Integration by Parts: “Undoing” the Product Rule for Derivatives  Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu    Or ∫udv =uv−∫vdu ∫vdu should be simpler that the original ∫udv The integral If two functions are not related by derivatives (substitution does not apply), choose one function to be the u (to differentiate) and the other function to be the dv (to integrate)
View full slide show




Integration by Parts: “Undoing” the Product Rule for Derivatives  Look at the derivative of a product of functions: d dv du uv = u +v ( ) dx dx dx  Let’s use the differential form: d (uv ) =udv+ vdu  And solve for udv udv =d (uv) −vdu  Integrating both sides, we get: ∫udv = ∫d (uv) −∫vdu
View full slide show




Integration by Parts Evaluate:  y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) Chapter 7.1 March 22, 2007
View full slide show




Integration by Parts Evaluate:  y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) y 5 u =y 3 y 3 5 y 2 5 dv =ye dy 3y 2 5e 5 6y 25e 6 125e 0 625e y 5 y 5 y 5 y 5 y 5 y 5 5y e −75 y e + 750 ye −3750 e =e ( 5y 3 − 75y 2 + 750y − 3750 )
View full slide show




Integration by Parts ∫udv =uv−∫vdu 0 Evaluate:  ∫(2x + 6)e −x dx −1 Choose u (to differentiate (“du”)) dv (to integrate (“v”))
View full slide show




Integration by Parts Back to:  ∫udv =uv−∫vdu ∫x ln(x)dx Choose u (to differentiate (“du”)) dv (to integrate (“v”)) u =ln(x) 1 du = dx x dv = xdx x2 v= 2 x2 ⎛ x 2⎞⎞ ⎛ 1 ⎞ ∫x ln(x)dx = 2 ln(x) −∫ ⎜⎝ 22⎟⎠⎟⎠dx⎜⎝ x ⎟⎠ dx x2 x2 = ln(x) − + C 2 4 This second integral is simpler than the first
View full slide show




8.1 Integration by parts d dv du  uv  u  v Product Rule: dx dx dx d dv du dx  uv dx u dxdx  v dx dx uv udv  vdu udv  uv  vdu  
View full slide show




Integration by Parts d(uv) = udv + vdu udv = uv - vdu Show that xnexdx = xnex - nxn-1exdx + C let u = xn; dv = exdx then du = nxn-1dx; v = ex + C Thus, xnexdx = xnex - nxn-1ex dx + C 03/22/2019 rd 12
View full slide show




Bibliography Laredo Morning Times Online. March 2009. 21 March 2009. . Wikipedia. March 2009. 21March 2009. . Wikipedia. March 2009. 21 March 2009. . Wikipedia. March 2009. 21 March 2009. . Wikipedia. March 2009. 21 March 2009.
View full slide show




3 Integrate:  ∫x −1 x + 1 dx Choose u (to differentiate (“du”)) dv (to integrate (“v”))
View full slide show




Integration by Parts ∫udv =uv−∫vdu Integrate: ∫cos (ln(x))dx ∫cos (ln(x))dx =x cos ( ln(x)) + ∫ sin ( ln(x)) dx u =sin (ln(x)) 1 du = cos(ln(x))dx x dv =dx v =x ∫cos (ln(x))dx =x cos ( ln(x)) +x sin ( ln(x)) −∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx + ∫ cos ( ln(x)) dx 2 ∫cos (ln(x))dx =x cos ( ln(x)) + x sin ( ln(x)) 1 cos ln(x) dx = ( ) ( x cos ( ln(x)) + x sin ( ln(x))) ∫ 2
View full slide show




Integration by Parts Integrate: ∫cos (ln(x))dx ∫udv =uv−∫vdu
View full slide show




Course Plan  4 assignments 8 points each # 1 Given Feb 1, Due Feb 15 – Due Feb 17 – Returned #2 Given Feb 15, Due March 1 – Due March 8 - Returned #3 Given March1, Due March 29 – Given March 10 – due March 31 #4 March 29 Due April 12 – Given March 31, Due April 14  Two exams – 16 points each Exam #1: March 3 or March 10 – March 10 – will return March 31 - returned Exam #2: May 3 or May 10 – May 10  2 term papers – 10 points each - #1: Due March 3 or March 10 (depending on exam) - Returned - #2: April 19  Programming project: 16 points April 26  Total 100 points
View full slide show




DCAF From Executor From Scheduler M&C CMCS Correlation Results CBE Archive Pointers to Correlation Results DCAF Time Metadata Metadata Metadata Integration 1 Integration 2 ... Integration m Integration 1 Integration 2 ... Integration m Integration 1 Integration 2 ... SubScan 1 SubScan 2 Scan 1 SubScan 3 Exec Block 1 Integration m ... ... ... ... Integration 1 Integration 2 ... Metadata SubScan M Scan N Integration m 14 June, 2004 EVLA Overall Design Subsystems II Tom Morgan 21
View full slide show




Integration by Parts ∫udv =uv−∫vdu Examples: 3 x2 ∫x e dx ∫tan x ln(cos x)dx x x e cos(e )dx ∫ x e ∫ cos(4 x)dx
View full slide show




Or you can use Integration by Parts ⎛t ⎞ ∫cos ⎜⎝ 2 ⎟⎠ dx 2 ∫udv =uv−∫vdu
View full slide show




LO2 Supply Chain Integration Relationship RelationshipIntegration Integration Firm-to-Firm Firm-to-Firm Social Social Interactions Interactions Measurement Measurement Integration Integration Technology Technologyand and Planning PlanningIntegration Integration Operational Operational Planning Planningand and Control Control Material Material and and Service Service Supplier SupplierIntegration Integration Internal InternalOperations Operations Integration Integration Customer Customer Integration Integration Customer CustomerIntegration Integration Chapter 14 Copyright ©2012 by Cengage Learning Inc. All rights reserved 9
View full slide show




Data Returned: Parts List • Each resource is identified as a URL • Parts list has links to get each part’s detailed info • Key feature of REST design pattern – Client transfers from one state to next by examining and choosing from alternative URLs in the response document SOP Basics Slide 28
View full slide show




Budget Time Line Date set by each vice president or president Budget Managers will submit proposed budgets to divisional vice president or president Budget Office will submit revenue projections and salary and benefit projections to the Vice President for Financial & Administrative Affairs March 2, 2007 March 16, 2007 VPs will submit proposed budgets to VP for FAA Last date revisions for estimated budget will be accepted in the Accounting Office ** March 16, 2007 Budget Review Meetings March 19 – March 23, 2007 Evaluation of Budget Review Meetings by F&AA staff, recommendations by the vice president for F&AA to president, and president’s approved requests March 26 – March 30, 2007 ** Accounting Office will input budget data and prepare budget draft Final review of proposed budget by the vice presidents and president Proposed budget submitted to the TBR April 2 – April 6, 2007 ** April 9 - April 17, 2007 April 18, 2007 4
View full slide show




• Component Interfaces A component interface is a set of application programming interfaces (APIs) that you can use to access and modify PeopleSoft database information programmatically.  Integration Tools • Integration Broker PeopleSoft Integration Broker performs asynchronous and synchronous messaging among internal systems and third-party systems, exposes PeopleSoft business logic as web services to PeopleSoft and third-party systems, and consumes and invoke web services from third-party and PeopleSoft systems. • Integration Broker Administration Integration Broker Administration performs system administration tasks in PeopleSoft Integration Broker such as: • Set up and configure integration system components, such as messaging servers, nodes, integration gateways, listening and target connectors, and so on. • Configure the integration system to handle services, including specifying namespaces, setting up UDDI repositories, and so on. • Secure the integration environment by applying security at the web server, gateway, application server, node and service operation level. • Fine tune integration system performance by employing failover, master/slave processing, load balancing, and so on. • And more. • MultiChannel Framework PeopleSoft MultiChannel Framework (MCF) provides the tools that are required to support multiple channels of communication between customers (users) and agents. Some PeopleSoft applications, such as an email response management system (ERMS) from PeopleSoft CRM, use PeopleSoft MultiChannel Framework or you
View full slide show




Integration by parts udv  uv  vdu   Let dv be the most complicated part of the original integrand that fits a basic integration Rule (including dx). Then u will be the remaining OR Let u be a portion of the integrand whose derivative is a function simpler than u. Then dv will be the remaining factors (including dx).
View full slide show




Integration by parts udv  uv  vdu   x xe dx  x x x x u=x dv= exdx du = dx v = ex x xe dx  xe  e dx   x xe dx  xe  e  C 
View full slide show




Integration by parts udv  uv  vdu   2 x ln xdx  3 x x ln xdx  3 ln x  2 u = lnx dv= x2dx du = 1/x dx 3 3 x 1 x 3 x dx  3 ln x  3 3 x x x ln xdx  3 ln x  9  C 2 v = x3 /3 2 x 3 dx
View full slide show




Integration by parts udv  uv  vdu   arcsin xdx  u = arcsin x du  1 1 x 2 v=x dx x arcsin xdx x arcsin x   1  arcsin xdx x arcsin x  dv= dx x2 dx 2 1 x C
View full slide show




Integration by parts 2 u = x2 dv = sin x dx du = 2x dx v = -cos x x sin xdx  x 2 udv uv  vdu 2 sin xdx  x cos  2 x cos xdx u = 2x dv = cos x dx du = 2dx v = sin x 2 2 x sin xdx  x cos x  2 x sin x  2sin xdx  x 2 2 sin xdx  x cos x  2 x sin x  2cos x  C
View full slide show