chapter 5 – the definite integral. 5.1 estimating with finite sums example finding distance...
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Chapter 5 – The Definite Integral
5.1 Estimating with Finite SumsExample Finding Distance
Traveled when Velocity Varies2A particle starts at 0 and moves along the -axis with velocity ( )
for time 0. Where is the particle at 3?
x x v t t
t t
Graph and partition the time interval into subintervals of length . If you use
1/ 4, you will have 12 subintervals. The area of each rectangle approximates
the distance traveled over the subint
v t
t
erval. Adding all of the areas (distances)
gives an approximation to the total area under the curve (total distance traveled)
from 0 to 3.t t
2
Continuing in this manner, derive the area 1/ 4 for each subinterval and
add them:
1 9 25 49 81 121 169 225 289 361 441 529 2300
256 256 256 256 256 256 256 256 256 256 256 256 2568.98
im
LRAM, MRAM, and RRAM approximations to the area under the graph of y=x2 from x=0 to x=3
p.270 (1-19, 26, 27)
5.2 Definite Integrals1 2 3 11
...n
k n nka a a a a a
5
1kk
3
11k
kk
2
1 1k
k
k 25
4 1k
k
k
Sigma notation enables us to express a large sum in compact form:
Ex)
Ex) Ex)
Ex)
The Definite Integral as a Limit of Riemann Sums
-1
0
Let be a function defined on a closed interval [ , ]. For any partition
of [ , ], let the numbers be chosen arbitrarily in the subinterval [ , ].
If there exists a number such that lim
k k k
P
f a b P
a b c x x
I 1
( )
no matter how and the 's are chosen, then is on [ , ] and
is the of over [ , ].
n
k kk
k
f c x I
P c f a b
I f a b
integrable
definite integral
All continuous functions are integrable. That is, if a function is
continuous on an interval [ , ], then its definite integral over
[ , ] exists.
f
a b
a b
We have that 1
limbn
k kn
k a
f c x f x dx
b
a
f x dxUpper limit
Integral sign
Lower limit
Integrand
Variable of Integration
th
2
1
The interval [-2,4] is partitioned into subintervals of equal length 6 / .
Let denote the midpoint of the subinterval. Express the limit
lim 3 2 5 as an integral.
k
n
k kn k
n x n
m k
m m x
Example Using the Notation
Area Under a Curve
If ( ) is nonnegative and integrable over a closed interval [ , ],
then the area under the curve ( ) from to is the
, ( ) .b
a
y f x a b
y f x a b
A f x dx
integral
of from to f a b
Area= ( ) when ( ) 0.
( ) area above the -axis area below the -axis .
b
a
b
a
f x dx f x
f x dx x x
Notes about Area
The Integral of a Constant
If ( ) , where is a constant, on the interval [ , ], then
( ) ( ) b b
a a
f x c c a b
f x dx cdx c b a
2
-1Evaluate numerically. sinx xdx
Evaluate the following integrals:2
2
2
4 x dx
2
1
xdx
x
FNINT( sin , ,-1,2) 2.04x x x
p.282 (1-27, 33-39) odd
5.3 Definite Integrals and
Antiderivatives
1 4 1
-1 1 -1
1
4
Suppose ( ) 5, ( ) 2, and ( ) 7 .
Find ( ) if possible.
f x dx f x dx h x dx
f x dx
1 4 1
-1 1 -1
4
1
Suppose ( ) 5, ( ) 2, and ( ) 7 .
Find ( ) if possible.
f x dx f x dx h x dx
f x dx
1 4 1
-1 1 -1
2
2
Suppose ( ) 5, ( ) 2, and ( ) 7 .
Find ( ) if possible.
f x dx f x dx h x dx
h x dx
Ex: Show that the value of 1
0
31 cos
2xdx
Average (Mean) Value
If is integrable on [ , ], its average (mean) value on [ , ] is
1( ) ( )b
a
f a b a b
avg f f x dxb a
2Find the average value of ( ) 2 on [0,4].f x x
The Mean Value Theorem for Definite Integrals
If is continuous on [ , ], then at some point in [ , ],
1( ) ( ) .b
a
f a b c a b
f c f x dxb a
1, 1
1
1
nn xx dx C n
n
dx dx x C
cossin
kxkxdx C
k
sincos
kxkxdx C
k
2sec tanxdx x C
2csc cotxdx x C
sec tan secx xdx x C csc cot cscx xdx x C
Integral Formulas
This is known as the indefinite integral. C is a constant.
5x dx 1dxx
sin 2xdxcos
2
xdx
Evaluate:
p. 290 (1 – 29) odd19 – 29 noteDo (31-35)After 5.4
5.4 Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus – Part 1
If is continuous on [ , ], then the function ( ) ( )
has a derivative at every point in [ , ], and
( ) ( ).
x
a
x
a
f a b F x f t dt
x a b
dF df t dt f x
dt dx
Evaluate the following:
cosxd
tdtdx
2
0
1
1
xddt
dx t
2
1
cosx
y tdt
Find dy
dx
Find dy
dx
5
3 sinx
y t tdt2
2
1
2
x
tx
y dte
Find a function y = f(x) with derivative
That satisfies the condition f(3) = 5.
tandy
xdx
The Fundamental Theorem of Calculus, Part 2
If is continuous at every point of [ , ], and if is any antiderivative
of on [ , ], then ( ) ( ) - ( ).
This part of the Fundamental Theorem is also called the
.
b
a
f a b F
f a b f x dx F b F a Integral
Evaluation Theorem
3 2
-1Evaluate 3 1 using an antiderivative.x dx
How to Find Total Area AnalyticallyTo find the area between the graph of ( ) and the -axis over the interval
[ , ] analytically,
1. partition [ , ] with the zeros of ,
2. integrate over each subinterval,
3. add the absolute values o
y f x x
a b
a b f
f
f the integrals.
Find the area of the region between the curve y = 4 – x2, [0, 3] and the x-axis.
Look at page 301 example 8.
p.302 (1-57) odd
5.5 Trapezoidal Rule
0 1 11 2
0
1 2 1
0 1 2 1
0 1 1 1 1
( ) ...2 2 2
...2 2
2 2 ... 2 ,2
where ( ), ( ), ..., ( ), ( ).
b n n
a
n
n
n n
n n n
y y y yy yf x dx h h h
y yh y y y
hy y y y y
y f a y f x y f x y f b
The Trapezoidal Rule
0 1 2 1
To approximate ( ) , use
2 2 ... 2 ,2
where [ , ] is partitioned into n subintervals of equal length
( - ) / .
LRAM RRAMEquivalently, ,
2where LRAM and RRAM are the Rienamm
b
a
n n
n n
n n
f x dx
hT y y y y y
a b
h b a n
T
sums using the left
and right endpoints, respectively, for for the partition.f
Use the trapezoidal rule with n = 4 to estimate . Compare with fnint.
Ex: An observer measures the outside temperature every hour from noon until midnight, recording the temperatures in the following table.
What was the average temperature for the 12-hour period?
22
1
x dx
Time N 1 2 3 4 5 6 7 8 9 10 11 M
Temp 63 65 66 68 70 69 68 68 65 64 62 58 55
Simpson’s Rule
0 1 2 3 2 1
To approximate ( ) , use
4 2 4 ... 2 4 ,3
where [ , ] is partitioned into an even number subintervals
of equal length ( - ) / .
b
a
n n n
f x dx
hS y y y y y y y
a b n
h b a n
Ex: Use Simpson’s rule with n = 4 to approximate 2
4
0
5x dx
p.312 (1-18)