# Answer to Question #3458 in Calculus for louis

Question #3458

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

Expert's answer

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 = {[x

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(e

P = {[x

_{0}, x_{1}), [x_{1}, x_{2}), ... [x_{n-1}, x_{n}]} be a partition of I, where a = x_{0}< x_{1}< x_{2}... < 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(e

^{6}+ e^{4}+ e^{2}+ e^{0}+ e^{-2}+ e^{-4}) = 933.139Need a fast expert's response?

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