[5-4] Riemann Sums and the Definition of Definite Integral Yiwei Gong Cathy Shin

Before… To find the Area… [0,2] using 4 subintervals.

Before… Using trapezoid rule…

Now we have Riemann Sum!!

Definitions: Riemann Sum and Its Parts I A partition of the interval [a,b] is an ordered set of n+1 discrete points a = x 0 < x 1 < x 2 < x 3 <…< x n = b The partition divides the interval [a,b] into n subintervals of width △ x 1, △ x 2, △ x 3,…, △ x n

Definitions: Riemann Sum and Its Parts II Sample points for a partition are points x = c 1, c 2, c 3,…, c n, with one point in each subinterval. A Riemann sum R n for a function f on the interval [a, b] using a partition with n subintervals and a given set of sample points in a sum of form R n =

Five Different Riemann Sums… Left Riemann Sum Right Riemann Sum Midpoint Riemann Sum Lower Riemann Sum Upper Riemann Sum

Left Riemann sum

Right Riemann sum

Midpoint Riemann sum

Lower & Upper Riemann Sum If the lower sums, L n, and the upper sums, U n, for a function f on the interval [a, b] approach the same limit as △ x approaches zero ( or as n approaches infinity in the case of equal-width subintervals), then f is integrable on [a,b]. This common limit is defined to be the definite integral of f(x) with respect to x from x = a to x = b. The numbers a and b are called lower and upper limits of integrations, respectively.

Lower & Upper Riemann Sum = lim L n = lim U n Provided the two limits exist and are equal.

Example of Left Riemann Sum Evaluate: 0 ∫ 2 (-3x 4 -4x 3 +x 2 -x+10)dx

Solution Equation to use: Substitute:

Riemann sums computed using left hand endpoints

Example of Right Riemann Sum Evaluate: 0 ∫ 2 (-3x 4 -4x 3 +x 2 -x+10)dx

Solution Equation to use: Substitute:

Riemann sums computed using right hand endpoints

Example of Midpoint Riemann Sum Evaluate: 0 ∫ 2 (-3x 4 -4x 3 +x 2 -x+10)dx

Solution Equation to use: On a general interval [a,b], the midpoints are For this example, the midpoint approximation is Substitute:

Riemann sums computed using midpoints

Example of Lower Riemann Sum Approximate the area under f(x) = x 2 over [1, 4] by the lower Riemann sum of order 6 on a regular partition.

Solution Let A be the required area. The common length of the sub-intervals is (4 – 1)/6 = 1/2. See Fig Since f is increasing and continuous a lower Riemann sum is obtained by selecting the left endpoints of the sub-intervals.

Solution cont… Approximation Of Area:

Solution cont… We have:

Example of Upper Riemann Sum The below graph, s(t), represents the rate at which snow falls, in centimeters per hour, over the six hour period from 12 a.m. to 6 a.m. Approximate the value of f(3) using upper Riemann sums with 6 rectangles.

Solution Upper sums are found by splitting the x- axis into 6 parts (since the problem specific 6 rectangles). As we're approximating the area beneath s(t) on [0,3], each rectangle must be 1/2 unit wide. On each of the 6 half-unit wide intervals, you have to pick the point at which the curve reaches it's highest value, and use that as the height of the rectangle.

Solution cont.. In essence, you're using these rectangles:

Solution cont… ☺ You may not have chosen the same approximations as I did, so your answer may vary slightly. Since this is an upper sum, it's value is larger than the actual snowfall.