Antierivtives 0. Introuction So fr much of the term hs been sent fining erivtives or rtes of chnge. But in some circumstnces we lrey know the rte of chnge n we wish to etermine the originl function. For emle, meters or t loggers often mesure rtes of chnge, e.g., miles er hour or kilowtts er hour. If you know the velocity of n object, cn you etermine the osition of the object. This coul hen in cr, sy, where the seeometer reings were being recore. Cn the osition of the cr be etermine from this informtion? Similrly, cn the osition of n irlne be etermine from the blck bo which recors the irsee? More generlly, given f 0 () cn we fin the function f (). If you think bout it, this is the sort of question I hve ske you to o on lbs, tests, n homework ssignments where I gve you the grh of f 0 () n si rw the grh of f (). Or where I gve you the number line informtion for f 0 () n f 00 () n ske you to reconstruct the grh of f (). Cn we o this sme thing if we strt with formul for f 0 ()? Cn we get n elicit formul for f ()? We usully stte the roblem this wy. DEFINITION 0.. Let f () be function efine on n intervl I. We sy tht F() is n ntierivtive of f () on I if F 0 () = f () for ll I. EXAMPLE 0.. If f () =, then F() = is n ntierivtive of f becuse F 0 () = = f (). But so is G() = + or, more generlly, H() = + c. Are there other ntierivtives of f () = besies those of the form H() = + c? We cn use the MVT to show tht the nswer is No. The roof will require three smll stes. THEOEM 0. (Theorem ). If F 0 () =0 for ll in n intervl I, then F() =k is constnt function. This mkes lot of sense: If the velocity of n object is 0, then its osition is constnt (not chnging). Here s the Proof. To show tht F() is constnt, we must show tht ny two outut vlues of F re the sme, i.e., F() =F(b) for ll n b in I. So ick ny n b in I (with < b). Then since F is ifferentible on I, then F is both continuous n ifferentible on the smller intervl [, b]. So the MVT
lies. There is oint c between n b so tht This mens F(b) b F() = F 0 (c) ) F(b) F() =F 0 (c)[b ] =0[b ] =0. F(b) =F(), in other wors, F is constnt. THEOEM 0. (Theorem ). Suose tht F, G re ifferentible on the intervl I n F 0 () =G 0 () for ll in I. Then there eist k so tht G() =F()+k. Proof. Consier the function G() F() on I. Then (G() F()) = G0 () F 0 () =0. Therefore, by Theorem so G() F() =k G() =F()+k. THEOEM 0.3 (Theorem 3: Fmilies of Antierivtives). If F() n G() re both ntierivtives of f () on n intervl I, then G() =F()+k. This is the theorem we wnt to show. Proof. If F() n G() re both ntierivtives of f () on n intervl I then G 0 () = f () n F 0 () = f (), tht is, F 0 () =G 0 () on I. Then by Theorem G() =F()+k. DEFINITION 0.. If F() is ny ntierivtive of f (), we sy tht F()+c is the generl ntierivtive of f () on I. Nottion for Antierivtives Antiifferentition is lso clle inefinite integrtion. is the integrtion symbol f () is clle the integrn inictes the vrible of integrtion f () = F()+c. F() is rticulr ntierivtive of f () n c is the constnt of integrtion. We refer to f () s n ntierivtive of f () or n inefinite integrl of f.
mth 30, y 0 introuction to ntierivtives 3 Here re severl emles. = + c cos tt= sin t + c e z z = e z + c = rctn + c + Antiifferentition reverses ifferentition so F 0 () = F()+c n ifferentition unoes ntiifferentition le f () = f (). Differentition n ntiifferentition re reverse rocesses, so ech erivtive rule hs corresoning ntiifferentition rule. Differentition (c) =0 (k) =k Antiifferentition 0 = c k= k + c (n )=n n n = n+ n+ + c, n 6= (ln ) = = = ln + c (sin ) =cos cos = sin + c (cos ) = sin sin = cos + c (tn ) =sec sec = tn + c (sec ) =sec tn sec tn = sec + c (e )=e e = e + c (rcsin ) = = rcsin + c (rctn ) = = rctn + c + + Vritions n Generliztions Notice wht hens when we use inste of in some of these functions. We multily by when tking the erivtive, so we hve to ivie by when tking the ntierivtive. Differentition (e )=e (sin ) = cos (tn ) = sec (rcsin( )= (rctn( )) = + Antiifferentition e = e + c cos = sin + c sec = tn + c = rcsin( )+c = + rctn( )+c Try filling in the rules for sin n sec() tn().
EXAMPLE 0.. Here re few emles. 0. Problems cos() = sin()+c e z/ z = e z/ + c 6 + = rctn + c. Determine ntierivtives of the following functions. Tke the erivtive of your nswer to confirm tht you re correct. Why shoul you +c to ny nswer? Bsics: () 7 6 (b) 6 (c) 6 () e (e) e (f )e (g) e (h) (j) cos (k) cos (l) cos() (m) cos() (n) (o) sec tn () sin (q) sec (r) (s) (t) 8 (u) 6 ln 6 (v) 6 6 ln 6 (w) 6 () 6+ sec (y). Now try these. Think it through. (i) 6 + (z) 8 () 8 + (b) e 8 c (c) sin () cos q sec q (e) + 3 (f ) cos (g) 5/ (h) 3/5 (i) (j) 3 + (k) ( + 5) 6 (l) e cos(e ) (m) e +9 + cos (n) 7 (ln )5 Answers. Answers () 7 + c (b) 7 7 + c (c) 7 7 + c () e + c (e) e + c (f ) e + c (g) e + c (h) ln + c (i) 8 ln + c (j) sin + c (k) sin + c (l) sin()+c (m) sin()+c (n) ln + sin + c (o) sec + c () cos + c (q) tn + c (r) + c (s) + c (t) + c (u) 6 + c (v) 3 6 + c (w) 6 + c () 6 + tn + c (y) 6 rctn()+c (z) rcsin()+c. Answers. () 9 9 + + c (b) 8 e8 c + (c) cos + c () sin q tn q + c (e) 3 3 + 3 + c (f ) sin + c (g) 7 7/ + c (h) 5 /5 + c (i) 3 3/ (j) ln( + )+c (k) ( + 5) 7 + c (l) sin(e )+c (m) 6e +9 + c (n) (ln ) 6 + c
Generl Antierivtive ules The key ie is tht ech erivtive rule cn be written s n ntierivtive rule. We ve seen how this works with secific functions like sin n e n now we emine how the generl erivtive rules cn be reverse. FACT. (Sum ule). The sum rule for erivtives sys (F() ± G()) = (F()) ± (G()). The corresoning ntierivtive rule is ( f () ± g()) = f () ± g(). FACT. (Constnt Multile ule). The constnt multile rule for erivtives sys (cf()) = c (F()). The corresoning ntierivtive rule is cf() = c f (). Emles 8 3 7 = 8 3 7 / = 8 3 7 / = 8 7 3/ 3/ + c = 3/ + c 3 6 cos 7 + /3 = 6 cos 7 + /3 = 6 /3 sin 7 ln + /3 + c = 3 sin 7 ln + 3/3 + c. 3e / 8 9 = 3 e / 8 9 = 3 e/ + 8 rcsin 3 + c. = 6e/ + 8 rcsin 3 + c. ewriting ewriting the integrn cn gretly simlify the ntierivtive rocess. 5 t 6 sec tt= + = 6 ( ) = 6 6 3 t = t 5 t /5 6 sec tt= t7/5 7/5 6 tn t + c = 0t7/5 7 + = 3 3 + + c = 3 3 + c. 6 = 67 7 + 63 3 + c = 67 7 + 3 + c. t 5/3 t /3 t = /3 + c = 3t /3 + c. 6 tn t + c. 8 + 7 = 8 3/ + 7 / = 85/3 5/3 + 7/ / + c = 5/3 + / + c. 5