Riemann Sums and The Definite Integral. time velocity After 4 seconds, the object has gone 12 feet....
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Transcript of Riemann Sums and The Definite Integral. time velocity After 4 seconds, the object has gone 12 feet....
Riemann Sums and The Definite Integral
time
velocity
After 4 seconds, the object has gone 12 feet.
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 ftsec
3t d
Estimating with Finite Sums
If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.
21 18
V t Example:
We could estimate the area under the curve by drawing rectangles touching at their left corners.
This is called the Left-hand Rectangular Approximation Method (LRAM).
1 118
112
128
t v
10
1 118
2 112
3 128
Approximate area: 1 1 1 31 1 1 2 5 5.758 2 8 4
Estimating with Finite Sums
We could also use a Right-hand Rectangular Approximation Method(RRAM).
118
112
128
Approximate area: 1 1 1 31 1 2 3 7 7.758 2 8 4
3
21 18
V t
Estimating with Finite Sums
Another approach would be to userectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).
1.031251.28125
1.78125
Approximate area:6.625
2.53125
t v
1.031250.5
1.5 1.28125
2.5 1.78125
3.5 2.53125
In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.
21 18
V t
Estimating with Finite Sums
21 18
V t
Approximate area:6.65624
t v
1.007810.25
0.75 1.07031
1.25 1.19531
1.382811.75
2.25
2.753.253.75
1.63281
1.945312.320312.75781
13.31248 0.5 6.65624
width of subinterval
With 8 subintervals:
The exact answer for thisproblem is .6.6
Estimating with Finite Sums
Circumscribed rectangles are all above the curve:
Inscribed rectangles are all below the curve:
Estimating with Finite Sums
We will be learning how to find the exact area under a curve if we have the equation for the curve. Rectangular approximation methods are still useful for finding the area under a curve if we do not have the equation.
Estimating with Finite Sums
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
21 18
V 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.
Definite Integrals
21 18
V 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
Definite Integrals
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 ax
n
The definite integral is then given by: 1
limn
kn k
f c x
Definite Integrals
1
limn
kn k
f c x
Leibnitz introduced a simpler notation for the definite integral:
1
limn b
k an k
f c x f x dx
Note: the very small change in x becomes dx.
Definite Integrals
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.
Definite Integrals
b
af x dx
We have the notation for integration, but we still need to learn how to evaluate the integral.
Definite Integrals
Definite IntegralsDefinition -- a Definite IntegralIf y = f(x) is non negative and integrable over a closed interval[a,b], then the DEFINITE INTEGRAL (area under the curve) y = f(x) from a to b is the integral of f(x) from a to b.
b
af x dxA =
0 1
limn
k kP k
f c x
b
af x dx
F x F a
Area
Where F(x) is the antiderivative of f(x)!
Definite Integrals
( ) An indefinite integral
is a family of functions
f x dx
( ) A definite integral
is a number
b
a
f x dx
( )f x dx F x C
( ) 7.6b
a
f x dx
Using integrals to find area works extremely well as long as we can find the antiderivative of the function.
Sometimes, the function is too complicated to find the antiderivative.
At other times, we don’t even have a function, but only measurements taken from a real-life object.
What we need is an efficient method to estimate area when we can not find the antiderivative.
Trapezoid Rule
Trapezoidal Rule:
0 1 2 12 2 ... 22 n nhT y y y y y
where [a,b] is partitioned into n subintervals ofequal lengthh = (b – a)/n
This gives us a better approximation than either left or right rectangles.
Trapezoid Rule
2nn RRAMLRAMT
lyEquivalent
1 9 1 9 3 1 3 17 1 171 32 8 2 8 2 2 2 8 2 8
T
1 9 9 3 3 17 171 32 8 8 2 2 8 8
T
1 272 2
T
274
6.75
Trapezoid Rule
Consider the function f whose graph is shown below. Use theTrapezoid Rule with n = 4 to estimate the value of
9
1f x dx
A. 21 B. 22 C. 23 D. 24 E. 25
1 3 2 1 4 2 5 222
2 B
X
X
X
X
X
Trapezoidal Rule:
1 altitude sum of bases2 1 2 3 n
1 x y 2 y y ... y2
Error in Trapezoidal Rule:
3b2
2a
2
M b af x dx Trap n
nwhere M is the maximum value of
12 f" x
Midpoint Rule midpt. altitude sum of bases
Error in Midpoint Rule:
3b2
2a
2
M b af x dx Mid n
nwhere M is the maximum value of
24 f" x
Rules for definite integrals
( ) ( ) a b
b a
f x dx f x dx
( ) 0a
a
f x dx
( ) ( )b b
a a
kf x dx k f x dx
( ( ) ( )) ( ) ( )a a a
b b b
f x g x dx f x dx g x dx
( ( )) ( ) ( ) ,b c b
a a c
f x dx f x dx f x dx a c b
5. b c c
a b af x dx f x dx f x dx
Intervals can be added(or subtracted.)
a b c
y f x
Definite Integrals and Antiderivatives
6.
Definite Integrals and Antiderivatives
b
aabfdxxfabf )(max)()(min
a b
Example
a) (x2 x) 1
4
dx x3
3
x2
2
1
4
43
3
42
2
( 1)3
3
( 1)2
2
643
162
13
12
643
8 13
12
1416
Example:b) ex dx
0
3
ex 0
3 e3 e0
e3 1
Example:c) 1 2x
1x
1
e
dx x x2 ln x 1
e
(e e2 lne) (112 ln1) (e e2 1) (11 0) e e2 1 1 1 e e2 3
Example:
x3 3x 1 dx 1
2
x4
4
32
x2 x
1
2
24
4
32
22 2
1 4
4
32
1 2 1
4 6 2 14
32
1
0 214
214
Definite Integrals and Antiderivatives
Suppose that f and g are continuous functions and that
4
2
2
2
4
26)(,3)(,5)( dxxhdxxfdxxf
Find
2
4)( dxxf
4
2)( dxxf
2
2)(2 dxxf
3
3)( dxxf
2
4)( dxxf
4
2)](2)(3[ dxxhxf
3 2*5=10 -5 – -3 = -2
-3 + 5 = 2 0 3*(-3)- 2 * 6 = -21
Example: Northeast Airlines determines that the marginal profit resulting from the sale of x seats on a jet traveling from Atlanta to Kansas City, in hundreds of dollars, is given by
Find the total profit when 60 seats are sold.
P (x) x 6.
Example (continued): We integrate to find P(60).
P 60 P x dx0
60
x 6 dx
0
60
23
x3 2 6x
0
60
23
603 2 6 60
23
03 2 6 0
50.1613
The average value of a function is the value that would give the same area if the function was a constant:
212
y x
3 2
0
12
A x dx3
3
0
16
x276
92
4.5
4.5Average Value 1.53
Area 1Average Value Width
b
af x dx
b a
1.5
Definite Integrals and Antiderivatives
Average Value of a Function Over an Interval
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1.
Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to
.111
1 1
1
dxx
An antiderivative of 1 – x is . Therefore,2
2xx
.123
21
21
211
211
21
2211
21 221
1
21
1
xxdxx
So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.
Average Value of a FunctionSuppose f is integrable on [a,b]. Then the average value of f over [a,b] is
b
a
dxxfab
)(1
Find the average value of over the interval [0,4] xxf )(
34
6846
1
0461
32
41
041
3
23
40
234
0
xdxx
For what value(s) in the interval does the function assume the average value?
212
y x
4.5Average Value 1.53
1.5
Definite Integrals and Antiderivatives
5.121 2 x 32 x 3x
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
(Average Temperature) During a certain 12-hour period the temperature at time
t (measured in hours from the start of the period) was degrees. What was the average temperature during that period?
2
31447 tt
The average temperature during the 12-hour period from t = 0 to t = 12 is 12
0
3212
0
2
9247
121
31447
0121
tttdttt
9002047
9121221247
121 3
23
2
.degrees 550660121
The mean value theorem for definite integrals says that for a continuous function, at some point on the interval the actual value will equal the average value.
Mean Value Theorem (for definite integrals)
If f is continuous on then at some point c in , ,a b ,a b
1 b
af c f x dx
b a
Definite Integrals and Antiderivatives
If you were being sent to a desert island and could take only one equation with you,
x
a
d f t dt f xdx
might well be your choice.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus, Part 1
If f is continuous on , then the function ,a b
x
aF x f t dt
has a derivative at every point in , and ,a b
x
a
dF d f t dt f xdx dx
Fundamental Theorem of Calculus
a
xd f t dtx
f xd
2. Derivative matches upper limit of integration.
First Fundamental Theorem:
1. Derivative of an integral.
Fundamental Theorem of Calculus
a
xd f t dt f xdx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
First Fundamental Theorem:
Fundamental Theorem of Calculus
x
a
d f t dt f xdx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
New variable.
First Fundamental Theorem:
Fundamental Theorem of Calculus
cos xd t dt
dx cos x1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
sin xd tdx
sin sind xdx
0
sind xdx
The long way: First Fundamental Theorem:
Fundamental Theorem of Calculus
20
1 1+t
xd dtdx 2
11 x
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
Fundamental Theorem of Calculus
2
0cos
xd t dtdx
2 2cos dx xdx
2cos 2x x 22 cosx x
The upper limit of integration does not match the derivative, but we could use the chain rule.
Fundamental Theorem of Calculus
53 sin
x
d t t dtdx
The lower limit of integration is not a constant, but the upper limit is.
53 sin
xd t t dtdx
3 sinx x
We can change the sign of the integral and reverse the limits.
Fundamental Theorem of Calculus
2
2
1 2
x
tx
d dtdx e
Neither limit of integration is a constant.
2 0
0 2
1 1 2 2
x
t tx
d dt dtdx e e
It does not matter what constant we use!
2 2
0 0
1 1 2 2
x x
t t
d dt dtdx e e
2 2
1 12 222 xx
xee
(Limits are reversed.)
(Chain rule is used.)
2 2
2 222 xx
xee
We split the integral into two parts.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus, Part 2
If f is continuous at every point of , and if
F is any antiderivative of f on , then
,a b
b
af x dx F b F a
,a b
(Also called the Integral Evaluation Theorem)
Fundamental Theorem of Calculus
Definite Integral
Substitution
The substitution rule for definite integrals
If g’ is continuous on [a,b], and f is continuous on the range of u=g(x) then
( )
( )
( ( )) '( ) ( )g bb
a g a
f g x g x dx f u du
240
tan sec x x dx
Let tanu x2sec du x dx
0 tan 0 0u
tan 14 4
u
1
0 u du
Find new limits
new limit
new limit
12
0
12
u
12
Example
Example e
1
lnCompute .x dxx
1e 1 2
1 0 0
ln 1Hence .2 2
x tdx tdtx
The substitution ln is suggested bythe function to be integrated. We have
1 0 and e 1.
t x
x t x t
1 2 3
13 x 1 x dx
3Let 1u x 23 du x dx
1 0u
1 2u
122
0 u du
232
0
23
u Don’t forget to use the new limits.
322 2
3
2 2 23
4 23
Example
Integrals of Even and Odd Functions
Assume that f is an odd function, i.e., that
f f . Then f 0.a
a
x x x x dx
0
Assume that f is an even function, i.e., that
f f . Then f 2 f .a a
a
x x x x dx x dx
Theorem
Theorem
Integrals of Even and Odd Functions
Problem 5 7Compute cos .x xdx
Solution5 7
5 7
Observe that the function cos is odd.Since the interval of integration is symmetric with
respect to the origin, cos 0.
x x
x xdx
An odd function is symmetric with respect to the origin. The definite integral from -a to a, in the case of the function shown in this picture, is the area of the blue domain minus the area of the red domain. By symmetry these areas are equal, hence the integral is 0.
-aa