5.2 definite integrals greg kelly, hanford high school, richland, washington
TRANSCRIPT
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5.2 Definite Integrals
Greg Kelly, Hanford High School, Richland, Washington
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When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
211
8V t
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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211
8V t
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.
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.
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time
velocity
After 4 seconds, the object has gone 12 feet.
In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance: 3t d
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
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If the velocity varies:
11
2v t
Distance:21
4s t t
(C=0 since s=0 at t=0)
After 4 seconds:1
16 44
s
8s
1Area 1 3 4 8
2
The distance is still equal to the area under the curve!
Notice that the area is a trapezoid.
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211
8v t What if:
We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example.
It seems reasonable that the distance will equal the area under the curve.
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211
8
dsv t
dt
31
24s t t
314 4
24s
26
3s
The area under the curve2
63
We can use anti-derivatives to find the area under a curve!
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Let’s look at it another way:
a x
Let area under the
curve from a to x.
(“a” is a constant)
aA x
x h
aA x
Then:
a x aA x A x h A x h
x a aA x h A x h A x
xA x h
aA x h
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x x h
min f max f
The area of a rectangle drawn under the curve would be less than the actual area under the curve.
The area of a rectangle drawn above the curve would be more than the actual area under the curve.
short rectangle area under curve tall rectangle
min max a ah f A x h A x h f
h
min max a aA x h A x
f fh
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min max a aA x h A x
f fh
As h gets smaller, min f and max f get closer together.
0
lim a a
h
A x h A xf x
h
This is the definition
of derivative!
a
dA x f x
dx
Take the anti-derivative of both sides to find an explicit formula for area.
aA x F x c
aA a F a c
0 F a c
F a c initial value
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min max a aA x h A x
f fh
As h gets smaller, min f and max f get closer together.
0
lim a a
h
A x h A xf x
h
a
dA x f x
dx
aA x F x c
aA a F a c
0 F a c
F a c aA x F x F a
Area under curve from a to x = antiderivative at x minus
antiderivative at a.
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0
1
limn
k kP
k
f c x
b
af x dx
F x F a
Area
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Area from x=0to x=1
Example: 2y x
Find the area under the curve from x=1 to x=2.
2 2
1x dx
23
1
1
3x
31 12 1
3 3
8 1
3 3
7
3
Area from x=0to x=2
Area under the curve from x=1 to x=2.
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Example: 2y x
Find the area under the curve from x=1 to x=2.
To do the same problem on the TI-89:
^ 2, ,1,2x x
ENTER
72nd
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Example:
Find the area between the
x-axis and the curve
from to .
cosy x
0x 3
2x
2
3
2
3
2 2
02
cos cos x dx x dx
/ 2 3 / 2
0 / 2sin sinx x
3sin sin 0 sin sin
2 2 2
1 0 1 1 3
On the TI-89:
abs cos , ,0,3 / 2x x 3
If you use the absolute value function, you don’t need to find the roots.
pos.
neg.