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


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


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


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


Circumscribed rectangles are all above the curve:

Inscribed rectangles are all below the curve:


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.


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.


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


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


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


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


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


a

xd f t dtx

f xd

2. Derivative matches upper limit of integration.

First Fundamental Theorem:

1. Derivative of an integral.


a

xd f t dt f xdx



3. Lower limit of integration is a constant.



x

a

d f t dt f xdx




New variable.



cos xd t dt

dx cos x1. Derivative of an integral.



sin xd tdx

sin sind xdx

0

sind xdx

The long way: First Fundamental Theorem:


20

1 1+t

xd dtdx 2

11 x





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.


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.


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.


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)


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

riemann sums and the definite integral. time velocity after 4 seconds, the object has gone 12 feet....

Documents