Download - The Area Problem
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Lets take a trip back in time…to geometry. Can you find the area of the following? If so, why?
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Now, lets take a trip back to Advanced Algebra. Can you find the area of the region bounded by the line x=0, y=0 , y = 4 and y = 2x+3? If so, how?
3
04
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04
16
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When we find the area under a curve by adding rectangles, the answer is called a Riemann Sum.
subinterval
partition
The width of a rectangle is called a subinterval.
The entire interval is called the partition.
Subintervals do not all have to be the same size.
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subinterval
partition
If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by .P
As gets smaller, the approximation for the area gets better. Why?
P
0
1
Area limn
k kP
k
f c x
if P is a partition of the interval ,a b
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0
1
limn
k kP
k
f c x
is called the definite integral of
over .f ,a b
If we use subintervals of equal length, then the length of a
subinterval is:b a
xn
The definite integral is then given by:
1
limn
kn
k
f c x
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1
limn
kn
k
f c x
Leibnitz introduced a simpler notation for the definite integral:
1
limn b
k ank
f c x f x dx
Note that the very small change in x becomes dx.
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b
af x dx
IntegrationSymbol
lower limit of integration
upper limit of integration
integrandvariable of integration
(dummy variable)
It is called a dummy variable because the answer does not depend on the variable chosen.
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b
af x dx
We have the notation for integration, but we still need to learn how to evaluate the integral.
This will be another day.
We will master the Riemann Sum work first! Onwards!!!
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time
velocity
After 4 seconds, the object has gone 12 feet. Why?
Consider an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance:
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
ft3 4 sec 12 ft
sec
3t d d’ ?
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If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.
(The units work out.)
211
8V t Example:
We could estimate the area under the curve by drawing rectangles touching at their left corners.
We call this the Left-hand Rectangular Approx. Method (LRAM).
Approx. area:
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We could also use a Right-hand Rectangular Approximation Method(RRAM).
Approx. area:
211
8V t
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Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).
1.031251.28125
1.78125Approx area: 2.53125
In this example there are four subintervals. As the number of subintervals increases, so does the accuracy.
211
8V t
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211
8V t
Approximate area:6.65624
t v
1.007810.25
0.75 1.07031
1.25 1.19531
1.382811.75
2.25
2.75
3.25
3.75
1.63281
1.94531
2.32031
2.75781
13.31248 0.5 6.65624
width of subinterval
With 8 subintervals:
The exact answer for thisproblem is .6.6
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Circumscribed rectangles are all above the curve:
Inscribed rectangles are all below the curve:
What Riemann Method?
Over or under estimate?
Concave up or down?
What Riemann Method?
Over or under estimate?
Concave up or down?
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Riemann Sums Exercise Handout