f(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that

Transcription

1 Uiversity of Illiois t Ur-Chmpig Fll 6 Mth 444 Group E3 Itegrtio : correctio of the exercises.. ( Assume tht f : [, ] R is cotiuous fuctio such tht f(x for ll x (,, d f(tdt =. Show tht f(x = for ll x [, ] ; c you use the fudmetl theorem of clculus to prove this result? ( Use this to show tht if f is cotiuous o [, ] d f(tdt = the there must exist t (, such tht f(t =. Correctio. ( First,otice tht, sice f is cotiuous, provig tht f(t = for ll t [, ] is the sme s provig tht f(t = for ll t (,. Now, let us prove the cotrposite of the result we re iterested i ; i other words, let us prove tht if f(x > for some x (,, f(x for ll x (, d f is cotiuous o [, ] the f(tdt >. To prove this, otice tht sice f is cotiuous t x there exists δ > such tht f(y f(x for ll y [, ] such tht y x δ. If δ is smll eough, [x δ, x + δ] [, ] ; ut the x+δ x δ f(tdt δ f(x = δf(x >. Sice f(y for ll y [, ], we ow tht x δ f(tdt d f(tdt ; thus the dditivity theorem x+δ shows tht f(tdt >, which is wht we wted. Oe c ideed prove this result usig the fudmetl theorem of clculus (d the me vlue theorem : Set F (x = x f(tdt ; the F is dieretile d F (x = f(x, hece F is icresig. We hve F ( = y deitio, d the ssumptio tht f(tdt = gives F ( =. Sice F is icresig, this mes tht ctully F is costt o [, ], thus its derivtive is equl to o [, ], d this gives f(x = for ll x [, ]. ( First, otice tht if f does't te the vlue o (, the f is either lwys > or lwys < o [, ] (ecuse of the itermedite vlue theorem. But the the precedig quesio shows tht oe cot hve f(tdt =. Hece there must exist t (, such tht f(t =.. Use the result of the precedig exercise to solve the followig questios. ( Fid ll the cotiuous fuctios f : [, ] R such tht f(tdt = ( sup{ f(x : x [, ]}. ( Assume f : [, ] R is cotiuous fuctio such tht f(tdt = ; prove tht there exists (, such tht f( =. (c Show tht if f, g re cotiuous o [, ] d f(tdt = g(tdt the there must exist some c [, ] such tht f(x = g(c. Correctio. ( The fuctio g deed o [, ] y g(x = sup{ f(x : x [, ]} f(x is cotiuous, d g(x for ll x [, ]. The ssumptio f(tdt = ( sup{ f(x : x [, ]} is equivlet to g(tdt =, which i tur is equivlet to g(x = for ll x [, ]. This mes tht the fuctios tht stisfy the equlity we re iterested i re the fuctios f which re costt o [, ], d oegtive. ( Assume tht for ll t (, oe hs f(t > t ; the we ow tht (f(t tdt >, d this is the sme s syig tht f(tdt >. Similrly, if f(t < t for ll t (, oe gets f(tdt >. Thus, it is oly possile tht f(tdt = if there exist t, t (, such tht f(t t, f(t t. If either f(t = t or f(t = t we re doe ; otherwise the fuctio x f(x x chges sig o (,. Sice this fuctio is cotiuous, the me vlue theorem esures tht it must hve zero o (,, which shows tht there exists (, such tht f( =. (c This is direct cosequece of questio ( (pplied to the cotiuous fuctio f g.

2 3. Usig Riem sums, compute the limits (whe + of the followig sequeces : + ; + ; 3 ; ( si( π π si(( l( + si( π ; ; (. Correctio. Here, the tric is to recogize Riem sums : the rst oe is + =, d this + is Riem sum for the fuctio x for tgged prtitio of [, ] with mesh /. Thus, we oti + x lim ( dt = + + t = l(. The secod oe is similr : + = + (, hece lim( + = dt + t = rct( rct( = π 4. The third oe is more of the sme : 3 = (, ( hece lim = 3 t dt = 3. The fourth oe loos sty, ut gi it is Riem sum for the cotiuous fuctio t l( + t o [, ],, si ( π ], si ( π with regrd to the tgged prtitio { [ si ( ( π (use the me vlue theorem to prove this. Hece whe + the sum coverges to is computle usig itegrtio y prts : l( + tdt = [ (t + l(t + ] t= t= }, the mesh of which is smller th l( + tdt, which dt = l(. The lst oe does't loo lie Riem sum ; there is some wor to e doe efore oe c see Riem sum pper. Assume rst tht = p ; oe hs ( u p = = + = + Hece, whe = p, oe hs ( = Riem sum for the cotiuous fuctio t +x =p+ = p + which is p (c you see why?. So, we see tht u p coverges to o [, ] d the tgged prtitio {[. The sum o the right is ctully p, p ], p }, the mesh of dt +t = l(. Sice u p+ u p coverges to, we see tht oe lso hs lim(u p+ = l(. A theorem we sw i clss esures tht (u is coverget d lim(u = l(. 4. Let f, g : [, ] R e cotiuous fuctios. Show tht lim f ( ( g = f(tg(t dt. Correctio. This is tricier th it loos : if we hd f ( ( g i the sum, the it would just e usul Riem sum d we could pply the results see i clss. Ufortutely, this is ot wht we hve ; how c we del with this? Oe c proceed s follows : rst, write tht f ( ( g = f ( ( g + f ( ( g ( ( g. The rst term coverges to f(tg(tdt, so we wt to prove tht the secod term coverges to. For tht, we use the fct tht g is uiformly cotiuous o [, ] ; give ε >, there exists δ ε such tht x y δ ε

3 f(x f(y ε for ll x, y [, ]. Hece if is ig eough oe hs g ( ( g ε, so tht f ( ( g ( ( g ε f (. Sice f is Riem-itegrle f lso is Riem-itegrle, hece f( coverges to f(t dt. So if is ig eough oe hs f( f(t dt +. Puttig ll this together, we get tht for y ε there exists K N such tht f( ( ( g ( g ε( f(t dt + for ll K. This proves tht f( ( ( g ( g coverges to (whe +, which is wht we eeded to prove. 5. Let f : [, ] R e cotiuous fuctio such tht f(uu du = for ll {,..., }. Show tht f hs t lest + distict zeros i (,. Hit : prove the result y iductio usig itegrtio y prts d Rolle's theorem. Correctio. Followig the hit, let us prove the result y iductio. For = the result is direct cosequece of exercise ; ssume the result is true for. The pic cotiuous fuctio f such tht for ll {,..., + }, d set F (x = Also, for y =,...,, oe hs Thus we oti x f(uu du = f(tdt. The ssumptio o f for = yields F ( = F ( =. u du = [ u ] u F (udu u F (udu = for ll =,...,, which yields (ecuse of our iductio hypothesis tht F hs t lest distict zeros i (,. Sice F ( = F ( =, F must hve t lest + distict zeros o [, ]. Ad F = f hs zero etwee y two zeros of F, which shows tht f hs t lest + distict zeros o (, We eed to use the deitio of Riem itegrl ; ssume the poits c,..., c re idexed i such wy tht c < c <... < c < c, d set M = mx{ f(c i : i =,..., }. The pic tgged prtitio Ṗ = {[x i, x i ], t i } i=,...m of [, ]. Oe hs S(f, Ṗ = m (x i x i f(t i m m (x i x i f(t i Ṗ f(t i. i= i= Sice there re oly poits i the itervl t which f(x, d t ech of these poits oe hs f(x M, we see tht S(f, Ṗ Ṗ..M (ecuse there c e t most two t i with the sme vlue, d t most poits t which f is ozero, so t most of them c pper i the sum. But the (sice, M re costt ε M, wht we hve proved implies tht for y prtitio Ṗ with mesh less th we re doe : if oe sets δ ε = δ ε oe hs S(f, Ṗ ε. This is exctly wht we eeded to prove tht f R([, ] d i= f(xdx = This is cosequece of the precedig exercise : ideed, the fuctio f g stises the coditio of exercise 7..3, hece f g R[, ] d (f(t g(tdt =. But the f = (f g + g is the sum of two Riem-itegrle fuctios, so f R([, ] d f(tdt = (f(t g(tdt + g(tdt = g(tdt Let us follow the hit : pic ε >, set δ ε = ε 4α d pic tgged prtitio Ṗ = {[x i, x i ], t i } i=,..., with mesh δ ε. The y deitio oe hs S(ϕ, Ṗ = i= (x i x i ϕ(t i. There re two possiilities for ϕ(t i : either it is equl to, or it is equl to α. Oly the t i 's tht elog to [c, d] cotriute to the sum. Let I = {i: t i [c, d]}. The S(ϕ, Ṗ = α i I (x i x i. Sice t i [x i, x i ], t i c oly e i [c, d] if (x i < d x i, or (x i, x i re oth i [c, d], or (x i d d x i > d. The rst d third coditio c ech

4 e stised t most y oe idex, d the remiig [x i, x i ] from prtitio of suitervl of [c, d], so tht S(ϕ, Ṗ α(d c + δ εα. Similrly, the "chu" of [x, d] tht c e missed y the t i 's is t most δ ε log, hece S(ϕ, Ṗ α(d c δ εα. This shows tht wheever Ṗ is tgged prtitio with mesh less th δ ε = ε 4α oe hs α(d c ε S(ϕ, Ṗ α(d c + ε. This is eough to show tht ϕ R([, ] d ϕ(tdt = α(d c. { M if x [, c 7... Let's follow the hit d dee (give ε > α ε, ω ε y α ε (x = d ω ε (x = f(x if x [c, ] { M if x [, c (c is to e specied lter. The oe hs α ε (x f(x ω ε (x for ll x [, ]. Also, α ε, ω ε f(x if x [c, ] re oth Riem-itegrle o [, ] ecuse of the Additivity theorem. Filly, oe hs (ω ε(t α ε (tdt = c Mdt = M(c y deitio of α ε, ω ε. Hece if oe sets c = + ε M we get tht (ω ε(t α ε (t ε. So we mged to prove tht the ssumptios of the Squeeze Theorem re stised, hece f R([, ]. The sice f(x M for ll x [, ] we see tht c f(tdt c M(c, so limc f(tdt =. Thus the dditivity theorem gives lim c c f(tdt = f(tdt This is cosequece of the precedig exercise : g(x for ll x [, ], d g is cotiuous o [c, ] for ll c (,. Hece it is Riem-itegrle o [, ] Set F (x = x f(tdt ; sice f is cotiuous o [, ], the fudmetl theorem of clculus esures tht F is dieretile o [, ], hece it stises the ssumptios of the Me Vlue theorem o this itervl, so there exists c (, such tht F ( F ( = F (c(. This is the sme s syig tht there exists c (, such tht f(tdt = f(c(. Oe c lso solve this exercise dieretly : oe hs f([, ] = [m, M] y the theorems out cotiuous fuctios, from which we get m( R f(tdt M( But the m f(tdt M, hece there exists R c such tht f(c = f(tdt, which is the sme s syig tht ( f(x = f(tdt We c pply similr method to the oe i the exercise ove : deote gi f([, ] y [m, M]. The oe hs f(tg(t m g(tdt = (f(t mg(tdt (ecuse f(t m d g(t for ll t [, ]. Similrly, oe ds tht f(tg(tdt M g(tdt. Put together, this yields m g(tdt f(tg(tdt M. Ths to the itermedite vlue theorem, we c ow coclude : there exists c [, ] such tht f(c = R g(tdt f(tg(tdt, which is the sme s f(tg(tdt = f(c g(tdt. This result is clerly flse if oe o loger ssumes tht g tes oegtive vlues ; for istce, let =, =, f(t = t d g(t = t. The oe hs f(tg(tdt = ut f(c g(tdt = for ll c [, ] Here oe eeds to pply the Chi Rule (d the fudmetl theorem of clculus, which yields : ( I this cse F (x = G(x, where G (x = + x ; hece 3 F x (x = + (x 3 = x + x. 6 ( This time F (x = G(x G(x, where G (x = + x. Hece F (x = G (x xg (x = + x x + x Set rst F (x = x f(tdt. The we ow tht F is dieretile d F (x = f(x. By deitio, we hve g(x = F (x + c F (x c, hece g is compositio of dieretile fuctios. Thus g is dieretile o R, d the Chi Rule yields g (x = F (x + c F (x c = f(x + c f(x c.

5 First otice tht the ssumptio o f implies tht f(tdt = (te x =. Set F (x = x f(tdt. The the ssumptio o F ecome F (x = F ( F (x for ll x [, ], d sice F ( = this yields F (x = for ll x [, ]. Sice f is cotiuous the fudmetl theorem of clculus gives F = f, hece f(x = for ll x [, ] ( The fuctios x (tf(x + g(x d x (tf(x g(x re oth Riem-itegrle o [, ] d te oegtive vlues, hece (tf(u ± g(u dt. ( We hve : (tf(u+g(u du = (t f (u+tf(ug(u+g(u du = t f(u du+t f(ug(udu+ g(u du. Sice the qutity o the left is positive, we oti t f(ug(udu t f(u du + g(u du. Hece for y t > we hve f(ug(udu t f(u du + t g(u du. Similrly, usig the fct tht (tf(u g(u du, oe otis f(ug(udu t f(u du + t g(u du. The two iequlities together yield f(ug(udu t f(u du + t g(u du. (c If f (udu = the the result ove implies tht f(ug(udu t g(u du for ll t >. This is oly possile if f(ug(udu =. (d Sice oe hs oth fg fg d fg fg, it is true tht oth f(ug(udu f(ug(u du d f(ug(udu f(ug(u du. This mes tht f(ug(udu f(ug(u du, which is equivlet to the iequlity o the left. To prove the iequlity o the right,recll tht we ow from ( (pplied to f, g tht t f (udu + t f(ug(u du+ g u du for ll t R. This mes tht the polyomil fuctio t t f (udu+ t f(ug(u du + g u du eeps costt sig o R, d this is possile oly if its discrimit 4 ( f(ug(u u 4 f (udu g(u du is. I other words, oe must hve ( f(ug(u du f(u du g(u du. To get the iequlity we re sed to prove, pply this iequlity to the fuctios f(t = /t d g(t = : ( dt this yields t dt t dt = ( ( ( =. Tig the squre root, oe hs dt t (