Calculus 1- Question involving Rieman Sums

Question

Find expressions for the Riemann sums for the following functions on the given intervals

with the given number of equal intervals.

( a ) f ( x ) = 5 − x , [ 1 , 3 ] , n = 2

Answer 1

The width of each rectangle is (3-1)/n = 2/n.

For right hand sum:

x₁ = 1 + 1(2/n) : start at 1 and add on 1 width

x₂ = 1 + 2(2/n): start at 1 and add on 2 widths

In general x_i = 1 + i (2/n) = 1 + 2i/n

The area of the ith rectangle:

A_i = (2/n) [f(x_i)] = (2/n)[5 - x_i] = (2/n)[5 - (1 + 2i/n)] = (2/n)[4 - 2i/n] = 8/n - 4i/n²

The sum of the areas of all the rectangles is:

∑_i=1ⁿ (8/n - 4i/n²), where in this case n = 2. The post indicates you probably should replace n with 2 and stop here to get an approximate area.

But continuing:

Using summation properties (where ∑ implies from i = 1 to n):

(1/n)∑ 8 - 4/n²∑ i

(1/n)(8n) - 4/n² [(n²+n)/2]

8 - 2 [(n² + n)/n²] = 8 - 2[ 1 + 1/n]

Taking the limit of that expression as x -> ∞ gives 8 - 2[1 + 0] = 6.

This graph reveals that the region in question is a trapezoid and the area can be found using formula for area of a trapezoid rather than using a Riemann sum.

desmos.com/calculator/cqfvh2u5i8

Follow similar patterns for left hand sum and midpoint sum, but the x_i will probably be a bit more complicated.