Riemann Sums and Definition of Definite Intregal

1. Riemann Sums and Definition of Definite Intregal Book reference section 5-5, Pages 195-202 By Shoshanna Craig

2. Introduction I chose this topic to deeper understand a concept I was previously unclear about. Real World Applications: Finding the definite integral using this simple method is useful in many ways, such as if you had a graph of velocity and time, the definite integral would be the distance traveled.

3. Topics of Discussion You will learn what a definite integral is, and how to calculate it using a method called the Riemann Sum, and comparing it to the Trapezoidal Rule learned in Chapter 1.

4. The Riemann Sum in comparison to the Trapezoidal Rule First, lets remember the Trap Rule, which allows you to find an approximate value of a definite integral using trapezoids using the formula: 2(½n1+n2+n3…+½nlast) Example X2 from 1 to 4 2(½ (1)+4+9+½ (16))=21.5

5. What the Riemann sum is... The Riemann sum uses the method of, as the trap rule did, using the area of increments, except, instead of the area’s being traps, they are rectangular. This allows the height to be varied by deciding where the rectangle is drawn ( on the upper, lower, midpoint, or somewhere in the middle of the increments) For example, to evaluate the graph of X2 in the interval of 1 to 4 with 3 increments as in the trap rule example, using the midpoints to find the height, the graph would look as seen on the next slide.

6. Relation to trap rule continued... Midpoint chart X Y 1.5 2.25 2.5 6.25 3.5 12.25 Riemann Sum 1*2.25+1*6.25+1*12.25= 20.75

7. How the Trap Rule and Riemann Sum's compare The trap rule gives a value that is slightly less than the actual value, the Riemann sum’s differ, the upper is slightly more than the actual area, and the lower is slightly higher, the midpoint is usually, but not always, the most accurate, that is a tricky part, because in trig function graphs, or X2+ the midpoint interval is not always most accurate. Like the trap rule, the value becomes more accurate as more intervals are used, but both are approximations, and do not give the exact value.

8. A Typical Problem Calculate approximately the definite integral for (1/X)dx by using a Riemann sum with six increments, using U6, L6, M6, and show that M6 is between L6 and U6. Evaluate the Upper Riemann sum… X=C 1/X (1/X)( X) 1.0 1 .5 1.5 .666 .333 2.0 .5 .25 2.5 .4 .2 3.0 .333 .166 3.5 .028 .142 U6=1.59

9. Evaluate the Lower Riemann Sum… X=C 1/X (1/X)( X) 1.5 .067 .333 2.0 .5 .25 2.5 .4 .2 3.0 .333 .167 3.5 .286 .143 4.0 .25 .125 L6=1.2178 Using the midpoint sum… X=C 1/X 1.25 .8 1.75 .571 2.25 .444 2.75 .363 3.25 .307 3.75 .267 Sum=2.753 M6=0.5(2.753)=1.377

10. Real Life Examples Finding the definite integral is useful in examples such as finding the distance of a graph of velocity and time. Since velocity is distance divided by time, when velocity is multiplied by time, the times cancel out, leaving distance. Since area is the Y (velocity) multiplied by X (time), the area of definite integral, would be X*Y or *T=D. This graph represents the time and speed of Greer, who is training to run the marathon next spring. Time Velocity Distance (hours) (mph) (miles) 1h 5.54mph 11.08m 3h 2.03mph 4.06m 5h 6.42mph 12.84m Total Distance =27.98m Actual Distance=28.28m Average Speed 4.66 mph

11. Conclusion The Riemann sum is a way to calculate an area by evaluating the area of rectangles in the graph using the intervals, and either the midpoint, upper, lower, or another point in the interval to determine the height, and adding them together to find an approximate area of the section. This is useful in ways such as finding the distance traveled when given a graph of velocity and time. The area of a graph in an interval is known in fancy calculus terms as the definite integral.

12. Director: Shoshanna Craig Editor: Shoshanna Craig Producer: Shoshanna Craig Designer: Shoshanna Craig Created by: Shoshanna Craig Special thanks to Mr. Beck and Florence the Flower