Transcript Document

Book reference section 5-5, Pages 195-202
By Shoshanna Craig
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.
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.
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
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.
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
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.
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
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
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
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.