Chapter 6 Techniques of Integration

Save this PDF as:

Size: px
Start display at page:

Download "Chapter 6 Techniques of Integration"

Transcription

1 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln + C d = + C ln sin d = cos + C cos d = sin + C sec tn csc cot d = + C d = + C sec tn d= sec + C csc cot d= csc + C tn d = ln sec + C cot d = ln sin + C sec d = ln sec + tn + C d = sin + C d tn C = + + d = sin C d tn C + = + + d = sec + C

2 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6. Integrtion y Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Sustitution Rule for integrtion corresponds to the Chin Rule for differentition. The rule tht corresponds to the Product Rule for differentition is clled the rule for integrtion y prts. The Product Rule sttes tht if f nd g re differentile functions, then d f g = f g + g f d ( ) ( ) ( ) ( ) ( ) ( ) In the nottion for indefinite integrls this eqution ecomes or ( ) ( ) + ( ) ( ) = ( ) ( ) f g g f d f g ( ) ( ) + ( ) ( ) = ( ) ( ) We cn rerrnge this eqution s f g d g f d f g ( ) ( ) = ( ) ( ) ( ) ( ) f g d f g g f d This formul is clled the formul for integrtion y prts. It is perhps esier to rememer in the following nottion. Let u= f ( ) nd v g( ) = ( ) nd dv g ( ) du f d prts ecomes =. Then the differentils re =, so, y the Sustitution Rule, the formul for integrtion y udv = uv vdu Emple : Find cos d.

3 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Our im in using integrtion y prts is to otin simpler integrl thn the one we strted with. Thus in Emple we strted with cos dnd epressed it in term of the simpler integrl sin d. If we hd insted chosen u = cos nd dv = d, then du = sin d nd v= /, so integrtion y prts gives Although this is true, cos d= cos + sin d sin dis more difficult integrl thn the one we strted with. In generl, when deciding on choice for u nd dv, we usully try to choose u f ( ) = to e function tht ecomes simpler when differentited (or t lest not more complicted s long s ( ) dv = g d cn redily integrted to give v. Emple : Evlute ln d. Emple 3: Find cosπ d.

4 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 4: Evlute e cos d.

5 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 If we comine the formul for integrtion y prts with Prt of the Fundmentl Theorem of Clculus, we cn evlute definite integrls y prts. Assuming f nd g re continuous, nd using the Fundmentl Theorem, we otin If we let u= f ( ) nd v g( ) ( ) ( ) = ( ) ( ) ( ) ( ) f g d f g g f d =, then we hve = udv uv vdu Emple 5: Evlute 0 cos d

6 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 Section 6. Trigonometric Integrls In this section we use trigonometric identities to integrte certin comintions of trigonometric functions. Strtegy for Evluting sin m cos n d () If the power of cosine is odd ( n k ) cos = +, sve one cosine fctor nd use = sin to epress the remining fctors in terms of sine: Then sustituteu = sin. k ( ) ( ) m k+ m sin cos = sin cos cos () If the power of sine is odd ( n k ) sin d d m = sin sin cos d = +, sve one sine fctor nd use = cos to epress the remining fctors in terms of cosine: k ( ) ( ) k+ n n sin cos = sin cos sin d d k n = cos cos sin d Then sustituteu = cos. (Note tht if the powers of oth sine nd cosine re odd, either () or () cn e used.) (c) If the powers of oth sine nd cosine re even, use the hlf-ngle identities ( ) ( ) sin = cos cos = + cos It is sometimes helpful to use the identity sin cos = sin k

7 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 3 Emple : Evlute sin cos d. Emple : Evlute the integrl. ( π ) ( π ) ( ) sin cos 3 d ( ) 5 ( ) π 4 ()sin() c 0 sin d ( ) sin cos d d d

8 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 Strtegy for Evluting tn m sec n d () If the power of secnt is even ( n, k k ) =, sve the fctor of sec = + tn to epress the remining fctors in terms of tn : Then sustituteu = tn. k ( ) ( ) m k m tn sec d = tn sec sec d () If the power of tngent is odd ( n k ) k m = tn + tn sec d sec nd use = +, sve the fctor of sec tn nd use tn sec = to epress the remining fctors in terms of sec : Then sustituteu = sec. k ( ) ( ) k+ n n tn sec = tn sec sec tn d d k n sec sec sec tn = d Emple 3: Evlute the integrl () () π /4 0 sec tn sec tn d d

9 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 For other cses, the guidelines re not s cler-cut. We my need to use identities, integrtion y prts, nd occsionlly little ingenuity. We will sometimes need to e le to integrte tn y using the following formul: tn d= ln sec + C We will lso need the indefinite integrl of secnt: Emple 4: Find () sec d () 3 tn d sec d= ln sec + tn + C

10 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 0 Integrls such s the one the preceding emple my seem vry specil ut they occur m n frequently in pplictions of integrtion. Integrls of the form cot csc d cn e found y similr methods ecuse of the identity+ cot = csc. Finlly, we cn mke use of nother set of trigonometric identities: To evlute the integrl () sin m cos n d, () sin msin n d, or (c) cos m cos n d, Use the corresponding identity: ( )sin Acos B= sin ( ) sin ( ) A B + A+ B ( )sin Asin B= cos( ) cos( ) A B A+ B ( c)cos Acos B= cos( A B) + cos( A+ B) π / Emple 5: Evlute sin 4 cos d. 0

11 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6.3 Trigonometric Sustitution In finding the re of circle or n ellipse, n integrl of the form > 0. If it were stnds, d, the sustitution d rises, where u = would e effective ut, s it d is more difficult. If we chnge the vrile from to θ y the sustitution = sinθ, then the identity sin θ = cos θ llows us to get rid of the root sign ecuse ( ) ( ) = = = = sin θ sin θ cos θ cosθ In generl we cn mke sustitution of the form g( t) = y using the Sustitution Rule in reverse. To mke our clcultions simpler, we ssume tht g hs n inverse function; tht is, g is one-to-one. In this cse, if we replce u y nd y t in the Sustitution Rule, we otin ( ) ( ) ( ) = ( ) f d f g t g t dt This kind of sustitution is clled inverse sustitution. We cn mke the inverse sustitution = sinθ provided tht it defines one-to-one function. This cn e ccomplished y restricting θ to lie in the intervl[ π /, π /]. In the following tle we list trigonometric sustitutions tht re effective for the given rdicl epressions ecuse of the specified trigonometric identities. In ech cse the restriction on θ is imposed to ensure tht the function tht defines the sustitution is one-toone.

12 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi TABLE OF TRIGONOMETRIC SUBSTITUTIONS Epression Sustitution Identity π π = sin θ, θ + π π = tn θ, < θ < π 3π = sec θ, 0 θ < or π θ < sin θ = cos θ + tn θ = sec θ sec θ = tn θ Emple : Evlute 9 d

13 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Emple : Evlute + 4 d

14 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 3: Evlute d

15 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 Emple 4: Evlute 6+ 3 d

16 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 Section 6.4 Integrting Rtionl Functions y Prtil Frctions In this section we show how to integrte ny rtionl function ( rtio of polynomils) y epressing it s sum of simpler frctions, clled prtil frctions, tht we lredy know how to integrte. To illustrte the method, oserve tht y tking the frctions ( + 3) nd ( ) to common denomintor we otin ( ) ( ) ( )( ) = = If we now reverse the procedure, we see how to integrte the function on the right side of this eqution: function 7 d = d = ln + 3 ln + C To see how the method of prtil frctions works in generl, let s consider rtionl f ( ) ( ) ( ) P = Q where P nd Q re polynomils. It s possile to epress f s sum of simpler frctions provided tht the degree of P, denoted s deg(p), is less thn the degree of Q, denoted s deg(q). If f is improper, tht is, deg( P) deg( Q), then we must tke the preliminry step of dividing Q into P (y long division) until reminder R() is otined such tht deg( R) < deg( Q). The division sttement is ( ) ( ) where S nd R re lso polynomils. Emple : + 6 d + 4 ( ) ( ) P R f ( ) = = S( ) + Q Q

17 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 The net step is to fctor the denomintor Q( ) s fr s possile. It cn e shown tht ny polynomil Q cn e fctored s product of liner fctors (of the form + ) nd irreducile qudrtic fctors (of the form 4 ( ) Q = 6, we could fctor it s + + c, where 4 ( ) = 6 = ( + 4)( 4) = ( + 4)( + )( ) Q The third step is to epress the proper rtionl function R( ) / ( ) prtil frctions of the form A A+ B or i ( + ) ( + + c) CASE I: The denomintor Q( ) is product of distinct liner fctors. This mens tht we cn write ( ) = ( + )( + ) ( + ) Q k k j 4c< 0). For instnce, if Q s sum of where no fctor is repeted (nd no fctor is constnt multiple of nother). In this cse the prtil frction theorem sttes tht there eist constnts A, A,..., A such tht k ( ) ( ) R A A Ak = Q k k () Emple : Evlute + d 3 + 3

18 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 CASE II: Q( ) is product of liner fctors, some of which re repeted. Suppose the first liner fctor ( + ) is repeted r times; tht is, ( ) fctoriztion of Q( ). Then insted of the single term A ( ) would use r + occurs in the + in Eqution (), we A A Ar ( ) ( ) r () By wy of illustrtion, we could write Emple 3: Evlute A B C D E = ( + ) + ( + ) ( + ) d

19 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 CASE III: Q( ) contins irreducile qudrtic fctors, none of which is repeted. If Q( ) hs the fctor + + c, where 4c< 0, then, in ddition to the prtil frctions in Eqution () nd (), the epression for R( ) Q( ) will hve term of the form A + B + + c (3) where A nd B re constnts to e determined. For instnce, the function given y ( 3)( + )( + ) hs prtil frction decomposition of the form A B+ C D+ E = + + ( 3)( + )( + ) The term given in (3) cn e integrted y completing the squre nd using the formul d = tn + C + Emple 4: Evlute d

20 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 0 CASE IV: Q( ) contins repeted irreducile qudrtic fctor. Q hs the fctor ( c) If ( ) + +, where r 4c< 0, then insted of the single prtil frction (3), the sum A + B A + B A r + Br c + + c + + c ( ) ( ) r (4) occurs in the prtil frction decomposition of R( ) Q( ). Ech of the term in (4) cn e integrted y first completing the squre. Emple 5: Evlute + 3 d ( + )

21 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Section 6.5 Improper Integrls In defining definite integrl f ( ) d we delt with function f defined on finite intervl [, ] nd we ssumed tht f does not hve n infinite discontinuity. In this section we etend the concept of definite integrl to cse where the intervl is infinite nd lso to the cse where f hs n infinite discontinuity in [, ]. In either cse the integrl is clled n improper integrl. TYPE I: Infinite Intervls Definition of n improper integrl of type I t d eists for every numer t, then () If f ( ) ( ) = lim ( ) f d f d t t provided this limit eists (s finite numer). d eists for every numer t, then () If f ( ) t ( ) = lim ( ) f d f d t t provided this limit eists (s finite numer). The improper integrl f ( ) d nd f ( ) limit eists nd divergent if the limit does not eist. (c) If oth f ( ) d nd f ( ) d re clled convergent if the corresponding d re convergent, then we define ( ) = ( ) + ( ) f d f d f d In prt (c) ny rel numer cn e used.

22 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Emple : Determine whether the integrl d is convergent or divergent.

23 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 3 Emple : Evlute 0 e d.

24 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 4 Emple 3: Evlute d +

25 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 5 Emple 4: For wht vlues of p is the integrl d convergent? p

26 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 6 TYPE II: Discontinuous Integrnds Definition of n improper integrl of type II () If f is continuous on [, ) nd is discontinuous t, then ( ) = lim ( ) f d f d t t if this limit eists (s finite numer). () If f is continuous on (, ] nd is discontinuous t, then ( ) = lim ( ) + f d f d t if this limit eists (s finite numer). The improper integrl f ( ) divergent if the limit does not eist. t d is clled convergent if the corresponding limit eists nd (c) If f hs discontinuity t c, where c convergent, then we define c ( ) = ( ) + ( ) c < <, nd oth f ( ) d nd f ( ) f d f d f d c d re c Emple 5: Find 5 d

27 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 7 Emple 6: Determine whether π / sec dconverges or diverges. 0

28 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 8 d Emple 7: Evlute if possile 3 0

29 MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi 9 Emple 8: Evlute 0 ln d

The Product Rule state that if f and g are differentiable functions, then

The Product Rule state that if f and g are differentiable functions, then Chpter 6 Techniques of Integrtion 6. Integrtion by Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Substitution Rule for integrtion corresponds to the Chin Rule for differentition.

More information

MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations

MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx... Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting

More information

Chapter 8: Methods of Integration

Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

7. Indefinite Integrals

7. Indefinite Integrals 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find

More information

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into

More information

Math 3B Final Review

Math 3B Final Review Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

More information

Thomas Whitham Sixth Form

Thomas Whitham Sixth Form Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

More information

Formulae For. Standard Formulae Of Integrals: x dx k, n 1. log. a dx a k. cosec x.cot xdx cosec. e dx e k. sec. ax dx ax k. 1 1 a x.

Formulae For. Standard Formulae Of Integrals: x dx k, n 1. log. a dx a k. cosec x.cot xdx cosec. e dx e k. sec. ax dx ax k. 1 1 a x. Forule For Stndrd Forule Of Integrls: u Integrl Clculus By OP Gupt [Indir Awrd Winner, +9-965 35 48] A B C D n n k, n n log k k log e e k k E sin cos k F cos sin G tn log sec k OR log cos k H cot log sin

More information

Section 7.1 Integration by Substitution

Section 7.1 Integration by Substitution Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find

More information

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) = Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

More information

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?

How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

More information

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =. Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts

More information

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

Section - 2 MORE PROPERTIES

Section - 2 MORE PROPERTIES LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when

More information

Fundamental Theorem of Calculus

Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

More information

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

Chapters Five Notes SN AA U1C5

Chapters Five Notes SN AA U1C5 Chpters Five Notes SN AA U1C5 Nme Period Section 5-: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles

More information

Calculus AB. For a function f(x), the derivative would be f '(

Calculus AB. For a function f(x), the derivative would be f '( lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:

More information

Bridging the gap: GCSE AS Level

Bridging the gap: GCSE AS Level Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

5.5 The Substitution Rule

5.5 The Substitution Rule 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in

More information

Mat 210 Updated on April 28, 2013

Mat 210 Updated on April 28, 2013 Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

ntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x)

ntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x) ntegrtion (p) Integrtion by Inspection When differentiting using function of function or the chin rule: If y f(u), where in turn u f( y y So, to differentite u where u +, we write ( + ) nd get ( + ) (.

More information

RAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties

RAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IIT-JEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII

More information

Unit 5. Integration techniques

Unit 5. Integration techniques 18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd

More information

MA 124 January 18, Derivatives are. Integrals are.

MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

More information

Techniques of Integration

Techniques of Integration Chpter 8 Techniques of Integrtion 8. Integrtion by Prts Some Exmples of Integrtion Exmple 8... Use π/4 +cos4x. cos θ = +cosθ. Exmple 8... Find secx. The ide is to multiply secx+tnx both the numertor nd

More information

Spring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17

Spring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17 Spring 07 Exm problem MARK BOX points HAND IN PART 0 5-55=x5 0 NAME: Solutions 3 0 0 PIN: 7 % 00 INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE

More information

Chapter 9 Definite Integrals

Chapter 9 Definite Integrals Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

More information

Disclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.

Disclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know. Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Lesson 1: Quadratic Equations

Lesson 1: Quadratic Equations Lesson 1: Qudrtic Equtions Qudrtic Eqution: The qudrtic eqution in form is. In this section, we will review 4 methods of qudrtic equtions, nd when it is most to use ech method. 1. 3.. 4. Method 1: Fctoring

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

MATH 1080: Calculus of One Variable II Fall 2017 Textbook: Single Variable Calculus: Early Transcendentals, 7e, by James Stewart.

MATH 1080: Calculus of One Variable II Fall 2017 Textbook: Single Variable Calculus: Early Transcendentals, 7e, by James Stewart. MATH 1080: Clculus of One Vrile II Fll 2017 Textook: Single Vrile Clculus: Erly Trnscendentls, 7e, y Jmes Stewrt Unit 2 Skill Set Importnt: Students should expect test questions tht require synthesis of

More information

Chapter 8.2: The Integral

Chapter 8.2: The Integral Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in

More information

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous Anti-Derivtive : An nti-derivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

Techniques of Integration

Techniques of Integration 7 Techniques of Integrtion Shown is photogrph of Omeg Centuri, which contins severl million strs nd is the lrgest globulr cluster in our gl. Astronomers use stellr stereogrph to determine the ctul densit

More information

Taylor Polynomial Inequalities

Taylor Polynomial Inequalities Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil

More information

10 Vector Integral Calculus

10 Vector Integral Calculus Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

More information

The practical version

The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

More information

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

More information

Math 100 Review Sheet

Math 100 Review Sheet Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

More information

The Evaluation Theorem

The Evaluation Theorem These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for in-clss presenttion nd should not

More information

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists. AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

More information

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x) Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should

More information

Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable.

Optimization Lecture 1 Review of Differential Calculus for Functions of Single Variable. Optimiztion Lecture 1 Review of Differentil Clculus for Functions of Single Vrible http://users.encs.concordi.c/~luisrod, Jnury 14 Outline Optimiztion Problems Rel Numbers nd Rel Vectors Open, Closed nd

More information

Bob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk

Bob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions

More information

MATH 222 Second Semester Calculus. Fall 2015

MATH 222 Second Semester Calculus. Fall 2015 MATH Second Semester Clculus Fll 5 Typeset:August, 5 Mth nd Semester Clculus Lecture notes version. (Fll 5) This is self contined set of lecture notes for Mth. The notes were written by Sigurd Angenent,

More information

7.5 Integrals Involving Inverse Trig Functions

7.5 Integrals Involving Inverse Trig Functions . integrls involving inverse trig functions. Integrls Involving Inverse Trig Functions Aside from the Museum Problem nd its sporting vritions introduced in the previous section, the primry use of the inverse

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

Total Score Maximum

Total Score Maximum Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question 4 5 6 7 Totl Score Mimum 6 4 8 9 4 No clcultors or formul sheets. Check tht you hve 6 pges.. Find

More information

Jim Lambers MAT 169 Fall Semester Lecture 4 Notes

Jim Lambers MAT 169 Fall Semester Lecture 4 Notes Jim Lmbers MAT 169 Fll Semester 2009-10 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of

More information

NUMERICAL INTEGRATION

NUMERICAL INTEGRATION NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C

Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C Integrtion Mth 07H Topics for the first exm Bsic list: x n dx = xn+ + C (provided n ) n + sin(kx) dx = cos(kx) + C k sec x dx = tnx + C sec x tnx dx = sec x + C /x dx = ln x + C cos(kx) dx = sin(kx) +

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

FINALTERM EXAMINATION 9 (Session - ) Clculus & Anlyticl Geometry-I Question No: ( Mrs: ) - Plese choose one f ( x) x According to Power-Rule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information

Mathematics. Area under Curve.

Mathematics. Area under Curve. Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding

More information

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009 Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm-6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No

More information

Stuff You Need to Know From Calculus

Stuff You Need to Know From Calculus Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you

More information

Indefinite Integral. Chapter Integration - reverse of differentiation

Indefinite Integral. Chapter Integration - reverse of differentiation Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

0.1 Chapters 1: Limits and continuity

0.1 Chapters 1: Limits and continuity 1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem 0.1.1 Sndwich Theorem(F 96 # 20, F 97

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.-3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This

More information

Topics Covered AP Calculus AB

Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.

More information

Introduction to Algebra - Part 2

Introduction to Algebra - Part 2 Alger Module A Introduction to Alger - Prt Copright This puliction The Northern Alert Institute of Technolog 00. All Rights Reserved. LAST REVISED Oct., 008 Introduction to Alger - Prt Sttement of Prerequisite

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

0.1 THE REAL NUMBER LINE AND ORDER

0.1 THE REAL NUMBER LINE AND ORDER 6000_000.qd //0 :6 AM Pge 0-0- CHAPTER 0 A Preclculus Review 0. THE REAL NUMBER LINE AND ORDER Represent, clssify, nd order rel numers. Use inequlities to represent sets of rel numers. Solve inequlities.

More information

1 Techniques of Integration

1 Techniques of Integration November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

LINEAR ALGEBRA APPLIED

LINEAR ALGEBRA APPLIED 5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nth-order

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

Mathematics Number: Logarithms

Mathematics Number: Logarithms plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Calculus II: Integrations and Series

Calculus II: Integrations and Series Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

Chapter 3 Single Random Variables and Probability Distributions (Part 2)

Chapter 3 Single Random Variables and Probability Distributions (Part 2) Chpter 3 Single Rndom Vriles nd Proilit Distriutions (Prt ) Contents Wht is Rndom Vrile? Proilit Distriution Functions Cumultive Distriution Function Proilit Densit Function Common Rndom Vriles nd their

More information

a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution

a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution I.B Trigonometric Substitution Uon comletion of the net two sections, we will be ble to integrte nlyticlly ll rtionl functions of (t lest in theory). We do so by converting binomils nd trinomils of the

More information

( β ) touches the x-axis if = 1

( β ) touches the x-axis if = 1 Generl Certificte of Eduction (dv. Level) Emintion, ugust Comined Mthemtics I - Prt B Model nswers. () Let f k k, where k is rel constnt. i. Epress f in the form( ) Find the turning point of f without

More information

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0) 1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

More information