Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics§6.2

NumericalIntegration



Review §

Any QUESTIONS About• §6.1 → Integration by Parts, Use of

Integral Tables Any QUESTIONS

About HomeWork• §6.1 → HW-01

6.1

https://ecourses.ou.edu/cgi-bin/view_anime.cgi?file=ma08121.swf&course=ma&chap_sec=08.1','yes','240','196



§6.2 Learning Goals Explore the trapezoidal rule and

Simpson’s rule for numerical integration Use error bounds for numerical

integration Interpret data using

numerical integration

https://ecourses.ou.edu/cgi-bin/view_anime.cgi?file=ma08423.swf&course=ma&chap_sec=08.4','yes','240','196



Why Numerical Methods? Numerical

Integration • Very often, the function f(x) to differentiate, or the integrand to integrate, is TOO COMPLEX to yield exact analytical solutions.

• In most cases in Real World testing, the function f(x) is only available in a TABULATED form with values known only at DISCRETE POINTS



Numerical Integration Game Plan:

Divide Unknown Area into Strips (or boxes), and Add Up

To Improve Accuracy the TOP of the Strip can Be• Slanted Lines

– Trapezoidal Rule• Parabolas

– Simpson’s Rule• Higher Order

PolyNomials



Strip-Top Effect

Parabolic (Simpson’s) Form

Trapezoidal Form

• Higher-Order-Polynomial Tops Lead to increased, but diminishing, accuracy.



Strip-Count Effect

Adaptive Integration → INCREASE the strip-Count in Regions with Large SLOPES• More Strips of Constant

Width Tends to work just as well

10 Strips 20 Strips



AUC by Flat Tops

xfy

*5xf

x

xxfA jj *WidthHeight

N

jj

N

jj xxfAA

1

*

1



Trapezoidal Area By the Diagram at Right

• Side Heights:

• Width: Now “Stack Up” for 2A Then

orx

1jxf jxf

A

A

jxf jjxf

x

1jxf jxf

xxfxfA jj 12

x

xfxfA jj

21



AUC by Trapezoids

xxfxf

AA jjtrap

21

xxfxfAAN

jjj

N

jj

1

01

1

0 21

xfy

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=GvcYXTLtYOBRpM&tbnid=4al8WTaDBJuu6M:&ved=0CAUQjRw&url=http://www.math24.net/definite-integral.html&ei=VoHQUtrbOcf-oQTEpoGYBA&bvm=bv.59026428,d.cGU&psig=AFQjCNFm1Qx0h4baUnSAcNjwG0PtB4o_xA&ust=1389482658165213



The Trapezoidal Rule To Find the APPROXIMATE Area Under

the Curve given by y = f(x), and divided into vertical strips of equal width, Δx

• Where:

b

a

N

jjj xxfxfdxxf

1

012

1AUC

bxxa N 0

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=GvcYXTLtYOBRpM&tbnid=4al8WTaDBJuu6M:&ved=0CAUQjRw&url=http://www.math24.net/definite-integral.html&ei=VoHQUtrbOcf-oQTEpoGYBA&bvm=bv.59026428,d.cGU&psig=AFQjCNFm1Qx0h4baUnSAcNjwG0PtB4o_xA&ust=1389482658165213



Trapezoidal Rule Error AUC by the

Trapezoidal Approximation incurs error in the amount of

Where• n ≡ the strip count• K ≡ the maximum

value of |d2y/dx2|

2

3

12nabKEn



Trapezoidal Rule Error Example The Function does NOT have a Closed

Form, Analytical Solution

Calculate the Area Under the Curve for this function between x=1 & x=3 using a 10-strip Trapezoidal Calculation

???dxxe

xexfy

xx

1 1.5 2 2.5 30

1

2

3

4

5

6

7

x

y =

ex /x

MTH16 • Bruce Mayer, PE

MTH15 Quick Plot BlueGreenBkGnd 130911.m



Trapezoidal Rule Error Example Calculate

Δx Then make Fcn T-Table using

ThenTheT-Table

2.0102

1013

n

abx

4.202.2.2 7j e.g; 81 xxxx jjj x f(xj) = ex/x ΔA = [½][f(xj)+f(xj+1)]•Δx1 1.0 2.718282 0.5485045932 1.2 2.766764 0.5663335513 1.4 2.896571 0.5992216674 1.6 3.095645 0.6456560525 1.8 3.360915 0.7055443316 2.0 3.694528 0.7796806917 2.2 4.102279 0.8695269028 2.4 4.592990 0.9771350949 2.6 5.178361 1.10514489210 2.8 5.873088 1.25682671111 3.0 6.695179

Σtotal = 8.053574484

Then theApproximation

0536.82

1 dx

xex



Trapezoidal Rule Error Example ReCall from

Error Equation Taking the Derivative Twice

Plot d2y/dx2 to EyeBall Max Value

2

2

maxdx

xfdK

322

2

2 22xe

xe

xe

dxyd

xe

xe

dxdy

xey

xxxxxx

Maximumat x = 3



Trapezoidal Rule Error Example Then

Thus, to 5 Sig-Figs: Finally the Maximum 10-Strip,

Trapezoidal Error

333

3

2

33

32

2

275

272

92

31

32

32

3eeeee

dxydK

x

7195.3K

%48.20248.01012

137195.312 2

3

2

3

nabKEn



Simpson’s Rule The Simpson Method

tops TWO Strips with successive 3-pt Curve-Fit Parabolas

A Parabola can befit EXACTLY to ANY 3 (x,y) points

a, b, cycbxaxycbxaxycbxax

yxyxyx

for Solve,,,

333

222

111

33

22

11



Simpson’s Rule Since 3-pts defines 2-strips Simpson’s

Rule requires an EVEN Strip-Count Then for an Even Counting Number, n

• if M = max(|d4y/dx4|)then the Error

11321 42243

AUC

nnn

b

a

xfxfxfxfxfxfx

dxxf

4

5

180nabMEn



Simpson’s Rule Example Use Simpson’s rule with

n = 10 strips to approximate: SOLUTION From the Trapezoidal example Δx = 0.2 Now the SideWays T-Table

3

1 AUC dx

xex

j 1 2 3 4 5 6 7 8 9 10 11

x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

f(xj) = ex/x 2.718282 2.766764 2.896571 3.095645 3.360915 3.694528 4.102279 4.592990 5.178361 5.873088 6.695179

f(xj) CoEff 1 4 2 4 2 4 2 4 2 4 1f(xj)·CoEff 2.718282 11.067056 5.793143 12.382581 6.721831 14.778112 8.204558 18.371961 10.356722 23.492353 6.695179

Σ[f(xj)·CoEff] = 120.5817763AUC ≈ (Δx/3)·Σ[f(xj)·CoEff] = 8.038785084



Find Precise Value by MuPAD The Integrand Function

• fOFx := E^x/x Plot the AREA under the Integrand Curve

• fArea := plot::Function2d(fOFx, x = 1..3):plot(plot::Hatch(fArea), fArea)

The Precise Value• AUCn = numeric::int(fOFx, x=1..3)



Simpson’s Error Find Fourth Derivative by MuPAD

• d4fdx4 := diff(fOFx, x $ 4)

Then the 4th Derivative Plot• plot(d4fdx4, x=1..3, GridVisible = TRUE)

• Max at x=1

4645.2494

4

edx

fd



Simpson’s Error Then the

Error Calc

The Error comparing to MuPAD Value

• Thus the TextBook Formula is Conservative

44

5

1035.410180139

eEn

038714754.8038714754.8038785084.8

ActualActualCalc

nE

6108.7488 nE



NO Equation Functions Often in REAL LIFE “functions” disguise

themselves as “Data Tables” When I was Research Tech at

Lawrence Berkeley Lab (1978) I made Ventilation-Duct Volume-Flow measurements. A typical Data Set

r (in) V1 (ft/S) V2 (ft/S) V3 (ft/S) V4 (ft/S) V5 (ft/S) V6 (ft/S) Vavg (ft/S)2.15 24.1 24.3 27.6 27.3 25.7 28.1 26.24.38 15.1 13.9 13.1 13.9 14.4 14.8 14.25.62 3.9 3.8 3.9 3.4 3.6 3.9 3.7

12 inch OD Round Duct FlowSpeed Traverse



NO-Equation Functions I then had to Calculate the Duct Volume

Flow, Q, from the Data Table using the integration

This type of Integration OccursFrequently in the Physical, Life, and Social Sciences, aswell as in the Business world

rrVQ avg



NO-Eqn Integration Example The Cylindrical Tank

shown at right has a bottom area of 130 ft2 . The tank is initially EMPTY. To Fill the Tank, Water Flows into the top at varying rates as given in the Tank-Table below.

Time(min) 0 1 3 5 6 9 11 12 13 15 18

FlowRate(ft3/min) 0 80 130 150 150 160 165 170 160 140 120



NO-Eqn Integration Example For this situation determine the water

height ,H, at t = 18 minutes SOLUTION Use the TRAPEZOIDAL

Rule to Integrate the WaterFlow to arrive at the the Total Water VOLUME• Use the Max No. of

strips permitted by Data 0 2 4 6 8 10 12 14 16 180

20

40

60

80

100

120

140

160

t (min)

Q =

(ft3 /m

in))





NO-Eqn Integration Example Make ΔV Calcs

for the 10 strips Then by GeoMetry

So Finally

t (min) Q (cfm) Qavg (cfm) ΔV= Qavg•Δt0 0 40.0 40.01 80 105.0 210.03 130 140.0 280.05 150 150.0 150.06 150 155.0 465.09 160 162.5 325.0

11 165 167.5 167.512 170 165.0 165.013 160 150.0 300.015 140 130.0 390.018 120

Σtotal = 2492.5tnkwtr

tnkwtr

AVHor

HAV

ft 17.19ft52

997ft 130ft 5.2492 23 H



NO-Eqn Integration Example Note that in

this case Δx is NON-constant• 10 Strips of

Varying Width Thus

SIMPSON’s Rule Can NOT be Used• Simpson’s Rule

Requires constant Δx 0 2 4 6 8 10 12 14 16 18

0

20

40

60

80

100

120

140

160

t (min)

Q =

(ft3 /m

in))





MatLab C

ode% Bruce Mayer, PE% MTH-15 • 01Aug13 • Rev 11Sep13% MTH15_Quick_Plot_BlueGreenBkGnd_130911.m%clear; clc; clf; % clf clears figure window%% The Domain Limitsxmin = -6; xmax = 6;% The FUNCTION **************************************x = [0 1 3 5 6 9 11 12 13 15 18]; y = [0 80 130 150 150 160 165 170 160 140 120];% ***************************************************% the Plotting Range = 1.05*FcnRangeymin = min(y); ymax = max(y); % the Range Limitsxmin = min(x); xmax = max(x); % the Range LimitsR = ymax - ymin; ymid = (ymax + ymin)/2;ypmin = ymid - 1.025*R/2; ypmax = ymid + 1.025*R/2% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ypmin*1.05 ypmax*1.05];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenplot(x,y, '-d', 'LineWidth', 4),grid, axis([xmin xmax ypmin ypmax]),... xlabel('\fontsize{14}t (min)'), ylabel('\fontsize{14}Q = (ft^3/min)'),... title(['\fontsize{16}MTH16 • Bruce Mayer, PE',]),... annotation('textbox',[.53 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'MTH15 Quick Plot BlueGreenBkGnd 130911.m','FontSize',7)hold onplot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)stem(x,y, '-r.', 'LineWidth', 2)hold off



WhiteBoard Work Problems From §6.2

• P40 → Consumer’s Surplus

http://www.wikihow.com/images/a/a6/Calculate-Consumer-Surplus-Step-2-Version-2.jpg

http://www.wikihow.com/images/e/ec/Calculate-Consumer-Surplus-Step-6.jpg



All Done for Today

TrackingTrapezoids

http://1.bp.blogspot.com/-FwWggUQojps/UPAysUAGOzI/AAAAAAAAA9I/jXTtZlWVe1c/s1600/Trapezoids+A.jpg

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=pWcvubTSFSYqKM&tbnid=d1lV6UadwRKkPM:&ved=0CAUQjRw&url=http://wwcsd.net/groups/vintilescu/wiki/ef6f9/&ei=F5_RUrr0PM2CogTQ5IG4Dw&psig=AFQjCNEunp27wsLIWK9whSWveevy86zX-Q&ust=1389555832711297



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22







P6.2-40 MatLAB Codex = [0 4 8 12 16 20 24]y = [49.1200 42.9000 31.3200 19.8300 13.8700 10.5800 7.2500]ps = y-yminM = [1 4 2 4 2 4 1]CS1 = ps.*MCS2 = (4/3)*CS1CS3 = sum(CS2)CS4 = sum(CS1)CStot = (4/3)*CS4

% Bruce Mayer, PE% MTH-16 • 11Jan14% MTH15_Quick_Plot_BlueGreenBkGnd_130911.m%clear; clc; clf; % clf clears figure window%% The FUNCTION **************************************x = [0:4:24]; y = [49.12 42.9 31.32 19.83 13.87 10.58 7.25];% ***************************************************% the Plotting Range = 1.05*FcnRangeymin = min(y); ymax = max(y); % the Range Limitsxmin = min(x); xmax = max(x); % the Range LimitsR = ymax - ymin; ymid = (ymax + ymin)/2;ypmin = ymid - 1.025*R/2; ypmax = ymid + 1.025*R/2ypmin =0% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ypmin*1.05 ypmax*1.05];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([1 1 1]); % Chg Plot BackGround to Blue-Greenplot(x,y, '-d', 'LineWidth', 4),grid, axis([xmin xmax ypmin ypmax]),... xlabel('\fontsize{14}q (kUnits)'), ylabel('\fontsize{14}p ($/Unit)'),... title(['\fontsize{16}MTH16 • Bruce Mayer, PE',]),... annotation('textbox',[.53 .05 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'MTH15 Quick Plot BlueGreenBkGnd 130911.m','FontSize',7)hold onplot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)stem(x,y, '-r.', 'LineWidth', 2)plot([xmin, xmax], [7.25 7.25], '-.m', 'LineWidth', 3)hold off



Example NONconstant ∆x Pacific Steel Casting Company (PSC) in

Berkeley CA, uses huge amounts of electricity during the metal-melting process.

The PSC Materials Engineer measures the power, P, of a certain melting furnace over 340 minutes as shown in the table at right. See Data Plot at Right.

0 50 100 150 200 250 300 3500

50

100

150

200

250

time, t (min)

Pow

er C

onsu

mpt

ion,

P (k

W)

Furnace Power Consumption



Example NONconstant ∆x The T-table at Right displays

the Data Collected by the PSC Materials Enginer

Recall from Physics that Energy (or Heat), Q, is the time-integral of the Power.

Use Strip-Integration to find theTotal Energy in MJ expended byThe Furnace during this processrun

Time (min)

Power (kW)

0 47 24 107 45 104 74 146 90 126

118 178 134 147 169 211 180 151 218 233 229 184 265 222 287 180 340 247



Example NONconstant ∆x GamePlan for Strip Integration Use a Forward Difference

approach• ∆tn = tn+1 − tn

– Example: ∆t6 = t7 − t6 = 134 − 118 = 16min → 16min·(60sec/min) = 960sec

• Over this ∆t assume the P(t) is constant at Pavg,n =(Pn+1 − Pn )– Example: Pavg,6 = (P7 − P6)/2 =

(147+178)/2 = 162.5 kW = 162.5 kJ/sec

Time (min)

Power (kW)

0 47 24 107 45 104 74 146 90 126

118 178 134 147 169 211 180 151 218 233 229 184 265 222 287 180 340 247



Example NONconstant ∆x The

GamePlan Graphically• Note the

VariableWidth, ∆x,of the StripTops

t (minutes)

P (k

W)

MTH15 • Variable-Width Strip-Integration

0 50 100 150 200 250 300 3500

25

50

75

100

125

150

175

200

225Bruce May er, PE • 25Jul13

4x

9x



MATLA

B C

ode% Bruce Mayer, PE% MTH-15 • 25Jul13% XY_Area_fcn_Graph_6x6_BlueGreen_BkGnd_Template_1306.m%clear; clc; clf; % clf is clear figure%% The FUNCTIONxmin = 0; xmax = 350; ymin = 0; ymax = 225;x = [0 24 24 45 45 74 74 90 90 118 118 134 134 169 169 180 180 218 218 229 229 265 265 287 287 340]y = [77 77 105.5 105.5 125 125 136 136 152 152 162.5 162.5 179 179 181 181 192 192 208.5 208.5 203 203 201 201 213.5 213.5]% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green% Now make AREA Plotarea(x,y,'FaceColor',[1 0.6 1],'LineWidth', 3),axis([xmin xmax ymin ymax]),... grid, xlabel('\fontsize{14}t (minutes)'), ylabel('\fontsize{14}P (kW)'),... title(['\fontsize{16}MTH15 • Variable-Width Strip-Integration',]),... annotation('textbox',[.15 .82 .0 .1], 'FitBoxToText', 'on', 'EdgeColor', 'none', 'String', 'Bruce Mayer, PE • 25Jul13','FontSize',7)set(gca,'XTick',[xmin:50:xmax]); set(gca,'YTick',[ymin:25:ymax])set(gca,'Layer','top')



Example NONconstant ∆x

The NONconstant Strip-Width Integration is conveniently done in an Excel SpreadSheet

The 13 ∆Q strips Add up to 3456.69 MegaJoules of Total Energy Expended

n Time, t Power ∆t = 60*(tn+1-tn) Pavg=(Pn+1−Pn)/2 ∆Q= Pavg*∆t

(cnt) (min) (kW) (Sec) (kW) (kJ)1 0 471 1440 77 1108802 24 1072 1260 105.5 1329303 45 1043 1740 125 2175004 74 1464 960 136 1305605 90 1265 1680 152 2553606 118 1786 960 162.5 1560007 134 1477 2100 179 3759008 169 2118 660 181 1194609 180 1519 2280 192 437760

10 218 23310 660 208.5 13761011 229 18411 2160 203 43848012 265 22212 1320 201 26532013 287 18013 3180 213.5 67893014 340 247

3456.69Total Energy in MJ = (∑∆Q)/1000 =

Documents

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]