Ekstrom Math 115b Mathematics for Business Decisions, part II Integration Math 115b.

Ekstrom Math 115b

Mathematics for Business Decisions, part II

Integration

Math 115b

Ekstrom Math 115b

Integration

Motivation

Revenue as an area under Demand function

.

-1.2

-10 q

D(q)

Demand Function

Revenue

qD(q)

qDqqR

Ekstrom Math 115b

Integration

Total Revenue

Total possible revenue is the revenue gained by charging the max price per customer

-1.2

-8

Demand Function

Total PossibleRevenue

Ekstrom Math 115b

Integration

Revenue

Consumer surplus – revenue lost by charging less Producer surplus – revenue lost by charging more

(i.e. “not sold” revenue)

-1.2

-8q

D(q)

Revenue

ConsumerSurplus

NotSold

Demand Function

Ekstrom Math 115b

Integration

Approx. area under curve

Counting rectangles (by hand) Using midpoint sums (by hand) Using Midpoint Sums.xlsm (using Excel) Using Integrating.xlsm (using Excel)

Ekstrom Math 115b

Integration

Counting Rectangles

Ex.

Approx. 9 rectangles

Each rectangle is 0.25 square units

Total area is approx. 2.25 square units0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

Ekstrom Math 115b

Integration

Midpoint Sums

given interval :,

givenfunction :

rectangles with sum :

ba

f

nSn

Notation

Meaning

bafSn ,,

Ekstrom Math 115b

Integration

Midpoint Sums

Process

Find endpoints of each subinterval

Find midpoint of each subinterval

nxxxxx ...,,,,, 3210

nmmmm ...,,,, 321

Ekstrom Math 115b

Integration

Midpoint Sums

Process (continued)Find function value at each midpoint

Multiply each by and add them all

This sum is equal to

nmfmfmfmf ...,,,, 321

imf x xmfxmfxmfxmf n ...321

bafSn ,,

Ekstrom Math 115b

Integration

Midpoint Sums

5.1

0.1

5.0

0

3

2

1

0

x

x

x

x

25.1

75.0

25.0

3

2

1

m

m

m

Ex. Determine where .

5.1,0,3 fS 246 xxxf

Ekstrom Math 115b

Integration

Midpoint Sums

Ex. (Continued)

375.2

5.025.15.025.25.025.1

5.025.15.075.05.025.0

5.1,0, 3213

fff

xmfxmfxmffS

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. (Continued)

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

Ekstrom Math 115b

Integration

Midpoint Sums.xlsm

x f (x ) a b= =

n x

6 0.00000

stuvw

31

Constants

Plot IntervalFormula for f (x )

0.00000

Intervals Midpoint Sum

Definition Computation

Number of subintervals (n) :

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

x

f(x)

n

ii xmf

1

)(

Ekstrom Math 115b

Integration

Midpoint Sums.xlsm

x f (x ) a b0 0 0 1.5

n x

3 0.50000

31


2.37500




0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

x

f(x)

n

ii xmf

1

)(

246 xxxf

Ekstrom Math 115b

Integration

Midpoint Sums.xlsm

x f (x ) a b0 0 0 1.5

n x

13 0.11538

31


2.25666




0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

x

f(x)

n

ii xmf

1

)(

x f (x ) a b0 0 0 1.5

n x

500 0.00300

31


2.25000




0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

x

f(x)

n

ii xmf

1

)(

Ekstrom Math 115b

Integration

Integrating.xlsm

File is similar to Midpoint Sums.xlsm

Notation: or or…. b

adxxf

b

adttf

Ekstrom Math 115b

Integration

Integrating.xlsm

x f (x ) A B a b= =

stu

vw

3132

#VALUE!

Constants

Formula for f (x )

Integration IntervalDefinition Computation Plot Interval

FUNCTION

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

x

f (x )

dxxfb

a )(

Ekstrom Math 115b

Integration

Integrating.xlsm

Ex. Use Integrating.xlsm to compute 4

1

6/61 dxe x

x f (x ) A B a b0.16667 0.16667 0 7 1 4

stuvw

3132

0.3331

Constants

Formula for f (x )Integration IntervalDefinition Computation Plot Interval

FUNCTION

-0.05

0

0.05

0.1

0.15

0.2

0 2 4 6 8

x

f (x )

dxxfb

a )(

Ekstrom Math 115b

Integration

Integrating.xlsm

Ex. (Continued) So . Note that

is the p.d.f. of an exponential random variable with parameter . This area could be calculated using the c.d.f. function

3331.04

1

6/61 dxe x 6/

61 xe

6 aFbF XX

Ekstrom Math 115b

Integration

Integrating.xlsm

Ex. (Continued)

3331.0

1535182751.0486582881.0

11

146/16/4

ee

FFaFbF XxXX

Ekstrom Math 115b

Integration

Signed Area

Values from Midpoint Sums.xlsm can be positive, negative, or zero.

Values from Integrating.xlsm can be positive, negative, or zero.

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. Suppose a demand function was found to be:

Determine the consumer surplus at a quantity of 400 units produced and sold.

196.321225.00001392. 2 qqqD

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. (Continued)

x f (x ) A B a b321.196 321.196 0 910 0 400

stuvw

3132

107508.8000

Constants

Formula for f (x )Integration IntervalDefinition Computation Plot Interval

FUNCTION

-50

0

50

100

150

200

250

300

350

0 200 400 600 800 1000

x

f (x )

dxxfb

a )(

Total Revenue at 400 units produced and sold

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. (Continued)

313233

FUNCTION

-50

0

50

100

150

200

250

300

350

0 200 400 600 800 1000

x

f (x )

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. (Continued)

Calculate Revenue at 400 units:

60.83569$

924.208400

400400

D

qDqqR

Ekstrom Math 115b

Integration

Consumer Surplus

Ex. (Continued) Take total revenue possible and subtract

revenue at 400 units

$107,508.80 - $83,569.60 = $23,939.20

So the consumer surplus is $23,939.20

Ekstrom Math 115b

Integration

Consumer Surplus

Formula for consumer surplus:

0

0 0

qqRdqqD

Ekstrom Math 115b

Integration

Integration Application

Income Stream revenue enters as a stream take integral of income stream to get total

revenue/income

Ekstrom Math 115b

Integration

Fundamental Theorem of Calculus

The derivative of with respect to x is

applies to p.d.f.’s and c.d.f.’s

aFbFdxxf XX

b

a X

x

duuf0 xf

Ekstrom Math 115b

Integration

Project (What to do)

Calculate the consumer surplus to answer Question #5

Use Integrating.xlsm (watch units)

= 459.99 - 360.86 = $99.13 million

1262.27391414.5344440.034403020.000053491262.273

0

2 Rdqqq

Ekstrom Math 115b Mathematics for Business Decisions, part II Integration Math 115b.

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