apcalcRiemann+Sums

Riemann Sums in Calculus

What is a Riemann Sum?  A Riemann Sum is an approximation of the total area under the curve on a graph; also known as a [|definite integral] of a function f, on the interval of a and b.

How do you find Riemann Sums?

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There are four ways to calculate Riemann Sums. These four ways include right hand, left hand, mid-point, and trapezoidal. The basic concept of calculating the area under the curve is to divide the area into n rectangular segments. The rectangles have a width of delta X, which is chosen, and a height that is derived from the function in question, f(X). You use these measurements to calculate the area of each rectangle. With the calculations of each rectangle, you sum them together to find the total area under the curve. As the width of the rectangles becomes smaller, the approximation of the area is better.

· Find the value of f(X) at the first X value. · Multiply the height, as found in the previous step, by delta X. This will give you the area of the first rectangle. · Add delta X to the first X value. This will give you the X value at the left side of the second rectangle. · Repeat the above steps for the second rectangle. · Continue in this way until you have reached the final X value. · Add the area of all the rectangles. This is the Riemann sum.

In the calculation of Riemann Sums it is assumed that ﻿ and  where a is the start of the interval, b is the end of the interval, and n is the number of rectangles you are using. This will need to be determined before calculations are made.

** Le ﻿ ft Hand ** // Left // (//n//) = (//f//(//x//0) + //f//(//x//1) + //f//(//x//2) + ... + //f//(//xn// − 1))Δ//x//

The [|left hand] Riemann Sum is an under estimate (as seen in the graph). For the left hand sum you systematically sample the height of the function at the left hand endpoints of all of the intervals. // Right // (//n//) = (//f//(//x//1) + //f//(//x//2) + //f//(//x//3) + ... + //f//(//xn//))Δ//x//
 * Right Hand **

The [|right hand] Riemann Sum is an over estimate (as seen in the graph). For the right hand sum we systematically sample the height of the function at the right hand endpoints of all the intervals.

**Midpoint **

The [|midpoint] Riemann Sum is the most accurate way to estimate the area under the curve. The midpoint sum is an adaptation of the left and right hand sums. You add up the areas of many small rectangles where the height of the rectangles is obtained by frequently sampling the height of the function at many points on the interval. **Trapezoidal ** or

With the [|trapezoidal] rule we approximate the area underneath the curve by a collection of trapezoids. It can also be shown that Trap(n) is the average of the Left(n) and Right(n) approximations. Examples of Riemann Sums 1.) f(x) = x3 - 6x2 + 9x + 2

We examine how the process of Riemann Sums works with this [|cubic function] between x = 0 and x = 5. First, the region is divided into five rectangles with width one and height taken to be the midpoint in each interval. The red line is the plot of the function. The green boxes underneath the curve are the rectangles underneath the curve. If we sum the areas of each of these rectangles, we obtain a representation of the area underneath. Notice that the height of each rectangle is found by evaluating f(x) at the midpoint of each subinterval. Here we have 5 rectangles, some of the area is not included and some boxes stick out into where the curve shouldn't be counting area. The regions each have a width of one. (f(1/2)+f(3/2)+f(5/2)+f(7/2)+f(9/2)) = 28.125 With 10 rectangles, the area is better calculated. Now the widths of the rectangles are 0.5.



(f(1/4)+f(3/4)+f(5/4)....+f(19/4)) = 28.58375

With 40 rectangles, there is a very dense grid of rectangles. The width of the rectangles are 1/8. The more rectangles, the closer the actual approximation of the area under the curve.

(f(1/16)+f(3/16)+f(5/16)....+f(39/16)) = 28.74023

2.)  f(x) = 9 - x2.   Sketch a graph showing the area under the graph, then use the **Midpoint rule** and the **Trapezoid rule** to approximate the integral with **// n // =3 **and** // n //= <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;"> 6 **.**

<span style="display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Approximate the definite integral given by <span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 10pt; text-align: left;">For the <span style="font-family: Arial,Helvetica,sans-serif;">Midpoint rule, the midpoints are // c // 1 <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 27px;"> 0.5, so <span style="color: #0000af; font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 27px;">f(1/2) //8.75// //, and// c // 2 // <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 27px;"> 1.5 , so f(1.5)   //6.75// //, and// c // 3 // <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 27px;"> 2.5 , so f(2.5)  //2.75//  //<span style="font-family: Arial,Helvetica,sans-serif;">. The Midpoint rule gives // <span style="font-family: Arial,Helvetica,sans-serif; font-size: 10pt;">For the Trapezoid rule**,** the [|integral] approximation formula gives <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">For n // = 6, the six subintervals are [0, 0.5] , [0.5, 1] , [1, 1.5] , [1.5, 2] , [2, 2.5] , and [2.5, 3] , and the value of D // x //is 0.5 .// <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">For the Midpoint rule, the midpoints are <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> c1 0.25, so f(0.25)  8.9375 , c2  0.75 , so f(0.75)  8.4375 , c3  1.25 , so f(1.25)  7.4375 , c4  1.75 , so f(1.75)  5.9375 , c5  2.25 , so f(2.25)  3.9375 , and c6  2.75 , so f(2.75) 1  .4375. The Midpoint rule gives



<span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">For the Trapezoid rule, the integral approximation formula gives

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">In each of the cases above, we see that the midpoint rule is over estimating the integral and the trapezoid rule is under estimating the value. <span style="color: #3e7398; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 150%; text-align: left;">Real world applications of Riemann Sums <span style="color: #333333; font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Riemann sums help you approximate the area under a curve. In calculus, we typically see [|polynomial] or [|exponential] functions (which are pretty easy to integrate). In real life, however, we rarely see these pretty functions.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> For example, it's virtually impossible to find a function that exactly represents the contours of a leaf on a tree outside. If you wanted to find the area of a side of the leaf, you certainly can't rely on plain [|integration]. Riemann sums can help you approximate the area. Everything depends on what information you have about the underlying curve.

1.) **Exports** <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">From figures released by the U.S. Agriculture Department, the value of U.S. pork exports from 1985 through 1993 can be approximated by the equation

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Where q represents annual exports (in millions of dollars) and t the time in years, with t = 0 corresponding to January 1, 1985. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The chart below represents an estimation of data that was used to determine the function above. We can use a Riemann process to approximate the total U.S. exports of pork over the given period by summing the individual figures. Note that in this summation we would be adding the areas of rectangles in the figure (since each rectangle has width 1 unit).

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">If we add up the areas of the rectangles in the chart, we calculate <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">(60 + 80 + 125 + 250 + 330 + 320 + 335 + 450 + 460)(1)

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">= $2,410 million. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">If we now calculate the left- and right-hand sums of q(t) with n = 9, the number of subdivisions shown in the chart; rounding off to the nearest $1 million, we get

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Left-hand sum: $2,070 million <span style="font-family: Arial,Helvetica,sans-serif; font-size: 10pt;">Right-hand sum: $2,519 million **//. //** <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Their average is the trapezoidal sum <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 24px;">= $2,294.5 million.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Note that the Riemann sums are close to the true total, so that the model seems reasonable. One cannot expect the actual pork export figures to coincide with the abstract mathematical function. But the model does allow one to predict future exports.

2.) **Population Growth** <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Suppose the population growth rate at a time t of a certain rural locale is given by

p(t) = 200 +6t1/2, t ³ 0,

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">where t represents years from 1960. (For example, p(1)=206 tells us that at year one the rate of growth is about 206 people per year.) We now use a Riemann sum to estimate the population growth from 1960 to 1975 using five equal subintervals of time and using the left endpoint of each subinterval to estimate the population growth rate for the subinterval//.//

//Solution:** The interval from 1960 to 1975 corresponds to 0 £// t// £ 15. If we divide this interval into five equal <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">subintervals, each has width <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: 24px;">= 3; therefore, <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">the endpoints of the subintervals are 0, 3, 6, 9, 12, and 15.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The Riemann sum to estimate the population growth is as follows:

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> = 3[p(0) + p(3) + p(6) + p(9) + p(12)]

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">= 3 [ 200 + (200 + 6 ( 3)^1/2+ (200+6(6)^1/2 + (200+6(9)^1/2 + (200+6 ( 12)^1/2 ] = 3192.

<span style="color: #3e7398; font-family: Arial,Helvetica,sans-serif; font-size: 150%;">History of Riemann Sums



<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> The method was named after German mathematician [|Bernhard Riemann]. Riemann was born in Germany in September 1826, his father Friedrich a Lutheran minister. He was second born in a family of six children. The father tutored the children, and when the young Bernhard was ten, a local teacher helped. Starting at age 14, he continued in public education and took a particular liking to mathematics. He read a book of some 900 pages on [|number theory] in but six days. At age 19, Bernhard entered the [|University of Göttingen], moving two years later to the [|University of Berlin]. Then in 1849, he moved back to Göttingen. earning his doctorate under [|Carl Friedrich Gauss] in 1851. Eight years later, he was appointed to the chair of mathematics at Göttingen, and within a few days was elected to the [|Berlin Academy of Sciences]. Riemann (1826-1866) became one of the leading mathematicians of the nineteenth century. The man for whom Riemann Sums is named did far more than that. His work in differential geometry provided the mathematical basis for the general [|theory of relativity]. His most famous contribution to calculus is the Riemann sum. It is clear that calculus was developed in the 1660’s and 1670’s by [|Isaac Newton] and [|Gottfried Wilhelm Leibniz], but the view that integration was simply a process reverse to [|differentiation] prevailed until the nineteenth century. The familiar conception of the definite integral as the [|limit] of approximating sums was given by Riemann in a paper he submitted upon joining the faculty at Göttingen in 1854. It was not published until 13 years later, and then only after his untimely death. His formulation of what today is known as the “Riemann Sums.”

<span style="color: #3e7398; font-family: Arial,Helvetica,sans-serif; font-size: 150%;">Works Cited

[] [] [] [] [] http://www.hardycalculus.com/calcindex/IE_riemann.htm