. Let I = [a, b] be a closed interval contained in D, and let
P = {[x₀, x₁), [x₁, x₂), ... [x_n-1, x_n]} be a partition of I, where a = x₀ < x₁ < x₂ ... < x_n = b.
The Riemann sum of f over I with partition P is defined as

where xi-1 ≤ yi ≤ xi. The choice of yi in this interval is arbitrary. If yi = xi-1 for all i, then S is called a left Riemann sum.
Partitions of equal size(as in our task): the interval [a, b] is therefore divided into n subintervals, each of length Q = (b − a) / n. The points in the partition will then be
a, a + Q, a + 2Q, ..., a + (n−2)Q, a + (n−1)Q, b.

Left sum
For the left Riemann sum, approximating the function by its value at the left-end point gives multiple rectangles with base Q and height f(a + iQ). Doing this for i = 0, 1, ..., n−1, and adding up the resulting areas gives
Q [f(a) + f(a+ Q) + f(a +2Q) + ... + f(b-Q)]
a = -6, b = 6
So we have Q = (6-(-6))/6 = 2
Left Riemann sum
S = 2(f(-6) + f(- 6 + 2) + f(-6 + 4) + f(-6 + 6) + f(-6 + 8) + f(-6 + 10)) = 2(f(-6) + f(-4) + f(-2) + f(0) + f(2) + f(4))=
= 2(e⁶ + e⁴ + e² + e⁰ + e^-2 + e^-4 ) = 933.139