CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS

CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS LEARNING OBJECTIVES After stuying this chpter, you will be ble to: Unerstn the bsics of ifferentition n integrtion; Know how to compute erivtive of function by the first principle, erivtive of : function by the ppliction of formule n higher orer ifferentition; Apprecite the vrious techniques of integrtion; n Unerstn the concept of efinite integrls of functions n its properties. INTRODUCTION TO DIFFERENTIAL AND INTEGRAL CALCULUS (EXCLUDING TRIGONOMETRIC FUNCTIONS) DIFFERENTIAL CALCULUS 9.A. INTRODUCTION Differentition is one of the most importnt funmentl opertions in clculus. Its theory primrily epens on the ie of limit n continuity of function. To epress the rte of chnge in ny function we introuce concept of erivtive which involves very smll chnge in the epenent vrible with reference to very smll chnge in inepenent vrible. Thus ifferentition is the process of fining the erivtive of continuous function. It is efine s the limiting vlue of the rtio of the chnge (increment) in the function corresponing to smll chnge (increment) in the inepenent vrible (rgument) s the lter tens to zero. 9.A. DERIVATIVE OR DIFFERENTIAL COEFFICIENT Let y f() be function. If h (or ) be the smll increment in n the corresponing increment in y or f() be y f(+h) f() then the erivtive of f() is efine s lim h f(+h) - f() i.e. h f( ) f() lim This is enote s f () or y/ or f(). The erivtive of f() is lso known s ifferentil coefficient of f() with respect to. This process of ifferentition is clle the first principle (or efinition or binitio). Note: In the light of bove iscussion function f () is si to ifferentible t c if f()-f(c) lim eist which is clle the ifferentil coefficient of f() t c n is enote hc -c 9. COM M ON PROFICIENCY TEST

y by f (c) or c. We will now stuy this with n emple. Consier the function f(). By efinition f( ) f() ( ) f() lim lim lim ( ) lim ( ) Thus, erivtive of f() eists for ll vlues of n equls t ny point. Emples of Differentitions from the st principle i) f() c, c being constnt. ii) Since c is constnt we my write f(+h) c. So f(+h) f() f(+h)- f() Hence f'()lim lim h h h h So (c) Let f() n ; then f(+h) (+h) n let +h t or h t n s h, t Now f () lim h h f(+h)-f() h n n (+h) - lim h lim t (t n n ) / (t ) n n iii) Hence ( n ) n n f () e f( + h) e +h So f () lim h h f(+h)- f() h +h e - e lim h h e (e -) lim h h M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS iv) e lim h h e - Hence (e ) e Let f() then f(+h) +h f () lim h lim h log e Thus h - h h f(+h)- f() h ( ) log e e. +h - lim h h v) Let f(). Then f( + h) +h f () lim h f(+h)- f() h +h - lim h h ( +h - ) ( +h + ) lim h h( +h + ) lim h h ( ) h lim h +h- h( +h+ lim +h + h Thus ( ) vi) f() log f( + h) log ( + h) f () lim h lim h f(+h)- f() h log (+h)- log h 9.4 COM M ON PROFICIENCY TEST

lim h log h h Let h lim h h h log t i.e. ht n s h, t f () lim log(+t) lim log(+t) t t t t, since log + t lim t t Thus (log ) 9.A. SOME STANDARD RESULTS (FORMULAS) () (4) Note: (n ) n n () (constnt) (e ) e () (5) (e ) e (5) { c f()} cf () c being constnt. In brief we my write below the bove functions n their erivtives: Function f() Tble: Few functions n their erivtives ( ) log e (log ) Derivtive of the function f () n n n e log c ( constnt) e / log e M ATHS 9.5

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS We lso tbulte the bsic lws of ifferentition. (i) Function h() c.f() where c is ny rel constnt (Sclr multiple of function) (ii) h() f() ± g() (Sum/Difference of function) (iii) h() f(). g() (Prouct of functions) (iv) h() f() g() (quotient of function) (v) h() f{g()} Tble: Bsic Lws for Differentition Derivtive of the function {h()} c. {f()} {h()} {h()} {h()} {h()} f() ± {g()} 9.6 COM M ON PROFICIENCY TEST f() {g()}+g() {f()} g() {f()}-f() {g()} {g()} z f(z)., where z g() z It shoul be note here even though in (ii) (iii) (iv) n (v) we hve consiere two functions f n g, it cn be etene to more thn two functions by tking two by two. Emple: Differentite ech of the following functions with respect to : () + 5 (b) + + (c) () e log (e) 5 (f) (g) e / log (h). log (i) +7 Solution: () Let y f() + 5 (b) y () + 5 + 5. 6 + 5 Let h() + + {h()} () () ( + + ) ( ) + -5 +6-log+ e ( ) + ( ), is constnt

log + + log +. (c) Let f () () -5 +6-log+. -5.+6.-. + -+6-. Let y e log y e (log ) + log e + e log So y (e) y 5 y 5 e e (+ log ) ( + log ) ( ) + 5 log e + 5. 4 (f) let y e f() -5 +6-log+ (e ) (Prouct rule) (5 ) (Prouct Rule) 8 y e ( ) (e ) (Quotient Rule) (e ) e - e (-) (e ) e (g) Let y e / log so y (log) (log) (e ) e (log ) (Quotient Rule) e log - e / (log ) e log - e (log ) So y e ( log -) (log ) M ATHS 9.7

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS (h) Let h(). log The given function h() is ppering here s prouct of two functions f () n g() log. {h ()} (. log) (log)+log ( ). +log.( log) + loglog (i) Let h() [Given function ppers s the quotient of two functions] +7 f() n g() + 7 {h()} ( +7) ()- ( +7) ( +7). -.(9 +) ( +7) ( +7) ( +7)-9 (7-6 ). ( +7) ( +7) 9.A.4 DERIVATIVE OF A FUNCTION OF FUNCTION y y u If y f [h()] then f'(u) h'() u Emple: Differentite log ( + ) wrt. Solution: Let y log ( + ) log t when t + y y t (+) t t t (+ ) where u h() This is n emple of erivtive of function of function n the rule is clle Chin Rule. 9.A.5 IMPLICIT FUNCTIONS A function in the form f(, y). For emple y + y + y where y cnnot be irectly efine s function of is clle n implicit function of. In cse of implicit functions if y be ifferentible function of, no ttempt is require to epress y s n eplicit function of for fining out y with respect of n substitution of y solving the resulting eqution.. In such cse ifferentition of both sies y gives the result. Therefter y my be obtine by 9.8 COM M ON PROFICIENCY TEST

Emple: Fin y for y + y + y Solution: y + y + y Differentiting with respect to we see or or or (y ) + y ( ) + (y) y y + y + y y + y + y () (y + + ) y + y + y y (y + y) ( y + +) () + y + y, This is the proceure for ifferentition of Implicit Function. 9.A.6 PARAMETRIC EQUATION (y ) (), y y (chin rule) When both the vribles n y re epresse in terms of prmeter ( thir vrible), the involve equtions re clle prmetric equtions. For the prmetric equtions f(t) n y h(t) the ifferentil coefficient y is obtine by using y y t. t y t t Emple: Fin y if t, y / t Solution : t t ; y y t - 4 6 t t t t y t / t4 This is the proceure for ifferentition of prmetric functions. 9.A.7 LOGARITHMIC DIFFERENTIATION The process of fining out erivtive by tking logrithm in the first instnce is clle logrithmic ifferentition. The proceure is convenient to opt when the function to be ifferentite involves function in its power or when the function is the prouct of number of functions. M ATHS 9.9

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS Emple: Differentite w.r.t. Solution: let y Tking logrithm, log y log Differentiting with respect to, y log + y +log y or y ( + log ) ( + log ) This proceure is clle logrithmic ifferentition. 9.A.8 SOME MORE EXAMPLES () If - y show tht ( + ) y + y. Solution: Tking logrithm, we my write log y {log ( ) log ( + )} Differentiting throughout we hve y y {log ( ) log ( + )} By cross multipliction ( ) y y Trnsposing ( ) y + y. () Differentite the following w.r.t. : () log ( + + ) (b) log - + -b. Solution: () y log ( + + ) y (+ + ) () (+ + ) (+ + ) + 9. COM M ON PROFICIENCY TEST

(+ + ) (+ + ) + + (b) Let y log ( b) or y b b () If m y n (+y) m+n prove tht y y Solution : m y n (+y) m+n Tking log on both sies log m y n (m+n) log ( + y) or m log + n log y ( m+n) log (+y) so m n y y (m n) y ( y) b ( b ). ( b) b or n y m n y y m n ( y m or (n+ny-my-ny) y m+n-m-my y(+y) (+y) or or (n-my) y n-my y y y Prove. (4) If y e y Prove tht y Solution : y e y So y log ( y) log e or y log ( y) Differentiting w.r.t. we get y y + log y log (+log )...() M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS or ( + log ) y y or y (-y) (+log ), substituting y log, from () we hve or y y(log) (+log ).. (b) From () y( + log ) or y (+log ) From (b) y log (+log ) 9.A.9 BASIC IDEA ABOUT HIGHER ORDER DIFFERENTIATION Let y f() 4 + 5 + + 9 y f() 4 + 5 + 4 f () Since f() is function of it cn be ifferentite gin. Thus y f () (4 + 5 + 4) + + 4 y y is written s (re s squre y by squre) n is clle the secon erivtive y of y with respect to while is clle the first erivtive. Agin the secon erivtive y here being function of cn be ifferentite gin n f () 4 +. y Emple: If y e m + be m prove tht m y. Solution: y ( e m + be m ) me m bme m y y ( me m bme m ) m e m + bm e m m (e m + be m ) m y. 9. COM M ON PROFICIENCY TEST

9.A. GEOMETRIC INTERPRETATION OF THE DERIVATIVE R Let f() represent the curve in the Fig. We tke two jcent pir s P n Q on the curve Let f() represent the curve in the fig. We tke two jcent points P n Q on the curve whose coorintes re ( y) n ( +, y+y) respectively. The slope of the chor TPQ is given by y/ when, Q P. TPQ becomes the tngent t P n y lim The erivtive of f() t point represents the slope (or sometime clle the y grient of the curve) of the tngent to the curve y f() t the point. If lim eists for prticulr point sy n f() is finite we sy the function is ifferentible t n continuous t tht point. Emple : Fin the grient of the curve y 5 +4 t the point (, ). y Solution : y 5 + 4 6 5 so [y /] y 6. 5 6 5 Thus the grient of the curve t (, ) is. y M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS Eercise 9 Choose the most pproprite option () (b) (c) or (). The grient of the curve y +8 t is ) b) c) ) none of these. The grient of the curve y 5 t is ) b) c) / ) none of these. The erivtive of y + is ) / + b) / + c) / + ) none of these 4. If f() b c ) e 5. If f() +b+c e the f () is b c b) + - e then f () is ( +b) c) +b ) none of these ) 4 / ( ) b) 4 / ( ) c) / ( ) ) none of these 6. If y ( ) ( ) then y is ) 6 + b) 6 + c) + ) none of these 7. The grient of the curve y y + p + qy t the point (, ) is -. The vlues of p n q re ) (/, /) b) (, ) c) ( /, /) ) (/, /6) 8. The curve y u + v psses through the point P(, ) n y 4 t P. The vlues of u n v re ) (u, v 7) b) (u, v 7) c) (u, v 7) ) (, -) 9. The grient of the curve y + p +qy t (, ) is /. The vlues of p n q re ) (, ) b) (, ) c) (, ) ) (, -). If y then y + y/ is equl to ) b) c) ) none of these. The erivtive of the function + is ) + b) + c) ) none of these 9. 4 COM M ON PROFICIENCY TEST

. Given e -y 4y, y cn be prove to be ) y / b) y / c) / y ) none of these. If y - ) y, y cn be epresse s b) - c) - ) none of these 4. If log ( / y) + y, y ) y(-) (+y) b) y my be foun to be c) - +y ) none of these 5. If f(, y) + y y, y cn be foun out s ) y- y + b) y- y - c) y+ y + ) none of these 6. Given t, y t; y is clculte s ) t b) /t c) /t ) none of these 7. Given t + 5, y t ; y is clculte s ) t b) /t c) /t ) none of these 8. If y ) then y is equl to b) - 9. If t, y t t, then y ) t - 6t is equl to c) b) t c) t - 6t ) none of these ) none of these M ATHS 9. 5

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS. The slope of the tngent to the curve y 4- t the point, where the orinte n the bsciss re equl, is ) b) c) ) none of these. The slope of the tngent to the curve y t the point, where the line y cuts the curve in the Ist qurnt, is ) b) c) ) none of these. For the curve + y + g + hy, the vlue of y t (, ) is ) -g/h b) g/h c) h/g ) none of these. If y e -e e +e y, then is equl to ) e 5 b) /(e 5 + e ) c) e 5 /(e 5 + e ) ) none of these 4. If y. y M, where M is constnt then y is equl to ) -y -y(y+ log y) y+ log y b) c) ( log +) y log + ) none of these 5. Given t + t n y t t the vlue of y t t is ) /5 b) /5 c) 5/ ) none of these 6. If y + 5 +y 5 then y t, y is equl to ) 4/ b) 4/ c) /4 ) none of these 7. The erivtive of log is ) +log b) ( + log ) c) log ) none of these 8. The erivtive of -5 +5 is ) /( +5) b) /( +5) c) /( +5) ) none of these 9. Let y + then y is equl to ) (/ ) +. log e b) / c). log e ) none of these. The erivtive of?? log? e? is???? 4 «? - º??»?? +? ¼ ª ¹ 9. 6 COM M ON PROFICIENCY TEST

) + +4. The erivtive of b) - -4-6+ e is c) -4 ) ( 5) 5 b) ( 5 ) 5 c) 6( ) e. If y ) e e -e (e -) then y b) is equl to e (e -) - c) (e -) -6+ ) none of these ) none of these ) none of these. If f() ( + ) ( + ) + + the vlue of f () is ) + b) log ª ¹ c) log ) none of these y 4. If t, y t then t is equl to ) / b) c) / ) none of these 5. Let f() + then f () is equl to ) /4 b) / c) ) none of these 6. If f() 6+8 then f (5) f (8) is equl to ) f () b) f () c) f () ) none of these 7. If y + +m n then y/ is equl to ) ny b) ny/ +m c) ny/ +m ) none of these 8. If y + /m + m / then y y/ /m + m / is equl to ) b) c) ) none of these n 9. If y + +! +! +..+ y +... then y is prove to be n ) b) c) ) none of these 4. If f() k n f () the vlue of k is ) b) c) / ) none of these M ATHS 9. 7

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 4. If y +m then y y (where y y/) is equl to ) b) c) / ) none of these y 4. If y e + e then y 4 is equl to ) b) c) ) none of these 4. The erivtive of ( )/ is ) + / b) / c) / ) none of these 44. The ifferentil coefficients of ( +)/ is ) + / b) / c) / ) none of these 45. If y e then y is equl to. e e ) b) e c) 46. If y then y is equl to. ) none of these ) y y log b) y y log c) y log ) none of these 47. If ( t )/( + t ) y t/( + t ) then y/ t t is. ) / b) c) ) none of these 48. f() /e then f () is equl to. ) /e b) /e c) e ) none of these 49. If y ( + - )m then ( ) (y/) m y is prove to be ) b) c) ) none of these 4 5. If f() + + then the vlues of for which f () is ) ( ± 5 ) b) ( ± ) c) ) none of these INTEGRAL CALCULUS 9.B. INTEGRATION Integrtion is the reverse process of ifferentition. we know 9. 8 COM M ON PROFICIENCY TEST

Integrtion f() f () Differentition or n n n n n n n n () Itnegrtion is the inverse opertion of ifferentition n is enote by the symbol. Hence, from eqution (), it follows tht n+ n n+ i.e. Integrl of n with respect to vrible is equl to n+ n+ n+ Thus if we ifferentite we cn get bck n n+ n+ Agin if we ifferentite n+ + c n c being constnt, we get bck the sme n. n n ii.e. c n Hence n+ n + c n this c is clle the constnt of integrtion. n+ Integrl clculus ws primrily invente to etermine the re boune by the curves iviing the entire re into infinite number of infinitesiml smll res n tking the sum of ll these smll res. i) 9.B. BASIC FORMULAS n n+ n + c, n?? (If n-,? which is not efine) n n+ ii), since M ATHS 9. 9

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS iii) e e + c, since e e e iv) e + c, since e e v) log +c, since log vi) / log e +c, since žlog žÿ e Note: In the nswer for ll integrl sums we +c (constnt of integrtion) since the ifferentition of constnt is lwys zero. Elementry Rules: c f() c { f() ± g()} f() ± g() f() where c is constnt. Emples : Fin (), (b), (c) e () (e). Solution: () ò / + / (/ + ) / / + c / (b) + + c + c where c is rbitrry constnt. + (c) e () + c. log e + c e + c e (e). 5/ c 5 Emples : Evlute the following integrl: i) ( + /) + + /. 9. COM M ON PROFICIENCY TEST

+ ++ + + +c ii) ( + ) ÿ 7/ + / / 7/+ /+ /+ + 7/ + /+ /+ iii) 4 9 5 9/ 5/ / + + c e + e e + ÿ e 4 iv) + + 4 e e + 4 + - ) ( + + + e + c 4 4e v) ( ) + log( + ) + log( + ) + c + 5 ( + ) By simple ivision + 5 ( + ) «9 º» É ¼ 6 + 6+9log(+)+c 9.B. METHOD OF SUBSTITUTION (CHANGE OF VARIABLE) It is sometime possible by chnge of inepenent vrible to trnsform function into nother which cn be reily integrte. M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS We cn show the following rules. To put z f () n lso just z f () Emple: F{h()} h (), tke e z h() n to just z h () then integrte F ( z) z using norml rule. 7 Emple: ( + ) We put ( + ) t so t or t / ( + ) t t Therefore 8 6 6 This metho is known s Metho of Substitution Emple: We put ( ( + ) +) t so t or t /. t t t t t t t t 4 t 7 7 ( + ) ½ t t + t + ( + ) ( + ) + t 4 t t t +.. + c 4 ( + ) ( + ) IMPORTANT STANDARD FORMULAE () log + c + 8 8 8 c + b) log + c 9. COM M ON PROFICIENCY TEST

c) log + + ) log + c + c e) e {f() f ()} e f() + c f) log + c g) - - - log ( ) + c h) f' () f() log f() + c Emples :( ) e 4 z e z where ze z e e log 4 e +c (b) ( ) ( ) ( ) log( ) c, wheref() (c) e ( ) e f() f'() [by (e) bove)] e +c 9.B.4 INTEGRATION BY PARTS (u) uv u v [ v] where u n v re two ifferent functions of Evlute: i) e Integrting by prts we see M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS ii) e e () e e. e e e c log Integrting by prts, (log ) log. log.. iii) log. log 4 + c e e ( ) e e e. e.e e e () e e e e. e e + e +c 9.B.5 METHOD OF PARTIAL FRACTION Type I : Emple : Solution : let ( ) ( - ) ( - ) ( +) (-) (-) 9. 4 COM M ON PROFICIENCY TEST

A B + ( ) ( ) [Here egree of the numertor must be lower thn tht of the enomintor; the enomintor contins non repete liner fctor] Type II: so + A ( ) + B ( ) We put n get. + A ( ) + B ( ) > A 8 we put n get. + A ( ) + B ( ) > B ( ) 8 ( ) ( ) log( )+log( -)+c ( ) Emple : ( ) ( ) ( + ) A B C Solution : let + + (-) (-) ( ) ( ) ( ) or + A ( ) ( ) + B ( ) +C ( ) Compring coefficients of, n the constnt terms of both sies, we fin A+C (i) 5A +B 4C (ii) 6A B +4C.(iii) By (ii) + (iii) A B 5...(iv) (i) (iv) B + C 5.(v) From (iv) A 5+B From (v) C 5 B From (ii) 5 ( 5+B) + B 4 ( 5 B) or 5 B + B + + 8B or B 5 or B 8, A 5 6, from (iv) C A M ATHS 9. 5

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS ( ) Therefore ( - ) ( - ) ( ) ( ) 8 8 log ( ) + + log ( ) ( ) ( ) log Type III: Emple: ( ) ( ) ( ( - ) + 8 ( ) - 5) ( 5) + c +5 A B+C Solution: Let + ( ) ( +5) ( +5) so +5 A ( + 5 ) + (B +C) ( ) Equting the coefficients of, n the constnt terms from both sies we get A + B C B 5A C 5 by (i) + (ii) A + C by (iii) + (iv) 6A 6 or A (i) (ii).(iii) therefore B n C Thus ( ( - ) - 5) ( 5) Emple: ( ) Solution : ( ) 5 log ( ) + log ( + 5) log ( + 5) ( ) + c (iv) (v) 9. 6 COM M ON PROFICIENCY TEST

( ) we put z, z z z(z ) z z z log z log(z ) log c Emple : Fin the eqution of the curve where slope t (, y) is 9 n which psses through the origin. Solution : y 9 y or y 9 / +c Since it psses through the origin, c ; thus require eqn. is 9 y. 9.B.6 DEFINITE INTEGRATION Suppose F() f () b As chnges from to b the vlue of the integrl chnges from f () to f (b). This is s F() f (b) f () b is clle the upper limit n the lower limit of integrtion. We shll first el with inefinite integrl n then tke up efinite integrl. Emple : Solution : 5 5 6 5 6 6 6 M ATHS 9. 7

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 6 (6 ) 64/6 / Note: In efinite integrtion the constnt C shoul not be e Emple: ( Solution: 9.B.7 5 ) 5 5 ( 5 ). Now, ( 5 ) 5 5 9/6 IMPORTANT PROPERTIES Importnt Properties of Definite Integrl b (I) f () f (t) t (II) f () b b (III) f () f () f (), c (IV) f () c b o b c f ( ) (V) When f() f (+) f () n (VI) n b f () f ( ) f ( ) if f( ) f() Emple : ( if f( ) f() ) b b f () Solution : Let I ( ) ( - ) ( - ) [by prop IV] 9. 8 COM M ON PROFICIENCY TEST

( ) ( ) ( ) ( ) ( ) or I / Emple : Evlute ( ) 4 5 5 Solution : ( ) ( ) let 5 t so tht 5 4 t 4 Now 5 5 ( ) ( ) 4 5 5 5 5 ( ) ( ) t 5 5 ( ) t 4 4 log 5 5 5 5 5 (by stnr formul b) Therefore, 4 4 ( by prop. VI) 5 5 log 5 5 5-5 5 log 5 5 M ATHS 9. 9

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS EXERCISE 9 [K constnt] Choose the most pproprite option () (b) (c) or (). Evlute 5 : 5 () 5 / + k (b) +k (c) 5. Integrtion of 4 will become 5 () none of these () 5 / 5 (b) + k (c) + + k () none of these 5 5. Given f() 4 + + 5 n f () is () 4 + + 5 (b) 4 + + 5 + k (c) + 6 () none of these 4. Evlute ( - ) () 5 /5 / + + k (b) k (c) () none of these 5. ( )( + ) is equl to () (b) + (c) + k () none of these 6. [ / ] is equl to () / ½ +k (b) + k (c) + + k () none of these 7. The integrl of p + q + rk + w/ is equl to () p + q + r + k 5 (b) p / + q / + r (c) p + q w/ () none of these 8. Use metho of substitution to integrte the function f() ( 4 + 5 ) 6 n the nswer is () /8 ( 4 + 5 ) 7 + k (b) ( 4 + 5 ) 7 /7 + k (c) ( 4 + 5 ) 7 /7 () none of these 5 9. Use metho of substitution to evlute ( + 4) n the nswer is () ( + 4 ) 6 + k (b) / ( + 4 ) 6 + k (c) ( + 4 ) 6 / + k. Integrte ( + ) n n the result will be () ( + ) n + n+ + k () none of these (b) ( + ) n + n + (c) ( + ) n + () none of these 9. COM M ON PROFICIENCY TEST

. 8 /( ) is equl to 4 () 4/( + ) + k (b) ( ) + k 4 (c) k () none of these ( ). Using metho of prtil frction the integrtion of f() when f() - nswer is () log k (c)? -? log + k? +? (b) log ( ) log ( + ) + k () none of these n the. Use integrtion by prts to evlute e () e / e /9 + /7 e + k (c) e / e /9 + e + k 4. log is equl to (b) e e + e + k () none of these () log + k (b) log + k (c) log + k () none of these 5. e is () ( )e + k (b) ( ) e (c) e + k () none of these 6. (log ) n the result is () (log) log + + k (c) log + k (b) ( log ) + k () none of these 7. Using metho of prtil frction to evlute ( + 5) /( + ) ( + ) we get () 4 log ( + ) 4 log ( + ) + / + + k (b) 4 log ( + ) / + ) + k (c) 4 log ( + ) 4 log ( + ) () none of these 8. Evlute ( - ) n the vlue is () 4/ + k (b) 5/ (c) 4/ () none of these M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 4 9. Evlute ( - ) n the vlue is () 4 (b) (c) () none of these.. Evlute e n the vlue is () (b) (c) /9 () +. ( + log) is equl to () log + k (b) e + k (c). If f() + then f () is () ( + ) / + k (c) ( + ) / + k (b) + k () + c + + log (+ +) +k () none of these. ( + ) / + () + +k É 4. (e + e ) (e e ) is equl to (b) + + k (c) /( + ) / + k () none of these is () (c) (e. e ) +k e +k 5. [ f() + f( ) ] is equl to (b) (e e ) +k () none of these () f() (b) f() (c) () 6. e /( + ) is equl to f( ) () e /( + ) + k (b) e / + k (c) e + k () none of these 7. ( 4 + /) is equl to () 5 /5 + log ôô (c) /5 5 + k (b) /5 5 + log ôô + k () none of these 9. COM M ON PROFICIENCY TEST

8. Evlute the integrl É / n the nswer is equl to () log ôô + / + k (c) log + + k (b) log + + k () none of these 9. The eqution of the curve in the form y f() if the curve psses through the point (, ) n f () is () y (b) y y (c) y () none of these 4. Evlute ( + 5 ) n the vlue is () (b) (c) () none of these. + is equl to () log e (5/) (c) log e (/5) (b) log e 5 log e + k () none of these. 4 + 4 is equl to () 9/ (b) /9 (c) /9 () none of these +. + is () + log e (b) + log e (c) log e () none of these 4. Evlute e n the vlue is log É () / (b) / (c) 6/ () / (log e 5) 4 ( + )( + 4) 5. () is equl is 5 5 (b) 48/5 (c) 48 () 7 55 5 6. The eqution of the curve which psses through the point (, ) n hs the slope 4 t ny point (, y) is () y + 4 (b) y + 4 (c) y y + 4 () none of these M ATHS 9.

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 7. The vlue of (5 ) f f () is () (b) (c) () none of these 8. É e / is equl to () e / + k (b) e / + k (c) e / + k () none of these e (log + ) 9. is equl to () e log + k (b) e + k (c) log + k () none of these 4. log 4. is equl to () (log ) + k (c) (log ) + k log is equl to (b) (log ) + k () none of these () log (b) /4 (c) log ¾ () none of these - + 4. Evlute µ e. The vlue is () e ( e ) (b) e [ e ]+k (c) e e () none of these 4. is () 7 (b) 8 (c) 8 () none of these ( )e 44. Evlute n the vlue is ( ) e () - + k (b) e + k (c) + k () none of these 45. Using integrtion by prts () 4 /6 + k (c) 4 log + k log (b) 4 /6 ( 4 log ) + k () none of these 46. log ( log)/ is () log (log ) + k (b) log + k (c) [ log (log ) ] log + k () none of these 9. 4 COM M ON PROFICIENCY TEST

47. ( log ) is equl to () (log ) log + + k (b) (log ) log + + k (c) (log ) + + k () none of these 48. e e Evlute e + e n the vlue is () log e ôe + e ô (b) log e ôe + e ô + k (c) log e ôe e ô + k () none of these 49. By the metho of prtil frction is -- É () log e ô ô + log e ô + ô + k (b) log e ô ô log e ô + ô + k (c) log e ô ô + log e ô + ô + k () none of these 5. If f (), the eqution of curve y f() pssing through the point (, ) is given by () y + (b) y / + (c) y / + / () none of these M ATHS 9. 5

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS ANSWERS Eercise 9.. b. c 4. b 5. 6. 7. 8. b 9.. b. c.. 4. 5. b 6. c 7. 8. c 9... b.. 4. b 5. c 6. 7. b 8. c 9.. b. c.. b 4. 5. 6. b 7. b 8. 9. c 4. 4. b 4. c 4. 44. b 45. 46. b 47. c 48. b 49. c 5. Eercise 9. b. b. b 4. b 5. c 6. 7. 8. 9. b.. b. c. 4. 5. 6. 7. 8. b 9.... b. 4. 5. b 6. 7. b 8. 9.. c. b. b. b 4. 5. 6. b 7. b 8. 9. 4. b 4. c 4. 4. c 44. 45. b 46. c 47. 48. b 49. 5. c 9. 6 COM M ON PROFICIENCY TEST

ADDITIONAL QUESTION BANK Differentil Clculus. If. If y then y/ is 4 4 /4 - /4 (D) - / y then y/ is -/ 5/ 5/ (/) (/5) (-/5). If 4. If -8 y then y/ is -9-8 -9 8 9-8 (D) 9 8 y5 then y/ is 5 5. If y then y/ is 4 + (-) + (D) 6. If 7. If 8. If 4 y4-7 then y/ is (-4 +6) (4-6) (4 +6) 7 - y(4/) -(6/7) +4 then y/ is 6-4 4-6 - 6-4 4 +6-6 -4 4 +6 + 4 - - y9-7 +8-8 + then y/ is 9. If. If. If. If - -4 6 - +6+8 - - -4 6 - +6-8 + - -4 6 + +6+8 + y[(-)/] then y/ is - - - - - - ( + ) (- + ) ( - ) y( +) ( +) then y/ is 4 5 + + 5 + + 5 ++ y( +5) ( ++7) then y/ is 4 +9 +4+5 4 +9 +4 +5 4 +9 +4 +5 / / / y ( +) ( -) then y/ is / / / / 4+5(-6) 4+5(-) / / 4+5(-) M ATHS 9. 7

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS. If y( -)/( +) then y/ is - - 4( +) 4( +) 4( -) 4. If y(+)(-)/(-) then y/ is 5. If 6. If 7. If 8. If 9. If ( -6-)/(-) ( +6-)/(-) ( +6+)/(-) / / y( +)/ then y/ is -/ - -/ / / y( -7) then y/ is -/ -/ / ( -7) 6( -7) ( -7) y( -5 +8) then y/ is ( -5 +8) (9 -) ( -5 +8) ( -9) 5 -/ y(6-7 +9) then y/ is ( -5 +8) (9 +) 5-4/ 4 5-4/ 4 (-/)(6-7 +9) ( - ) (/)(6-7 +9) ( - ) 5 4/ 4 (-/)(6-7 +9) ( - ) / / - y[( + ) +( +b ) ] then y/ is - / / ( -b ) [( + ) -( +b ) ] - / / ( -b ) [( + ) -( +b ) ] - / / ( -b ) [( + ) +( +b ) ] (D) - / / ( -b ) [( + ) +( +b ) ]. If ylog5 then y/ is. If - - 5 -/ y then y/ is -/ -/ / (-/) (/) (/) (D) 5. If -7/ y- then y/ is -/ 7 -/ -/ -7 (-7/) 9. 8 COM M ON PROFICIENCY TEST

. If 4 y7 + -9+5 then y/ is 8 +9(+)(-) 8 +9(-) 8 +9(+) 4. If 5. If 6. If - -7 y+4 - then y/ is - -8-4 +4 - -8 +4-4 - -8 +4 +4 - y(- ) then y/ is - - - + + (D) - / -/ y( - ) then y/ is - -/ -4/ - + - - -/ -4/ + + - - -/ -4/ + + + 7. If y(+)(+b)(+c) then y/ is 8. If 9. If. If ++b+c+b+bc+c ++b+c+b+bc+c ++b+c+b+bc+c. If - y( +5)(7+4) then y/ is - - ( +4+)(7+4) ( ++4)(7+4) - ( +4+4)(7+4) - y(+)(+)(4+) then y/ is - - (4 ++)(4+) (4 ++)(4+) - (4 ++5)(4+) 4 y(5-6 -7+8)/(5-6) then y/ is 4 - (75 - - +7+)(5-6) 4 - (75 - + -7+)(5-6) 4 - (75 - - +7-)(5-6) / y( +b+c) then y/ is -/ (/)(+b)( +b+c) -/ (-/)(+b)( +b+c) -/ (/)(+b)( +b+c) M ATHS 9. 9

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS. If. If 4 -/ y( + -5+6) then y/ is 4-4/ (-/)( + -5+6) (8 +9-5) 4-4/ (/)( + -5+6) (8 +9-5) 4 4/ (/)( + -5+6) (8 +9-5) / / ylog[(-) -(+) ] then y/ is -/ (/)( -) -/ (-/)( -) / (/)( -) 4. If ylog + + then y/ is -/ -/ (/)( + ) (-/)( + ) / (/)( + ) 5. If t/(+t ), yt /(+t ), then y/ is 4 4 4 (t-t )/(-t ) (t-t )/(+t ) (t+t )/(+t ) 6. If 7. If / -/ ylog[e (5-) (4+) ] then y/ is +(/)[5/(5-)-4/(4+)] +(/)[5/(5-)+4/(4+)] y then the vlue of y/ is - [ +log. (+log)] - [ +log. (-log)] (D) 8. If y -y e then y/ is log/(-log) -(/)[5/(5-)-4/(4+)] - [ +log.(+log)] - [ +log.(-log)] log/(+log) log/(-log) (D) log/(+log) 9. If y(+)(+b)(+c)(+)/(-)(-b)(-c)(-) then the vlue of y/ is 4. If - - - - - - - - (+) +(+b) +(+c) +(+) -(-) -(-b) -(-c) -(-) - - - - - - - - (+) -(+b) +(+c) -(+) +(-) -(-b) +(-c) -(-) - - - - - - - - (-) +(-b) +(-c) +(-) -(+) -(+b) -(+c) -(+) / y( -4 ) ( - ) then y/ is 4 4 -/ -/ ( - +4 )( - ) ( -4 ) 9.4 COM M ON PROFICIENCY TEST

4. If 4 4 -/ -/ ( + -4 )( - ) ( -4 ) 4 4 -/ -/ ( + +4 )( - ) ( -4 ) / -/ y(-)(-) (+) then the vlue of [y/]/y is - - - (-) +(/)(-) -(/)(+) - - - (-) +(-) -(+) - - - (-) -(/)(-) +(/)(+) 4. If /4 ylog[e {-)/(+)} then y/ is - - +(/4)(-) -(/4)(+) - - +(/4)(-) +(/4)(+) - - -(/4)(-) +(/4)(+) 4. If 44. If 45. If 46. If 47. If 5/ / ye ( -) then the vlue of [y/]/y is - -/ ( - +5) ( -) - -/ ( + -5) ( -) - -/ ( -5 +) ( -) 5 -/ -/ y e (+) (-) then the vlue of [y/]/y is - - - 5+ -(/)(+) -(/)(-) - - - 5+ -(/)(+) -(/)(-) - - - 5+ -(/)(+) +(/)(-) / / -/4-4/5 y (5-) (4-) (7-4) then the vlue of [y/]/y is - - - - (/) -(4/)(5-) +(9/4)(4-) +(6/5)(7-4) - - - - (/) -(/4)(5-) +(9/4)(4-) +(6/5)(7-4) - - - - (/) +(4/)(5-) +(9/4)(4-) +(6/5)(7-4) y then the vlue of [y/]/y is log+ log- log(+) y(+) then the vlue of [y/]/y is - [(+) +log(+)] - [(+) -log(+)] - (+) +log(+) M ATHS 9.4

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 48. If / y then the vlue of [y/]/y is - (-log) (-log) - (+log) 49. If y( ) then y/ is + (+log) + (+log) + (-log) 5. If 5. If log y then y/ is log-.log log-.log log(log) y then the vlue of [y/]/y is given by log+.log - [+log(log)] [+log(log)] - [-log(log)] (D) [-log(log)] 5. If 5. If 54. If y + + + being constnt then y/ is - + log+ (log+) - + log- (log+) / / (+y) +y(+) then y/ is - + log+ (log-) - - - - -(+ ) (+ ) -(+ ) (D) (+ ) -y +-5y then y/ is - (+)(y+5) - (+)(y-5) - (-)(y-5) 55. If 56. If -y +y + then y/ is (y - )/[y (-)] (y - )/[y(+)] +hy+by +g+fy+c then y/ is -(+hy+g)/(h+by+f) (-hy+g)/(h-by+f) (y - )/[y (-)] (+hy+g)/(h+by+f) 57. If y... then y/ is y /[(-ylog)] y /[(+ylog)] (D) y /(-ylog) y /(+ylog)] 9.4 COM M ON PROFICIENCY TEST

58. The slope of the tngent t the point ( -) to the curve 59. If +y+y -4 is given by +y - then y/ is (-)/y (+)/y (-)/y 6. If +y+y -4 then y/ is -(+y)/(+y) -(+y)/(+y) (+y)/(+y) (D) (+y)/(+y) 6. If +5 y+y-5 then y/ is -( +y+y)/[(5+)] -( +y+y)/[(5-)] ( +y+y)/[(5+)] 6. If m+n m n (+y) - y then y/ is y/ -y/ /y (D) -/y 6. Fin the fourth erivtive of 64. If 65. If / log[(+4) ] -4-4 -4-4(+4) 4(+4) -4(4+) / m y[+(+ ) ] then the vlue of the epression (+ ) y/ +y/-m y is m n y e then y/ is m- n m- n m n m(m-) e +mn e +n e m- n m- n m n m(-m) e +mn e +n e m- n m- n m n m(m+) e +mn e +n e 66. If y(log)/ then y/ is (log-)/ (log-)/ (log+)/ 67. If m -m ye +be then y/ is m y my -m y (D) -my 68. If ye +be where n b re constnts the vlue of the epression y/ -4y/+4y is. M ATHS 9.4

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 69. If 7. If / n / n y[+( -) ] +b[-( -) ] the vlue of the epression ( -y) y/ +y/-n y is. by 7. If / / y(+) -(-) the vlue of the epression ( -) y/ +y/-y/4 is given / ylog[+(+ ) ] the vlue of the epression ( +) y/ +y/ is. 7. If t n yt then y/ is /(t ) -/(t ) t 7. If (-t)/(+t) n t(t)/(+t) then y/ is Integrl Clculus. Integrte w.r.t, 5 (5/) +k (/5) +k 5 +k (D). Integrte w.r.t 4, (-- ) 5 5 5 - - /5 +k + - /5 +k + + /5 +k. Integrte w.r.t, (4 + -+5) 4 + - +5 +k 4 - + -5 +k 4 + - +5 +k 4. Integrte w.r.t, ( -) 5 5 /5-(/) + +k /5+(/) + +k 5 /5+(/) + +k 5. Integrte w.r.t / -/, ( -/+ ) / / / / (/) -(/4) +4 +k (/) -(/4) +4 / / (/) +(/4) +4 6. Integrte w.r.t, (-)(+) ² ³+k ²+³+k +²+³+k 9.4 4 COM M ON PROFICIENCY TEST

7. Integrte w.r.t 4, ( +)/ ³/ /+k /- / +k /+/ +k 8. Integrte w.r.t -, ( +4 -+8) log-(4/) +(/) -8 +k log+(4/) +(/) +8 +k log+(4/) -(/) +8 +k 9. Integrte w.r.t, (-/) 4 - /4-(/) +log+ / +k 4 - /4-(/) +log+ / +k 4 - /4+(/) +log+ / +k. Integrte w.r.t / -/, ( -+ +7) 5/ / 5/6 / 5/ / 5/6 / (/5) - +(6/5) -4 +k (5/) - +(5/6) +4 +k 5/ / 5/6 / (/5) + +(6/5) +4 +k. Integrte w.r.t - -7, ( +b +c ) 4-4 (/4) +blog-(/4)c +k 4-4 4 +blog-4c +k 4-4 (/4) +blog+(/4)c +k. Integrte w.r.t 6/5, /5 /5 /5 (5/) +k (/5) +k (/5) +k. Integrte w.r.t 4/, 7/ 7/ / (/7) +k (7/) +k (/) +k 4. Integrte w.r.t -/, / / -/ +k (/) +k -(/) +k 5. Integrte w.r.t / -/, ( - ) / / / / (/) - +k (/) -(/) +k -/ -/ -(/) -(/) +k 6. Integrte w.r.t -/ - -, (7 -+8- + + ) / - (7/) -(/) +8- +log- +k / - (/7) -(/) +8-(/) +log+ +k / - (7/) +(/) +8+ +log+ +k M ATHS 9.4 5

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 7. Integrte w.r.t -, [ +b +c+] (/) +(/)b +c+log +k +b +c+log +k - +b- +k 8. Integrte w.r.t - 6 5 4, [4 + + + + +] 4 - + + ++log-(/) +k 4 - + + ++log+(/) +k 4 - + + ++log+ +k 9. Integrte w.r.t - - -/, [ +(/)e +4 - ] - / /log-(/)e +4log-(/) +k - / /log+(/)e +4log+(/) +k - / /log-e +4log-(/) +k. Integrte w.r.t 6, (4+5) 7 7 7 (/8)(4+5) +k (/7)(4+5) +k 7(4+5) +k. Integrte w.r.t 5, ( +4) 6 6 (/)( +4) +k (/6)( +4) +k 6 6( +4) +k. Integrte w.r.t n, (+) n+ n n- (+) /(n+) +k (+) /n +k (+) /(n-) +k. Integrte w.r.t, ( +) (/)( +) +k ( +) +k 4. Integrte w.r.t /, ( +) ( +) +k (D) 9 ( +) +k / (/9)( +) +k 5. Integrte w.r.t -, ( +) 8 / (/)( +) +k / (9/)( +) +k - - -(4/)( +) +k (4/)( +) +k 6. Integrte w.r.t -/4, ( +) - (/)( +) +k /4 (4/9)( +) +k /4 (9/4)( +) +k /4 (/4)( +) +k 9.4 6 COM M ON PROFICIENCY TEST

7. Integrte w.r.t -n, ( +) -n n- (/)( +) /(-n) +k (/)( +) /(-n) -n (/)( +) /(-n) +k 8. Integrte w.r.t -, ( +) -(/4)( +)/( +) +k -(/4)( +)/( +)+k 9. Integrte w.r.t, /[loglog(log)] (D) (/4)( +)/( +) +k (/4)( +)/( +) +k log[log(log)] +k log(log) +k log +k (D) -. Integrte w.r.t, /[(log) ] -/log +k /log +k log. Integrte w.r.t -, ( +) - -(/) ( +) +k. Integrte w.r.t -, (+7)( +-) - (/) ( +) +k (/4)log( +-)+(9/)log[(-)/{(+)}] +k (/4)log( +-)+ log[(-)/{(+)}] +k - ( +) +k (/4)log( +-)+(9/)log[(-)(+)] +k. Integrte w.r.t, /( --) (/)log[(-)/(+)]+c -(/)log[(-)/(+)] +c (/)log[(-)/(+)] 4. Integrte w.r.t -, (+)(+- ) -(/)log(+- )+(/)log[(+)/(-)]+c (/)log(+- )+(/)log[(+)/(-)] +c -(/)log(+- )+(/)log[(-)/(+)]+c 5. Integrte w.r.t -/, (5 +8+4) / (/ 5)log[{ 5+4/ 5+(5 +8+4) }] +c / 5log[{ 5+4/ 5+(5 +8+4) }] +c -/ (/ 5)log[{ 5+4/ 5+(5 +8+4) }] +c M ATHS 9.4 7

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS 6. Integrte w.r.t -/, (+)(5 +8-4) / / (/5)(5 +8-4) +[/(5 5)]log[5{+4/5+( +8/5-4/5) (/6)}] +c / -/ (/5)(5 +8-4) +[/(5 5)]log[5{+4/5+( +8/5-4/5) (/6)}] +c / / (/5)(5 +8-4) +[/(5 5)]log[5{+4/5+( +8/5-4/5) }]+c 7. Integrte w.r.t 4 -, ( -)( - +) [/( )]log[( - +)/( + +)]+c [/( )]log[( + +)/( - +)]+c [/( )]log[( - +)/( + +)]+c 8. Integrte w.r.t, e (/) ( e )-(/9)(e )+(/7)e +c (/) ( e )+(/9)(e )+(/7)e +c (/) ( e )-(/9)(e )+(/7)e +c 9. Integrte w.r.t, log (log-) +c (log+) +c log- +c (D) log++c 4. Integrte w.r.t n, log n+ - - (n+) [log-(n+) ] +c n+ - - (n+) [log+(n+) ] +c 4. Integrte w.r.t -, e (+) n- - - (n-) [log-(n-) ] +c - - - e (+) +c e (+) e (+) +c 4. Integrte w.r.t, e e (-) +k e (+) e (-) +k 4. Integrte w.r.t, e e ( -+) +k e ( ++) e (+) +k 9.4 8 COM M ON PROFICIENCY TEST

44. Integrte w.r.t, log (/4) log( /e) +k (/4) log(/e) +k (/) log( /e) +k 45. Integrte w.r.t, (log) (log) -log+ +k (log) -log+ +k (D) (log) +log+ +k (log) +log+ +k 46. Integrte w.r.t -, e (+)(+) - - - e (+) +k -e (+) +k (/)e (+) +k 47. Integrte w.r.t e log +k 48. Integrte w.r.t -, e (+log) -e log +k - e +k - -, (-) (+) (/)[log(-)+(/)log(+)]+k (/)[log(-)-(/)log(+)]+k 49. Integrte w.r.t -, (- ) (/)[log(-)+log(+)]+k (/)log[ /(- )] +k (/)log[ /(+) ] +k (/)log[ /(-) ] +k 5. Integrte w.r.t -, [(-)(-b)(-c)] given tht /A(-b)(-c)/, /B(b-)(b-c)/b, /C(c-)(c-b)/c +Alog(-)+Blog(- b)+clog(-c)+k Alog(-)+Blog(-b)+Clog(-c) +k +Alog(-)+Blog(-b)+Clog(-c)+k 5. Integrte w.r.t -, (5- ) from lower limit to upper limit 4 of (/4)log(/5) +k (/5)log(/4) (/5)log(4/) +k (D) (/4)log5 +k 5. Integrte w.r.t /, (+) from lower limit to upper limit of / 98/ M ATHS 9.4 9

BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS ANSWERS Differentil Clculus ) C ) A ) A 4) A 5) A 6) A 7) A 8) A 9) B ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) A 9) A ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) A 9) A ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) B 9) A 4) A 4) A 4) A 4) A 44) A 45) A 46) A 47) A 48) A 49) A 5) A 5) A 5) A 5) A 54) A 55) A 56) A 57) A 58) B 59) A 6) A 6) A 6) A 6) A 64) A 65) A 66) A 67) A 68) A 69) A 7) A 7) A 7) A 7) A Integrl Clculus ) A ) A ) A 4) A 5) A 6) A 7) A 8) B 9) A ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) A 9) A ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) A 9) A ) A ) A ) A ) A 4) A 5) A 6) A 7) A 8) A 9) A 4) A 4) A 4) A 4) A 44) A 45) A 46) A 47) A 48) A 49) A 5) A 5) B 5) C 9.5 COM M ON PROFICIENCY TEST