Mat 210 Updated on April 28, 2013


 Shannon Berry
 4 years ago
 Views:
Transcription
1 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 denomintor: Function: Given f ( ) Given f( ) Given f ( ) C then f ( ) C, f() C, f ( h) C. h h then f ( h) f ( ) h ( h) ( h) ( h) then f ( h) f ( ) h h h h h h h h h h( h ) h Given f ( ) then f ( h) f ( ) h h h h h h h h h h h h h h h
2 Mt Brief Clculus Section. A Dsh of Limits (Pge# ) In this section we will formulte some importnt rules for limits. e Emple. (). Use conjecture to determine the limit lim. e Solution. We use clcultor to produce the tle of vlues. Consider F()= F() From the tle it ppers tht s gets closer to, the function F() gets closer to. e Therefore we conclude lim e Emple. (). Algericlly determine the limit lim. Solution. e lim = / / 6 lim lim( / / 6 higher order terms) Section. More on Limits (Pge# 8) Limits t infinity: we sy tht f() hs the limit l s tends to infinity if f() cn e mde ritrrily close to l y mking sufficiently lrge. We write lim f ( ) l. Limits tht do not eists: Emple. (). Consider the emple lim Solution. We use clcultor to produce the tle of vlues. Consider F()= F() Oserve tht ut. Thus limit does not eist Emple. (). Determine the limit of the rtionl function lim Solution. 7 9 lim Emple. (). Determine the limit of the rtionl function lim
3 Mt Brief Clculus 5 5 Solution. lim 7 9 lim Emple. (). Determine the limit of the rtionl function Solution. lim lim lim DNE, tht mens the limit does not eist. Emple. (4). Determine the limit of the function Solution. lim lim lim. Test the following limits: Eercise. (). lim ( ) Eercise. (). lim ( ) Eercise. (). lim Eercise. (4). lim 7 4 Eercise. (5). lim Eercise. (6). lim Eercise. (7). lim e Section. Continuity (Pge# 5) Definition: Geometriclly, function is continuous on n intervl if its grph is connected tht mens, it hs no reks (no holes, no jumps). One sided limit of function: Right hnd nd left hnd limit: The right hnd limit of f() t point is written s lim f ( ) nd the left hnd limit t is lim f ( ). h h
4 Mt Brief Clculus The limit of function t point eits if lim f ( ) = lim f ( ) nd we write lim f ( ) l, where l is finite numer. h h h Continuity of function in terms of limits: The function f() is continuous t point in n intervl if lim f ( ) f ( ), tht mens the function hs limit t the point h nd is equl to f(), the vlue of the function t. Otherwise the function is discontinuous. Emple. (). Let the function f() e defined for ll vlues of y, for f ( ), for Drw the grph of the function f() nd test the continuity from the grph. The function f() is discontinuous t = Becuse lim f ( ) nd lim f ( ) nd f ( ). In this cse lim f ( ) does not h h h eist. Emple. (). Determine the vlue of t which the function f ( ) is continuous. Solution. The domin of the function is (,). The function is not defined t =. Thus the function is continuous for ll vlues of in the intervl (,). See the grph of f().
5 Mt Brief Clculus Section.4 Intermedite Vlue Theorem. (Pge# ) Theorem: (Intermedite Vlue Theorem) Let f e function continuous in [, ] nd ssume tht f() nd f() hve different signs. Then there is t lest one c in (, ) such tht f(c) =, tht mens the grph of f crosses the is. This theorem is importnt in ssuring the eistence of solutions to some equtions tht cnnot e solved eplicitly. Emple.4 (). Prove tht the eqution 5e ( ) hs solution in the intervl (, 5). Solution. Consider tht f() = 5e ( ). Verify tht f() = 5, nd f(5) = 9.. By the Intermedite Vlue Theorem, since we hve f() nd f(5) hve different signs, it hs solution in the intervl (, 5). Note tht for the function f() which is strictly incresing/decresing with f() nd f() different signs hs ectly one solution in (, ). Emple.4 (). Prove tht the eqution 5e ( ) intervl (, 5), which is unique. hs solution in the Solution. In emple, we hve seen tht f() hs solution in (, 5). Oserve tht f '( ) 5e ( ) e 5e ( ) 5e ( ) for ll. The function is incresing. Section.5 The Derivtive (Pge # ) When we study the grph of function, we would like to hve precise mesure of the steepness of the grph t point. For the liner eqution y m, m is the slope. If m is lrge nd positive, then the line rises steeply from left to right; if m is lrge nd negtive, then the line flls steeply. For n ritrry function f, the steepness of the curve t prticulr point is the slope of the tngent line to the curve t tht point. f ( h) f ( ) The slope mpq of the secnt PQ is defined y m PQ is lso clled the h Newton quotient of the function f. Note tht when h =, the frction ecomes /, which is undefined. But choosing h = corresponds to letting Q = P. When Q moves towrd P long the grph of the function f, the coordinte of Q, which is +h, must tend to, nd so h tends to. Simultneously, the secnt line PQ tends to the tngent line to the grph t P. f ( h) f ( ) Thus the definition f '( ) lim h h is clled the slope of the tngent line t P to the grph of the function f. The numer f '( ) is lso clled the derivtive of the function f t the point.
6 Mt Brief Clculus The eqution of the tngent line to the grph of y f () t the point (, f()) is y f ( ) f '( )( ) Def: f '( ) the slope of the tngent line to the curve y f () t the point (, f ( )) f ( ) f ( ) Rise Slope of line thru (, f ( )) nd (, f ( )) is Run Emple.5 (). Determine the slope of the lines given elow.. y 5. y 5 c. 4 y 8 Answer:. ,. , c. /8 Emple.5 (). The grph of f () is given elow. Find f '( ), f '(). 5 8 Answer: f '( ) 8. 75, f '() Emple.5 (). Compute f '( ) nd f '( ) of the given functions.. f ( ). f ( ) 6 c. f ( ) Emple.5 (4). Find the slope of the tngent line to the grph of f t the specified points nd lso find the eqution of the tngent line.. f ( ) t (, ). f ( ) 6 t (, ) c. f ( ) t (, ) Section.6 Rtes of Chnge nd Incresing nd decresing functions (Pge# ) The derivtive of function is defined s the slope of the tngent line to its grph t prticulr point. Suppose y is function of, which is defined s y f (). Let us consider two different vlues of s +h nd, the corresponding y vlues re f(+h) nd f(). In economics the chnge in y per unit chnge in hs prticulr nme,
7 Mt Brief Clculus The verge rte of chnge of f over the intervl from to +h is f ( h) f ( ) defined s h f ( h) f ( ) The instntneous rte of chnge of f t is defined s f '( ) lim h h f '( ) The reltive rte of chnge of f t is defined s f ( ) For the following functions C() = cost of producing units R() = p = revenue from selling units, where p is the selling price per unit P()= R() C() = profit from producing nd selling units Now we hve the following functions C'( ) R'( ) P'( ) C( h) C( ) lim = Mrginl cost t h h R( h) R( ) lim = Mrginl revenue h h P( h) P( ) lim = Mrginl profit h h Incresing nd decresing of functions: If f ( ) f ( ) whenever, then f is incresing If f ( ) f ( ) whenever, then f is strictly incresing If f ( ) f ( ) whenever, then f is decresing If f ( ) f ( ) whenever, then f is strictly decresing Emple.6 (). Emine where f ( ).5 is incresing/decresing. Solution. Using definition one cn esily find tht f '( ), which is lwys for ll nd for ll. Thus f() is incresing in [, ) nd decresing in (, ]. Oserve the grph of the function.
8 Mt Brief Clculus Emple.6 (). Emine where f ( ) / is incresing/decresing. Solution. Oserve tht f '( ) 4 ( )( ). Using sign digrm one cn see tht f '( ) 4 ( )( ) in the intervl [, ] nd f '( ) 4 ( )( ) in the intervl (,] [, ). Oserve the grph elow. Emple.6 (). If the cost function of firm is C ( ), give economic interprettion of the constnts. Solution. The constnt, is interpreted s the vrile cost or mrginl cost in dollrs per unit of production nd the constnt is interpreted s the fied cost in dollrs. Emple.6 (4). The totl cost of producing units of commodity is C ( ) 9 75,. Compute the mrginl cost function. For which vlue of is the mrginl cost the lest? Solution.. The mrginl cost is C '( ) 8 75,. The mrginl cost function C' ( ) decreses in [,] nd increses in the intervl [, ). Thus the mrginl cost is lest for =. See the grph of C '( ) 8 75,.
9 Mt Brief Clculus Section.7 Simple Rules for Differentition (Pge# 7) f ( h) f ( ) The derivtive of function f is defined y the formul f '( ) lim, if h h this limit eists, we sy tht f is differentile t. The process of finding the derivtive is clled the differentition. Other symols used in differentition. The derivtive of function f is lso represented df f ( h) f ( ) y Df f '( ) lim d h h Formul. The power rule. For the function of the type then f '( ). Derivtive of ny constnt is. f ( ), where is constnt, Emple.7 (). Determine the derivtive of the given power functions.. f ( ). f ( ) c. f ( ) +5 Solution.. f '( ). f '( ) / c. f '( ) 4 / Section.8 Sums, Products nd Quotients (Pge# 4) Following re the differentition rules for the functions with different opertions. F( ) F( ) F( ) f ( ) g( ), then F'( ) f '( ) g'( ) f ( ) g( ), then F'( ) f '( ) g( ) f ( ) g'( ) f ( ) g( ) f '( ) f ( ) g'( ), then F'( ) g( ) [ g( )] Emple.8 (). Differentite the functions (ddition/sutrction).. f ( ). f ( ) 5 c. f ( ) Solution.. f '( ). f '( ) c. f ( ) / ' 4 Emple.8 (). Differentite the functions (multipliction/division). 5. f ( ) ( ). f ( ) ( )(5) c. f ( ) 9 9 Solution.. f '( ) ( ) ( ). f '( ) (6 )5 ( ) 5 (6) ( 5) c. f '( )
10 Mt Brief Clculus Section.9 Chin Rule (Pge# 47) The chin rule is dy d dy du du d dy Emple.9 (). Find y chin rule. d. y 5 u nd u. Solution... c. dy d dy d dy dy du d du d f ( ) ( ) c. f ( ) u ( 5 ) 5 ( 9 9 ( ) (6 ) 6 ( Section. Higher Order Derivtives (Pge# 5) dy The symols of higher order derivtives. The first derivtive of y is or y ' d d y The second order derivtive is d d y The third order derivtive is d d y Emple. (). Find. d. y 5. f ( ) ( ) Solution.. df d. dy d ( ) d y d d y d ) , 9, 7 (), d f d ()( ) 9 ( ), d f d Section. Eponentil nd Logrithmic Functions (Pge# 56) Formul. The eponentil rule. For the function of the type constnt, then f '( ) ln. Derivtive of f ) 9 ( ) e is or y " or y nd so on. ()()(9)( ) f f 8 ( ( ), where is '( ) e. ) dy Emple. (). Find. d. f ( ). f ( ) e e 5
11 Mt Brief Clculus Solution.. df 9 df 5 6 ln,. e 5e d d Formul. For the logrithmic function f ( ) ln, Derivtive of f ( ) log is dy Emple. (). Find. d f '( ). ln f '( ).. f ( ) ln. f ( ) log df df Solution..,. d d ln Section. Implicit Differentition (Pge# 67) 5 Functions like y 5y 9y, where cnnot e written s independent vrile with respect to nother vrile y. In this cse direct differentition is either difficult or impossile. To find derivtive we will use implicit differentition using chin rule. Following emples will illustrte the sitution. dy Emple. (). Find or y'. d. y 5y 9. y log 5y 9 Solution.. 6yy' 5y 5y' 9 y' 6y 5. yy' y' ln y ln Emple. (). Find the slope of the tngent line to the curve y 7 t (, ). Also compute the second derivtive. 4 y 4() () 5 Solution. y ' y 4 y' () For the second derivtive we consider gin y ' y 4 4 6y' 6y 4 4 Tking derivtive we find y " y' y' 4 y" Derivtive formuls: d n n ( ) n d ( e ) d e d ( ) ln (ln ) d d d d d ( c ) d Rules: Product Quotient
12 Mt Brief Clculus ( fg) fg f g f f g g f f g fg g g f f g g d f g f g g d Chin Rule: ( ( ( )) ( ( )) ( ) Tngent line: y f ( ) f ( )( ) Derivtive Prctice Prolems: Set A d 9. ( ) d. d 4. ( ) d d d ( ) d e e e e d where (, f ( )) is the point on y f ( ) 8. d d ( ) () ln. (7 ) 7 ln 7() d d d d ( ) 6. d d ( ) d d ( ) d d.. ln ln e d 4 d ( ) d 8 d ( ) 4 4 d 9. e e d d 4 d ( ) 4 4 d 4 d 6. 5 e ( ) 5 e ( ) d d 6 d e d 7 7 d 7 7(ln ) e 8. ln 9. ln d e d d d 7 7 d d ln( ) ln. ln. e 7 e d d d Set B d 9 d. ( ) d. ( ) ln. ( e ) d d d d d d / 4. ( ) 5. ( ) / d d d d d / d d 6. ( ) ( ) 7. ( / d p ) 8. p p d d d d d d (9 7 ) (8 7) d
13 Mt Brief Clculus. p d p p p d ( ) d ln ln ln. d. d 6. d 5. ( ln ) 5 4 d ln d p p ln, p p d d d ln p 4. d ( ln ) ln d d d 4 4 ln d ( ). d d d d e e d ln ( ln ) d d. ln d d ln ln d. Section. Differentiting the Inverse (Pge# 74) If function y f ( ) is onetoone (strictly incresing or strictly decresing) function over n intervl I, it hs n inverse function gover ( ) the rnge of y f ( ). A function, which is strictly incresing (or strictly decresing) in n intervl I, it is onetoone. We hve the following importnt result: If the function y f ( ) is continuous nd strictly incresing (or strictly decresing) in n intervl I, then the function y f ( ) hs n inverse function y g( ), which is lso strictly incresing (or strictly decresing) in the intervl f() I, the rnge of y f ( ). If is n interior point of I nd y f ( ), then gis ( ) differentile t y f ( ) such tht g( y), where y f f ( ) ( ) y g( ) y ( y, ) y f ( ) (, y ) Emple. () Test the function f ( ) if it hs n inverse. Then find the inverse if ny. Consider point (,5) on the grph of the function f ( ). Find derivtive of the inverse function nd evlute t the point (5, ) on the inverse function.
14 Mt Brief Clculus Solution: The function f ( ) is strictly incresing; it is onetoone (Verify the grph y horizontl line test). It hs inverse. y g( ) To find inverse follow the steps: Step. Write y y Step. Interchnge vriles: y y f ( ) Step. Solve for y: y Step 4. The inverse is f ( ) g( ) Now f ( ) g( ) We now verify y the formul g( y) g(5) f ( ) f () Emple. () The function f ( ) ln, (.68, ) hs n inverse function g(). Eplin why. And find g () e Solution: Grph the function using your clcultor nd find tht it hs minimum t.68 nd it is incresing on (.68, ). Tht is why the function hs n inverse. To find g () e, we first solve for from e ln e. Finlly we hve g() e f ( ) ln e Emple. () The function f ( ) ln hs n inverse function g(). Eplin why. And find g( e ). Find eqution of the tngent line to y = g() t (e+, e). Solution: The given function is incresing, it hs inverse. The eqution of the tngent line is e( e ) y e g( e )( e ) e ( e ) e f ( e) e 5 Emple. (4) The function f ( ) 6 hs n inverse function g(). Eplin why. Find g (7). Find eqution of the tngent line to y = g() t (7, ). Solution: Oserve tht 5 f ( ) 6 is onetoone nd solving the eqution 5 7 6, we find. f() Thus (, 7) is point on the grph. g() Now g(7), f () (, 7) tngent line 4 where f ( ) (7,) The eqution of the tngent line to y = g() t (7, ) is
15 Mt Brief Clculus y ( 7) Note tht it is very esy to drw the grph for the inverse function, ut very difficult or impossile for us to find the function lgericlly. Look t the difficulty: Step. 5 y 6 5 Step. y y 6y Step. Solving for y is impossile lgericlly. But using the formul we re le to find derivtive of the inverse function t given point. Emple. (5) Given tht y f ( ) is onetoone nd f(4) 5, f(4) /, find ( f ) (5) Solution: Oserve tht (4, 5) is point of y f ( ). Now ( f ) (5) f (4) Section. Liner Approimtions (Pge# 76) f ( h) f ( ) In chpter, we hve used the rule of derivtive f '( ) lim t point h h f ( ) f ( ) (, f()). We now use nother modified rule f '( ) lim s generl rule of h derivtive t point (, f()). From this generl form we omit the limit nd use the result f ( ) f ( ) s n pproimtion like f '( ). After simplifiction we find f ( ) f ( ) f '( )( ), which is clled the liner pproimtion (lso clled tngent line pproimtion) of the function f out =, ( is close to ). Emple. (). Prove tht for close to, nd illustrte this pproimtion y drwing the grphs of y nd y on the sme screen. We consider f ( ) f '( ) f '(), s Now using liner pproimtion formul we find f ( ) f ( ) f '( )( ) f ( ) f () f '()( )
16 Mt Brief Clculus Emple. (). Prove tht to find pproimtion of ( ) m m for close to nd use this pproimtion / ( 6.95) Emple. (). Use the liner nd qudrtic pproimtion of f ( ) to / ( 6.95). pproimte the vlue of / Solution. From f ( ) we hve f '( ) nd f "( ) 9 The Liner Approimtion t =, is f ( ) f ( ) f '( )( ) Consider = 7, then (7) / f 7, f '(7) (7) 7 / Then (6.95) (6.95 7) Rules for Differentils. The nottion d( f ( )) is clled the differentil of the function f. We hve the following rules. d( f ( ) g( )) df ( ) dg( ) or simply we write. d( f. d( f g) df. d( fg) gdf / g) fdg dg gdf fdg g Emple. (4). Find the differentils (p, q, nd r re constnts):. d( ). d( ln ) c. d( p q) r Solution:.. c. d( ) 6d d d( ln ) 6ln d d ( ) r r d p q r( p q) pd 5 /
17 Mt Brief Clculus Section.4 Why We Use Elsticity (Pge # 8) The elsticity of function y f ( ) with respect to is defined s E El f ( ) f ( ) f( ) Interprettion: The mening of E % is tht n increse of % in the price would led to decrese of % of demnd, if ll other fctors tht influence the demnd remined constnt. A Emple.4 () Find the elsticity of y f ( ), A is constnt, nd interpret your nswer. A / A 5/ Solution: y f ( ) A f ( ) A 5/ Now E El f ( ) f ( ).5 f( ) A An increse of % in the price the demnd would decrese y.5%, when other fctors re unchnged. Emple.4 () The reltion etween trffic volume nd ependiture of uilding rods.6 re given s T( K).4K, K is ependiture. Find the elsticity of T with respect to K. Find lso the consequences if ependiture increses y %. K K.6 Solution: E El f ( ) T( ).4.6K.6.6 T( K).4K The increse in % of ependiture would led to.6% increse in trffic volume. The % increse in price would led to.6% increse in trffic volume. Emple.4 () Prove tht p Solution: p E El f ( ) pel f ( ), where p is constnt. p El f ( ) p( f ( )) f ( ) pelf ( ) p ( f( )) Emple.4 (4) Find E El 5 Solution: ( ) E El ( ) ( )( ) Section. Single Vrile Optimiztion (Pge# 95) Introduction: Mimum nd Minimum Vlues of function.
18 Mt Brief Clculus The points in the domin of function where it reches its lrgest nd its smllest vlues re usully clled mimum nd minimum vlues. These re sometimes clled etreme vlues. Suppose the function f() domin D, then c D is mimum point for f iff f ( ) f ( c) for ll in D d D is minimum point for f iff f ( ) f ( d) for ll in D f(c) is clled mimum vlue of f t = c, nd f(d) is clled minimum vlue of f t = d. 8 Emple. (). Determine the points where the function f ( ) hs mimum 4 or minimum vlues. 8 Solution. The function f ( ) hs mimum vlue when the denomintor is lest. 4 Note tht the denomintor 4 is positive for ll vlues of in the domin. The denomintor is lest only if =. Therefore f() hs mimum vlue t = nd f()=8 is the mimum vlue. Section. Simple Tests for Etreme Points (Pge# 98) In this section we will study the sign of first derivtive of f() to determine mimum nd minimum vlues of the function. Also we will confirm our results y grphing the function. First derivtive test for mimum/minimum vlues of f(). If If f '( ) for c nd f '( ) for c, then c is mimum point for f. f '( ) for c nd f '( ) for c, then c is minimum point for f. Emple. (). Find possile etreme points for the function nd verify your result. f ( ) ln for (, ). Grph Solution. We hve f ( ) ln f '( ) ln / e.7 Now using rel line test for the sign of f (), we get f '( ) for.7, nd f '( ) for.7. By the first derivtive test we see tht c e /. 7 is the minimum point for f().
19 Mt Brief Clculus Second derivtive test (Concve nd conve functions) for mimum/minimum vlues of f(). If f() is concve in n intervl I nd c is sttionry point (tht mens f '( c) ) for f, then c is mimum point of f. If f() is conve in n intervl I nd c is sttionry point (tht mens f '( c) ) for f, then c is minimum point of f. Rememer tht If f "( ), f is concve nd if f "( ), f is conve. When f "( ) then f ' decreses nd f is concve ( like concve down) When f "( ) then f ' increses nd f is conve ( like concve up) Emple. (). Find possile etreme points for f ( ) e e using second derivtive test. Grph the functions nd verify your result. Solution. We f ( ) e e f '( ) e e f "( ) e 4e for ll. The function f() is conve. It hve minimum vlue when ln f '( ) e e. c, which is clled the sttionry point. Thus we hve the minimum vlue of f ( ) e e =.89 for c =.. Section. Locl Etreme Points (Pge# ) Definition: The function f hs locl mimum (or minimum) t point c, if there eists n intervl (, ) out c such tht f ( ) f ( c), ( or f ( ) f ( c)) for ll in (, ) which re in the domin of f. Function vlues corresponding to locl mimum (minimum) points re clled locl mimum (minimum) vlues.
20 Mt Brief Clculus The First Derivtive Test: Suppose c (, ) is sttionry point for y f ().. If f '( ) on left of c nd f '( ) on right of c, (tht mens f is incresing on left of c nd decresing on right of c.) then = c is locl mimum point of f.. If f '( ) on left of c nd f '( ) on right of c, (tht mens f is decresing on left of c nd incresing on right of c.) then = c is locl minimum point of f. The Second Derivtive Test: Suppose f e twice differentile function nd c (, ) is sttionry point for y f ().. If f '( c) nd f "( c) then = c is strict locl mimum point.. If f '( ), nd f "( c) then = c is strict locl minimum point.. If f '( ), nd f "( c) nd there is chnge of concvity out the point c, then = c is n inflection point. Emple. (). Clssify the sttionry points of f ( ). 9 6 Solution. We hve f ( ) f '( ), nd lso 9 6 f "( ) with f "( ), nd f "(). We conclude tht the sttionry point c =  is locl mimum point nd c = is locl minimum point. Emple. (). Determine the possile etreme points nd etreme vlues of the 5 function f ( ) 5. Solution. Hints. The sttionry points re c =,, c = is point of inflection (verify), c is locl mimum point nd c is locl minimum point. Emple. (). A function f is given y the formul f ( ) 6 ) Find the domin of f nd the open intervl where f is positive ) Find the possile etreme points c) Emine f s, nd d) Determine the limit of f( ) s. Hs f mimum or minimum in the domin? Solution:
21 Mt Brief Clculus ) The domin of f is D [ 6,) (, ). One my look t the grph of f or check the sign of f using test points to find intervl where f is positive. We produce oth elow. f() is positive on ( 6, ) (, ) From grph Using test point: From clcultor y = ( / ) ( 6) grph nd trce t the test points: Sign of f: ) f ( ) 6 6 In clcultor input y = ( 6) / ( / ) ( 6) grph nd find zeros nd lso vlues for which the derivtive is undefined to get sttionry points. The sttionry points re 6, 4 nd 6 (Note tht is not sttionry point). Now use line test for intervl of increse or decrese. The point = 4 is point of mimum, the point = 6 is minimum nd the point = 6 is lso minimum. c) f s, nd f s nd f s. When, f M/min hs discussed in prt ). Section.4 The Etreme Vlue Theorem (Pge# 9) If f is continuous function over ounded closed intervl [, ], then there is point d [, ] where f hs minimum, nd point c [, ] where f hs minimum, such tht f ( d) f ( ) f ( c) for ll [, ]
22 Mt Brief Clculus The Men Vlue Theorem If f is continuous function over ounded closed intervl [, ], nd differentile in the open intervl (, ) then there eists t lest one interior point c (, ), such tht f ( ) f ( ) f '( c) Emple.4 (). Find the mimum nd minimum nd drw the grph of f ( ) 4 4 8, [, 8]. Solution. Since the function is polynomil, it is continuous. Using etreme vlue theorem we see tht the given function hs oth m nd min within the given domin. We pply second derivtive test to emine m/min of the function. f ( ) f '( ) 8 4 f "( ) 8. Note tht f "( ) 8, so the function hs minimum t c = 5, where 5 [, 8]. Emple.4 (). Find ll the numers c [, 8] for the function f ( ) f ( ) f '( c). Solution. We hve f ( ) f ( ) 64 f '( c) c 8 c 4 Section.5 Business nd Economic Models (Pge# 6) f ( ) such tht In this section we consider rel life prolems from usiness nd economics. Emple.5 (). The revenue nd cost function of product re given elow: R ( Q) Q Q /, nd C( Q) 5 Q, Q [, ]. Find the vlue of Q tht mimizes profit. Find lso the profit. Solution. The profit function is P( Q) R( Q) C( Q) Q Q / 5 Q P'( Q) 8 Q /5 P"( Q) /5 By the second derivtive test we see tht P(Q) hs mimum vlue t Q =4. The mimum profit is P(Q) =. Section.6 Inflection points (Pge# 5) Definition: The point c is clled n inflection point for the function f if there eists n intervl (, ) out c such tht:. f "( ) in (, c) nd f "( ) in (c, ) or
23 Mt Brief Clculus. f "( ) in (, c) nd f "( ) in (c, ) Emple.6 (). Find locl etreme points nd inflection points for the function ln f ( ). ln ln ln Solution: Hints. We hve f ( ) f '( ). Find 4 f "( ) nd verify tht the sttionry point is c = e. 65 is locl mimum point. It does not hve point of inflection. Emple.6 (). Find locl etreme points nd inflection points for the function f ( ) ( ) e. Solution. We hve f ( ) ( ) e f '( ) e ( ) e. The only sttionry point is c = . The point c =  is locl mimum point. The locl mimum vlue of the function is f ( ) ( ) e = e =.7. See the grph. Prctice Prolems for Chpter. Find ) f '( ) ) f "( ) c) All sttionry points d) Clssify sttionry points e) All locl mimum nd minimum points f) All locl mimum nd minimum vlues of the function g) All inflection points for the following functions:. f ( ) 5. f ( ). f ( ) f ( ) 5 5. f ( ) e 6. f ( ) ln f ( ) f ( ) 9. f ( ) 4 4. Verify etreme vlue theorem for the function f ( ) 4, [, ].. Verify etreme vlue theorem for the function f ( ) 6, [, 8]. Chpter 4. Integrtion (Pge# 4) Differentition hs een the min topic in the preceding three chpters. Some prolems in mthemtics require finding function from informtion out its derivtive. This cn e regrded s inverse of derivtives. The process of reconstructing function from its derivtives is clled the integrtion.
24 Mt Brief Clculus Emple. Assume tht the mrginl cost function is C'( ) 5 nd it is lso known tht the fied costs re $. Find the cost function. To solve this prolem we need to know the inverse of derivtive to construct the cost function. All the sections in this chpter will e discussing different rules nd methods of deriving such functions. Section 4. Indefinite Integrls (Pge# ) In this section we will use only the derivtive of functions to derive the most generl form of the integrl. Emple 4. (). Determine the generl integrl form of f '( ) y inspection. Solution. Rememer tht the derivtive of of of f ( ) C, where C is constnt is f '( ) is f ) C (. f ( ) is f '( ). Also the derivtive f '( ). Therefore the generl integrl Definition nd Symol: Suppose f() nd F() re two functions of hving the property tht f ( ) F'( ). Here F' ( ) is clled the derivtive of F(), so F() could e clled the ntiderivtive of F '( ). We write F '( ) d f ( ) d F( ). The most generl form is F '( ) d f ( ) d F( ) + C, where C is constnt, sometimes clled integrl constnt. Rules of Integrls. d, where d ln C e d e C d C, where, ln e e d C Emple 4. (). Use derivtive rule to prove tht ln d ln C, for >. We consider f ( ) ln C, then f '( ) ln ln. From definition we know tht the integrl of f '( ) ln is f ( ) ln C. In symol we write
25 Mt Brief Clculus ln d ln C. Emple 4. (). Find the generl integrl of the following functions. ) d ) d c) d Solution. ) d 5 / d ) d c) d C 7 / 7 / C d / d d) C / d) d C Emple 4. (4). Find the generl integrl of the following functions. ) e 4 ( ) d ) d c) d d) d 4 4 e Solution. ) e d C 4 ) d C ln ( ) c) d d 4 d 4 4ln C d) d d d C ln( ) Section 4. Riemnn Sums. (Pge# 8) Section 4. Are nd Definite Integrls. (Pge# 4) This section will show how the concept of integrl cn e used to clculte the re of mny plne regions. The symol f ( ) d is clled the definite integrl which represents the re under the grph of f() nd the ordintes =, nd =. Emple 4. (). Find the re under the prol The required re is More detils will e given in the clss. f ( ) over the intervl [, ]. d F() F(), where F( ) the integrl of f(). Section 4.4 Properties of Definite Integrls (Pge# 49)
26 Mt Brief Clculus From the definition of definite integrl numer of properties hve een derived in this section. If f is continuous function in n intervl tht contins the constnts,, c, then f ( ) d f ( ) d f ( ) d kf ( ) d k f ( ) d, where k is constnt ( ) d f ( ) d c f f ( ) d f '( ) e f ( ) d e c f ( ) f ( ) Differentition with respect to the Limits of Integrtion From the definition of integrl we hve tht, if F' ( t) f ( t) then f ( ) d F( t) F( ) t d Now f ( ) d F'( t) dt f ( t) In other words, the derivtive of the definite integrl w. r. to the upper limit of integrtion is equl to the integrnd evluted tht point. This is lso known s the Fundmentl Theorem of Clculus Prt I. More generl form of Fundmentl Theorem of Clculus Prt I is like u( t) d f ( ) d F'( u( t)) u'( t) dt Emple 4.4 (). Evlute the following definite integrls. ) d ) e d Solution. ) Then d 5 / d F() F() d 7 7 c) d 7 / C, where d) t F( ) 7 d 7 /
27 Mt Brief Clculus ) ( ) e d e e, using formul f '( ) e c) d ln( ) C, where F ( ) ln( ) Then d F() F() ln 9 ln ln 9 / d) d C Then d F() F( ), where f ( ) d e F( ) Emple 4.4 (). Find the following derivtive of the limit of integrls. d ) d dt t Solution. ) d t dt t d t ) dy dt t y d t d d d ) dy dy dy dt y dt y dt t t y d dt t dy y d dt t dy y t t / t f ( ) f ( ) Section 4.5 Integrtion y Sustitution (Pge# 5) In this section we shll see how the chin rule for differentition leds to n importnt method for evluting mny complicted integrls. We illustrte this with emples. Some of these types of prolems cn lso e solved y the method of integrtion y prts. For Integrtion y sustitution we do not use ny formul. One only need to choose the right form of sustitution. Emple 4.5 (). Evlute e d Solution. We consider u du d u u Now e d e du e C e C Emple 4.5 (). Show tht 8 ( ) Solution. We consider u du 9 d 8 89 ( ) C
28 Mt Brief Clculus Now 8 ( ) d 8 u 9 du Emple 4.5 (). Evlute 4 d Solution. We consider u Now 4 u C 8 89 ( ) C udu d d udu ( 4 4 d 4 d (4u u ) du ( 4 ) 9 5 ) 5 C Section 4.6 Integrtion y Prts (Pge# 57) There re some functions, which cnnot e integrted y the rules we hve discussed in erlier sections. Integrtion of such function pper s multipliction or division of two or more functions. In this section we will discuss the integrtion of such functions nd its ppliction in usiness nd economics. Formul of integrtion y prts. We consider the formul nd illustrte y emples. f ( ) g( ) d udv uv vdu where u f ( ) du f '( ) d nd dv g( ) d v g( ) d Emple 4.6 (). Evlute e d Solution. We consideru du d, nd Now e d udv uv vdu e dv e d v e e d e e C m m m Emple 4.6 (). Show tht ln d ln C, where m m ( m ) m m Solution. We consider u ln du d, nd dv d v m Now m m m m m ln d udv uv vdu ln d ln m m m ( m ) Emple 4.6 (). Evlute ln( ) d Solution. We consider u ln( ) du Now let us find d, nd dv d v C
29 Mt Brief Clculus F( ) ln( ) d udv uv vdu ln( ) d ln( ) d ln( ) ln( ) 4 Using definite integrl rule we hve ln( ) d F() F( ) Section 4.7 Infinite Intervls of Integrtion (Improper Integrtion) (Pge# 6) Suppose f is function tht is continuous for ll. Then f ( ) d is defined for ech. If the limit of this integrl s eists nd is finite, then we sy tht f is integrle over [, ), nd we define f ( ) d lim f ( ) d The improper integrl f ( ) d lim f ( ) d l, numer l. Otherwise the improper integrl is sid to diverge. is sid to converge to finite We define lso f ( ) d lim f ( ) d nd lim c c c f ( ) d lim f ( ) d lim f ( ) d Emple 4.7 (). Determine e d, if converges. Solution. e d lim e d lim ( e e ), the result converges to. Emple 4.7 (). Determine Solution. e e d, if converges d lim lim e d e e
30 Mt Brief Clculus Section 4.8 Business nd Economic Applictions (Pge# 7) This section will show the importnce of integrls in usiness nd economics. We will consider rel prolems from usiness nd economics. Consumer nd Producer Surplus Economists re interested in studying how consumer s nd producer s welfre re influenced y chnges in economic prmeters. Consumer s surplus nd Producer s surplus: The supply nd demnd curves re plotted considering price p long y is nd quntity q long is. The symol p indictes the equilirium price nd q represents the equilirium quntity, where demnd nd supply curve intersects. CS is for consumer s surplus nd PS is for producer s surplus. Using the concept of integrtion, one cn esily find the CS is equl to the re under the demnd curve with ordinte q minus the rectngle with sides equl to equilirium price nd equilirium quntity. On the other hnd the PS is equl to the re of the rectngle with sides equl to equilirium price nd equilirium quntity minus the re under the supply curve, with ordinte q. At equilirium point E, demnd is equl to supply. The corresponding equilirium price p is the one which induces the consumers to purchse (demnd) precisely the sme mount tht the producers re willing to offer (supply) t tht price. According to the demnd curve in the figure, there re consumers who re willing to py thn p per unit. In fct, even if the price is lmost s high s p, some consumers still wish to uy some units t tht price. The totl mount sved y ll the consumers uying the commodity t price lower thn wht they mimlly re willing to py, is clled the CONSUMER S SURPLUS. Most producers lso derive positive enefit from selling t the equilirium price ecuse they re willing to supply the commodity t price lower thn the equilirium price. The mount of this enefit is clled the surplus ccruing to the producer. In our figure, even if the price is lmost s low s p, some producers re still willing to sell it for, we cll the PRODUCER S SURPLUS. p p p CS PS E Supply Curve Demnd Curve q
31 Mt Brief Clculus We hve the following formuls: Suppose S(q) nd D(q) re supply nd demnd curves respectively, then PS = q p q S( q) dq nd CS = D( q) dq pq q 4 Emple 4.8 (). The demnd nd supply curves re D ( q) q nd S ( q). q 6 Find consumer s surplus nd producer s surplus. 4 Solution. For equilirium D( q) S( q) q q q nd q 6 p D q ) S( q ) ( Now PS = And CS = q pq S( q) dq ( q ) dq 6 4 q D( q) dq p q Income Strem Emple 4 8 dq 4ln q 6 6 Totl vlue Future vlue 4 5dt 5*4 5, 4.4(4 t) 5e dt
32 Mt Brief Clculus The rte is 5% (where r =.5).
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 informationSection 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 information2 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 informationChapter 6 Techniques of Integration
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
More informationChapter 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( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationDERIVATIVES NOTES HARRIS MATH CAMP Introduction
f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationCalculus 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 informationThe 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 informationImproper 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 informationThomas 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( ) where f ( x ) is a. AB/BC Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB/ Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 Find the intersection of f ( x) nd g( x). A3 Show tht f ( x) is even. A4 Show tht
More informationBob 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 informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationMA 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 informationA 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 informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide 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 informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationCalculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION
lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer..
More informationMATH SS124 Sec 39 Concepts summary with examples
This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationFundamental 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 informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationMA123, 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. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information( ) Same as above but m = f x = f x  symmetric to yaxis. 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 informationMath 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 informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationy = f(x) This means that there must be a point, c, where the Figure 1
Clculus Investigtion A Men Slope TEACHER S Prt 1: Understnding the Men Vlue Theorem The Men Vlue Theorem for differentition sttes tht if f() is defined nd continuous over the intervl [, ], nd differentile
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationGoals: 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 informationOptimization 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 informationMAT137 Calculus! Lecture 20
officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find
More information7. 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 informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationSection 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 informationThe 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 information0.1 THE REAL NUMBER LINE AND ORDER
6000_000.qd //0 :6 AM Pge 00 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 informationChapter 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 informationPolynomials 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 informationReview Exercises for Chapter 4
_R.qd // : PM Pge CHAPTER Integrtion Review Eercises for Chpter In Eercises nd, use the grph of to sketch grph of f. To print n enlrged cop of the grph, go to the wesite www.mthgrphs.com... In Eercises
More informationWhat Is Calculus? 42 CHAPTER 1 Limits and Their Properties
60_00.qd //0 : PM Pge CHAPTER Limits nd Their Properties The Mistress Fellows, Girton College, Cmridge Section. STUDY TIP As ou progress through this course, rememer tht lerning clculus is just one of
More informationIndefinite 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 information6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS
6. CONCEPTS FOR ADVANCED MATHEMATICS, C (475) AS Objectives To introduce students to number of topics which re fundmentl to the dvnced study of mthemtics. Assessment Emintion (7 mrks) 1 hour 30 minutes.
More informationcritical number where f '(x) = 0 or f '(x) is undef (where denom. of f '(x) = 0)
Decoding AB Clculus Voculry solute mx/min x f(x) (sometimes do sign digrm line lso) Edpts C.N. ccelertion rte of chnge in velocity or x''(t) = v'(t) = (t) AROC Slope of secnt line, f () f () verge vlue
More informationMathematics. 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 informationChapter 5 1. = on [ 1, 2] 1. Let gx ( ) e x. . The derivative of g is g ( x) e 1
Chpter 5. Let g ( e. on [, ]. The derivtive of g is g ( e ( Write the slope intercept form of the eqution of the tngent line to the grph of g t. (b Determine the coordinte of ech criticl vlue of g. Show
More informationTime in Seconds Speed in ft/sec (a) Sketch a possible graph for this function.
4. Are under Curve A cr is trveling so tht its speed is never decresing during 1second intervl. The speed t vrious moments in time is listed in the tle elow. Time in Seconds 3 6 9 1 Speed in t/sec 3 37
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More information0.1 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 informationARITHMETIC 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 informationAdvanced Algebra & Trigonometry Midterm Review Packet
Nme Dte Advnced Alger & Trigonometry Midterm Review Pcket The Advnced Alger & Trigonometry midterm em will test your generl knowledge of the mteril we hve covered since the eginning of the school yer.
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationKEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a
KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider
More informationAP Calculus AB First Semester Final Review
P Clculus B This review is esigne to give the stuent BSIC outline of wht nees to e reviewe for the P Clculus B First Semester Finl m. It is up to the iniviul stuent to etermine how much etr work is require
More information5.7 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 informationA. Limits  L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. 1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/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 information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationlim f(x) does not exist, such that reducing a common factor between p(x) and q(x) results in the agreeable function k(x), then
AP Clculus AB/BC Formul nd Concept Chet Sheet Limit of Continuous Function If f(x) is continuous function for ll rel numers, then lim f(x) = f(c) Limits of Rtionl Functions A. If f(x) is rtionl function
More informationCalculus 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 information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationThe area under the graph of f and above the xaxis 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 xxis etween nd is denoted y f(x) dx nd clled the
More informationInstantaneous Rate of Change of at a :
AP Clculus AB Formuls & Justiictions Averge Rte o Chnge o on [, ]:.r.c. = ( ) ( ) (lger slope o Deinition o the Derivtive: y ) (slope o secnt line) ( h) ( ) ( ) ( ) '( ) lim lim h0 h 0 3 ( ) ( ) '( ) lim
More informationAB Calculus Path to a Five Problems
AB Clculus Pth to Five Problems # Topic Completed Definition of Limit OneSided Limits 3 Horizontl Asymptotes & Limits t Infinity 4 Verticl Asymptotes & Infinite Limits 5 The Weird Limits 6 Continuity
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationBefore 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 PreChpter 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 information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More information5.1 How do we Measure Distance Traveled given Velocity? Student Notes
. How do we Mesure Distnce Trveled given Velocity? Student Notes EX ) The tle contins velocities of moving cr in ft/sec for time t in seconds: time (sec) 3 velocity (ft/sec) 3 A) Lel the xxis & yxis
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationMathematics Extension 1
04 Bored of Studies Tril Emintions Mthemtics Etension Written by Crrotsticks & Trebl. Generl Instructions Totl Mrks 70 Reding time 5 minutes. Working time hours. Write using blck or blue pen. Blck pen
More informationMA 15910, Lessons 2a and 2b Introduction to Functions Algebra: Sections 3.5 and 7.4 Calculus: Sections 1.2 and 2.1
MA 15910, Lessons nd Introduction to Functions Alger: Sections 3.5 nd 7.4 Clculus: Sections 1. nd.1 Representing n Intervl Set of Numers Inequlity Symol Numer Line Grph Intervl Nottion ) (, ) ( (, ) ]
More informationChapter 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 informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationEdexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks
Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1
More informationImproper 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 information1 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 informationAP * Calculus Review
AP * Clculus Review The Fundmentl Theorems of Clculus Techer Pcket AP* is trdemrk of the College Entrnce Emintion Bord. The College Entrnce Emintion Bord ws not involved in the production of this mteril.
More informationCalculus 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 informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationBridging 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 informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewiseliner function f, for 4, is shown below.
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationMath 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 informationMAT137 Calculus! Lecture 28
officil wesite http://uoft.me/mat137 MAT137 Clculus! Lecture 28 Tody: Antiderivtives Fundmentl Theorem of Clculus Net: More FTC (review v. 8.58.7) 5.7 Sustitution (v. 9.19.4) Properties of the Definite
More informationFINALTERM EXAMINATION 9 (Session  ) Clculus & Anlyticl GeometryI Question No: ( Mrs: )  Plese choose one f ( x) x According to PowerRule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+
More informationProperties 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 informationLINEAR 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 nthorder
More information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
More informationAP 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