MAT 403 NOTES 4. f + f =

Transcription

1 MAT 403 NOTES 4 1. Fundmentl Theorem o Clulus We will proo more generl version o the FTC thn the textook. But just like the textook, we strt with the ollowing proposition. Let R[, ] e the set o Riemnn integrle rel-vlued untions on [, ]. Proposition 1.1. Let,, R, < <, nd let e rel-vlued untion on [, ]. Then is integrle on [, ] i nd only i it is integrle on oth [, ] nd [, ], in whih se Even though this result is intuitively ler (keeping in mind tht the Riemnn integrls represent signed-re ), the proo is perhps more sutle thn you think. I reer you to the textook or the detils. The most useul orm o this result n e stted one we dopt the ollowing onventions: or R[, ], nd or ny [, ], = = = 0. Now we hve Corollry 1.2 (Additivity Theorem). I is rel-vlued untion on n intervl o R whih ontins the points,, nd i two o the ollowing quntities exist, then the third exists. Moreover,, Proo. With the onventions, the ssertions re trivil i ny two o,, re the sme. Hene we n ssume,, re ll distint. Moreover, i we write the lst equlity s (1.1), =., = 0, then it is ler tht the ssertions re unhnged under ny yli permuttion o,,. Thus, we n ssume either < < or < <. Note tht the ssertions in these two ses re equivlent under the onventions. In the irst se, nd hene in oth ses, they ollow rom Proposition

2 2 NOTES 4 Theorem 1.3 (Fundmentl Theorem o Clulus). Let R[, ]. Then the untion deined y F (x) = is ontinuous on [, ]. Moreover, i is ontinuous t point [, ] then F is dierentile t nd tht F () = (). Proo. Sine is integrle on [, ], it is ounded on [, ]. Let B 0 e numer suh tht B on [, ]. For ny [, ], x F (x) F () = = = B x 0, or x [, ], x. Sine [, ] is ritrry, we onlude tht F is ontinuous on [, ]. Suppose is ontinuous t some [, ]. By the dditive theorem, or x [.], F (x) F () = So the untion ϕ: [, ] R deined y 1 ϕ(x) = x () = x x = stisies F (x) F () = ϕ(x)(x ) or ll x [, ]. Sine is ontinuous t, so or ny ε > 0, there exists δ > 0, suh tht (t) () < ε or ll t [, ] with t < δ. Thereore or x [, ] ( δ, + δ), ϕ(x) () = 1 x (t) dt 1 () dt x x = 1 x (t) () dt 1 ε x = ε. x This shows tht ϕ(x) is ontinuous t nd hene F (x) is dierentile t nd F () = ϕ() = (). Certinly i is ontinuous on [, ], then F (x) deined s ove is dierentile on [, ] (with one-sided derivtive onsidered t the end points) nd F = on [, ]. We ll untion F n ntiderivtive o primitive o i F =. Thus we hve just shown tht i is ontinuous on [, ], then it hs n ntiderivtive on [, ]. I F is n ntiderivtive o then so is F + k (k R). On the other hnd, i F nd G re two ntiderivtives o on [, ] (note tht F, G re ontinuous on [, ]), then (F G) = F G = = 0 on [, ] hene F G is onstnt on [, ] y the MVT. (More generlly i the domin o is not onneted, then two ntiderivtives o dier y lolly onstnt untion). The next ouple exmples show tht i is merely integrle ut not ontinuous on [, ] then its indeinite integrl needs not e n nti-derivtive o..

3 MAT Exmple 1.1. Let e the signum untion, i.e. 1 x > 0; (x) = 0 x = 0; 1 x < 0. Sine is step untion, it is integrle on [ 1, 1]. Its indeinite integrl with se point 1 is F (x) := 1 = x 1. But F is not n nti-derivtive o on [ 1, 1] euse F does not exist t x = 0 while (0) = 0. Exmple 1.2. Let h: [0, 1] R e the Thome untion (.k. the rindrop untion). We know rom HW 3 tht h R[0, 1] nd tht 1 h = 0. Thus h is integrle on [0, x] or ny x [0, 1] nd tht H(x) := 0 0 h = 0 or every x [0, 1]. In other words, H nd hene H 0 on [0, 1]. Thus H nd h disgree on dense set o points, nmely the rtionls, in [0, 1]. The ollowing is useul orollry o the FTC whih turns the tsk o inding integrls into the tsk o inding nti-derivtives. Corollry 1.4. I F is rel-vlued untion on n open set U o R tht hs ontinuous derivtive nd, U, then Something muh more generl is true: = F () F (). Theorem 1.5. Suppose F, : [, ] R re untions nd E is inite suset o [, ] suh tht: (1) F is ontinuous on [, ]. (2) F (x) = (x) or ll x [, ] \ E. (3) R[, ]. Then = F () F (). Proo. We will prove the theorem in the se where E = {, }. The generl se n e otin y reking the intervl into the union o inite numer o suintervls. Consider prtition P = {x 0, x 1,..., x n } o [, ]. Sine F is ontinuous on [, ] nd F = on (, ), or eh 1 i n, y pplying the MVT to F on [x i 1, x i ], we onlude tht there exists u i (x i 1, x i ) suh tht F (x i ) F (x i 1 ) = F (u i )(x i x i 1 ) = (u i )(x i x i 1 ). Summing oth sides through 1 i n, we get n F () F () = (u i )(x i x i 1 ). i=1 Note tht the right-side o the eqution ove is Riemnn sum o orresponding to the prtition P. Sine is integrle on [, ], this sum n e ritrrily lose

4 4 NOTES 4 to the integrl i P is suiiently smll. Thereore, we onlude tht let-side must e the integrl. Exmple 1.3. Let A(x) = x or ll x [ 1, 1]. Sine A is not dierentile t x = 0, so Corollry 1.4 does not even pply. However, A nd the signum untion sgn gree on [ 1, 1] exept x = 0 nd s step untion, sgn is integrle. Thus y tking E = {0} in Theorem 1.5, we onlude tht 1 1 sgn(x) dx = A(1) A( 1) = 1 1 = 0. Exmple 1.4. Let F (x) = 2 x or x [0, 1], then F is ontinuous on [0, 1] nd F (x) = 1/ x or x (0, 1]. However, := F is unounded on (0, 1] so is not integrle on [0, 1] no mtter how we extend it t x = 0. Thereore, Theorem 1.5 does not pply, yet s you know we n hndle this y the theory o improper integrls. Exmple 1.5. Let F (x) = x 2 os(1/x 2 ) or nd F (0) = 0. It is not hrd to hek tht ( ) ( ) ( ) F (x) = 2x os x 2 + sin x x 2 nd tht F (0) = 0. Thus F nd F stisy the irst two ondition in Theorem 1.5. However, sine F is unounded on [0, 1], it is not Riemnn integrle nd so Theorem 1.5 does not pply. The ollowing onsequene o the FTC is known s the hnge o vrile theorem (ut I ll it the nti-hin-rule sometime) Corollry 1.6. Let I nd J e open intervls in R, g : I J dierentile untion with ontinuous derivtive, nd : J R ontinuous untion. Then, or ny, I Proo. Let F (y) = y g() g() g() (u) du = g(t)g (t) dt. (u) du (y J). Sine is ontinuous on J, y the FTC (Theorem 1.3), F = on J. The Chin rule then implies the untion is dierentile on I nd G(t) := F g(t) = g(t) g() (u) du, G (t) = F (g(t))g (t) = (g(t))g (t), (t I), (t I). Sine, g nd g re ontinuous, (g(t))g (t) is ontinuous on I nd so y the FTC, the untion H(x) := (g(t))g (t) dt, (x I) nd G(x) hve the sme derivtive nd so they dier y onstnt. But sine G() = 0 = H(), thus H(x) = G(x) or ll x I so in prtiulr H() = G() whih is the lst ssertion o the orollry.

5 Exmple 1.6. Consider the integrl MAT sin t t dt. We mke the sustitution u = g(t) = t or t [1, 4]. Sine du/dt = g (t) = 1/(2 t) is ontinuous on [1, 4] the hnge o vrile theorem pplies. From tht it ollows esily tht the integrl is 2(os(1) os(2)). 4 sin t Exmple 1.7. Consider the integrl dt. This time the sme sustitution 0 t will not work sine g (t) is not ontinuous on [0, 4]. However, we ould pply Theorem 1.5 to the untion F (t) := 2 os( t) with E = {0} to evlute this integrl. (Ans: 2(1 os(2)))