MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES CHARLIE COLLIER UNIVERSITY OF BATH These notes hve been typeset by Chrlie Collier nd re bsed on the leture notes by Adrin Hill nd Thoms Cottrell. These notes re rom the demi yer 2015/16, nd so they should not be used solely or revision o this ourse in subsequent yers. Contents 4. Limits nd Continuity o Funtions 1 4.1. Limits with Rel Domins 1 4.2. Limits on Dierent Domins 2 4.3. Continuity 2 4.4. Continuity on Bounded Intervls 3 4.5. Subsequene Arguments 4 5. Dierentition o Funtions 5 5.1. Dierentition t Point 5 5.2. Dierentition over n Intervl 6 5.3. Consequenes o the Men Vlue Theorem 7 5.4. De l Hospitl s Rule 7 6. Riemnn Integrtion 9 6.1. Upper nd Lower Riemnn Sums 9 6.2. Integrbility o Continuous Funtions 10 6.3. Properties o the Integrl 11 6.4. The Fundmentl Theorem o Clulus 11 6.5. Integrtion Tehniques 12 6.6. Improper Integrls 12

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 1 4.1. Limits with Rel Domins. 4. Limits nd Continuity o Funtions Deinition 4.1. Let : R R nd, l R. Then x (x) = l mens tht ε > 0, δ > 0 suh tht 0 < x < δ = (x) l < ε. Exmples 4.3. Show tht x (x) = l or the ollowing untions : R R. () (x) = 1 x R with = 0, l = 1. (x + 2) 1, x R\ 2} (b) (x) = 0, x = 2 () (x) = x 3 x R with = 1, l = 1. x + 3, x 2 (d) (x) = with = 2, l = 5. 7 x, x > 2 with = 0, l = 1/2. Lemm 4.4. Let : R R nd,, l R. Suppose tht x (x) = l nd l >. Then δ 0 > 0 suh tht 0 < x < δ 0 = (x) >. Corollry 4.5. Let : R R nd, L, M R nd d > 0. Suppose tht x (x) = L nd tht 0 < x < d = (x) M. Then L M. Theorem 4.6 (Sequentil hrteristion o the it). Let : R R nd let, L R. The ollowing re equivlent, () x (x) = L; (b) given sequene (x n ) n=1 R\}, then x n = (x n ) L. Lemm 4.7 (Uniqueness o its). Let : R R nd let R. Suppose tht onverges to it s x, then the it is unique. Theorem 4.8 (Algebr o its). Let, g : R R nd let, k, L, M R. Suppose tht x (x) = L nd x g(x) = M, then () x ( + g)(x) = L + M; (b) x (k)(x) = kl; () x (g)(x) = LM; (d) i M 0, then x (x)/g(x) = L/M. Corollry 4.9. Let R nd let p, q be rel polynomils. Then () x p(x) = p(); (b) i q() 0, then x p(x)/q(x) = p()/q(). Exmples 4.10 (Proving its do not exist). Show tht the ollowing untions do not hve it s x 0, sin(1/x), x R\0} () : R R deined by (x) = ; 0, x = 0 1, x Q (b) : R R deined by (x) = 0, x R\Q.

2 CHARLIE COLLIER UNIVERSITY OF BATH 4.2. Limits on Dierent Domins. 4.2.1. Open Intervls. Lemm 4.11. Let, b, R stisy < < b. Then δ 0 > 0 suh tht ( δ 0, + δ 0 ) (, b). Deinition 4.12. Let, b R stisy < b nd let : (, b) R, R nd L R. Then x (x) = L mens tht given ny ε > 0, δ > 0 suh tht x (, b), 0 < x < δ = (x) L < ε 4.2.2. End Tils o Closed Intervls. Lemm 4.15. Let, b R stisy < b. Then δ 0 > 0 suh tht 0 < x < δ 0 = x [, b]. Proo. Let x 0 = δ 0 /2. Then 0 < x 0 < δ 0, but x 0 < so x 0 [, b]. 4.2.3. One-Sided Limits. Deinition 4.17 (Funtions on the rels). Let : R R nd let, L R. Then x + (x) = L mens tht ε > 0, δ > 0 suh tht 0 < x < δ = (x) L < ε. Similrly, x (x) = L mens ε > 0, δ > 0 suh tht δ < x < 0 = (x) L < ε. Exmple 4.18. Let : R R be deined by (x) = Lemm 4.19. Let : R R nd let, L R. Then (x) = L nd x + 1, x 0 1, x < 0 ; ind x 0+ (x) nd x 0 (x). (x) = L (x) = L (4.1) x x Deinition 4.20 (Funtions on losed bounded intervls). Let, b, L R stisy < b. Let : [, b] R, then x + (x) = L mens x [, b] ε > 0, δ > 0 s.t. 0 < x < δ = (x) L < ε I δ 0 := b > 0, then ny δ (0, δ 0 ] stisies 0 < x < δ = < x < + δ + δ 0 = b, so x [, b] nd is well-deined. Theorem 4.21. Let : [, b] R, then the ollowing re equivlent, () (x n ) n=1 (, b], then n x n = = n (x n ) = L; (b) x + (x) = L. Deinition 4.22. Let, b, L R stisy < b nd : (, b) R. Then x + (x) = L mens ε > 0, δ > 0 suh tht 0 < x < δ = (x) L < ε. 4.3. Continuity. Deinition 4.23. Let : R R nd let R. Then is ontinuous t mens tht (x) = () x In other words, ε > 0, δ > 0 suh tht x < δ = (x) () < ε. This diers rom deinition 4.1 sine we inlude the se x = in x < δ sine () () = 0 < ε. Exmple 4.25. Let : R R stisy (x) x x R. Then is ontinuous t zero. Lemm 4.26. Let : R R nd let R. Then the ollowing re equivlent,

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 3 () is ontinuous t ; (b) (x n ) n=1 R, then n x n = = n (x n ) = (). Theorem 4.27 (Algebr o ontinuity). Let, g : R R nd let, K R. Let, g be ontinuous t, then () + g, (b) K, () g nd (d) /g re ll ontinuous t. Deinition 4.28. Let, g : R R, then the omposition o nd g is g : R R deined by ( g)(x) = (g(x)). Theorem 4.29. Let, g : R R nd let R. Assume tht g is ontinuous t nd is ontinuous t g(), then g is ontinuous t. 4.4. Continuity on Bounded Intervls. Deinition 4.30 (Open bounded intervls). Let, b R stisy < b nd let : (, b) R. Then is ontinuous on (, b) i (, b) nd ε > 0, δ(, ε) > 0 suh tht y < δ = y (, b) nd (y) () < ε. Exmple 4.32. Let : (0, 1) R be deined by (x) := 1/x. Then is ontinuous on (0, 1). 4.4.1. Closed Bounded Intervls. Deinition 4.34. Let, b R stisy < b nd let : [, b] R. Then is ontinuous on [, b] i [, b], (x n ) n=1 [, b], n x n = = n (x n) = () Deinition 4.35. Let, b R stisy < b. Then C([, b]) := : [, b] R is ontinuous on [, b]}. Exmple 4.36. Let : [0, 1] R be deined by (x) = x. Show tht C([0, 1]). Theorem 4.37 (Intermedite vlue theorem). Let, b R stisy < b. Let C([, b]) nd suppose tht () < (b), then y ((), (b)), (, b) suh tht () = y. Corollry 4.38. Let g : [0, 1] [0, 1] be ontinuous on [0, 1]. Then g hs ixed point ( point x [0, 1] suh tht g(x ) = x ). Deinition 4.39. Let, b R stisy < b nd : [, b] R be ontinuous on [, b]. Then is stritly inresing i x 1 < x 2 = (x 1 ) < (x 2 ) x 1, x 2 [, b]. Note tht is stritly deresing i is stritly inresing. Proposition 4.40. Let, b R stisy < b nd let C([, b]) be stritly inresing. Then : [, b] [(), (b)] is bijetion. Proposition 4.41. Proposition 4.40 n be extended to intervls suh s (, b), [, ) nd (, ]. Deinition 4.42. Let I nd J be intervls nd let : I J be bijetion. Then g : J I is the inverse untion o i y J, g(y) = x I suh tht (x) = y. We denote g = 1. Lemm 4.43. Let n N nd x 1, x 2 R. Then n 1 x n 1 x n 2 = (x 1 x 2 ) x k 1x2 n 1 k (4.2) Proposition 4.44. Let n N, then g : [0, ) [0, ) deined by g(y) = y 1/n is well-deined. Theorem 4.45. Let, b R stisy < b nd let : [, b] [(), (b)] be stritly inresing bijetion. Then the inverse untion g : [(), (b)] [, b] is stritly inresing nd ontinuous. k=0

4 CHARLIE COLLIER UNIVERSITY OF BATH 4.5. Subsequene Arguments. Lemm 4.46. Let, b, R stisy < b nd (x n ) n=1 [, b] with n x n =. Then [, b]. Corollry 4.48. Let, b R stisy < b nd let (x n ) n=1 [, b]. Then there exists subsequene (x nk ) k=1 (x n ) n=1 [, b] nd [, b] suh tht k x nk =. Theorem 4.50 (Weierstrss extreml theorem, WET). Let, b R stisy < b nd let C([, b]), then () is bounded; (b) p, q [, b] suh tht x [, b], (q) (x) (p) ( ttins its bounds). Deinition 4.52. Let I R be n intervl. Then : I R is uniormly ontinuous on I i ε > 0, δ > 0 suh tht x, y I, x y < δ = (x) (y) < ε. Lemm 4.53. Let, b R stisy < b nd let : (, b) R be uniormly ontinuous on (, b). Then is ontinuous on (, b). Exmple 4.54. Let : R R be deined by (x) = x 2. Show tht is ontinuous t, R but is not uniormly ontinuous on R. Theorem 4.55. Let, b R stisy < b nd let : [, b] R. Then the ollowing re equivlent, () C([, b]); (b) is uniormly ontinuous on [, b].

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 5 5.1. Dierentition t Point. 5. Dierentition o Funtions Deinition 5.1. Let : R R nd let R. Then is dierentible t i L R suh tht ( + h) () = L (5.1) h 0 h L is denoted s () nd is lled the derivtive o t. Note tht () is slr nd is not untion evluted t. Proposition 5.3. Eqution (5.2) is equivlent to sying tht ε > 0, δ > 0 suh tht x < δ = (x) [() + (x ) ()] ε x Lemm 5.4. Let : R R nd R. Then i is dierentible t, then is ontinuous t. Exmple 5.5 (Converse o lemm 5.4). Let : R R be deined by (x) := x. Then is ontinuous t 0; now onsider the ollowing nd similrly (0 + h) (0) h 0 h = = h 0+ h h 0+ h h 0+ h = 1 (0 + h) (0) = 1 h 0 h So L R suh tht (5.1) is stisied or = 0, hene is not dierentible t zero. Lemm 5.6. Let, L R, : R R nd F : R R be deined by ( + h) (), h R\0} F (h) = h L, h = 0 Then is dierentible t with derivtive L = () i, nd only i, F is ontinuous t zero. Theorem 5.7 (Algebr o dierentibility). Let R nd, g : R R be dierentible t. Then () + g is dierentible t nd ( + g) () = () + g (); (b) K R, K is dierentible t nd (K) () = K (); () g is dierentible t nd (g) () = ()g() + ()g (); (d) i g() 0, then 1/g is dierentible t nd (1/g) () = g ()/g() 2. Corollry 5.8. Under the ssumptions o theorem 5.7, (/g) () = ()g() ()g () g() 2 i g() 0. This ollows diretly rom theorem 5.7() nd (d). Lemm 5.9. Let n N nd let R. Then n : R R deined by n (x) := x n is dierentible t with () = n n 1. Theorem 5.10. Let n N nd let R, ( k ) n k=0 R. Let p : R R be deined by p(x) := n k=0 kx k. Then p is dierentible t nd p () = n k=1 k k k 1. Theorem 5.11 (Chin rule). Let R nd let, g : R R. Let g be dierentible t nd be dierentible t g(), then g is dierentible t with ( g) () = (g())g ().

6 CHARLIE COLLIER UNIVERSITY OF BATH 5.2. Dierentition over n Intervl. Deinition 5.12. Let I R be n open intervl nd let : I R. Then is dierentible over I mens tht is dierentible t, I. Deinition 5.14. Let I R be n open intervl nd let : I R be dierentible on I, i.e. I, L() R suh tht ( + h) () = L() h 0 h Then the derivtive untion : I R is deined by () = L() with I. I is dierentible t I, then we write () to denote the derivtive o t. Exmple 5.15. Let : R R be deined by (x) = x 2 or x 0 nd x 2 or x < 0. Then or > 0, () = 2 nd or < 0, () = 2. For = 0, (0 + h) (0) h, h > 0 (0 + h) (0) = = = 0 h h, h < 0 h 0 h Hene R, is dierentible. Hene we my deine the derivtive untion : R R whih is (x) = 2 x x R. For > 0, ( ) () = 2 nd or < 0, ( ) () = 2. We hve tht is not ontinuous t zero, hene is not dierentible t zero. So (0) does not exist. Deinition 5.16. Let I R be n intervl nd let p, q I nd let : I R. Then p is mx point (mximum) i (p) (x) x I nd similrly q is min point (minimum) i (q) (x) x I. An extremum is either mximum or minimum point; we let E := x I x is n extremum}. Exmple 5.18. Let I = [ 1, 1] nd : I R be deined by (x) = 2 x 2. Then (0) = 2 2 x 2 = (x) x I, hene 0 is mximum point. Deinition 5.19. Let I R be n intervl, let I nd let : I R. Then is ritil point mens tht is dierentible t nd () = 0. We deine C := x I x is ritil point}. Theorem 5.20. Let I R be n intervl, p I nd : I R be dierentible t p, nd let p be n extremum. Then p is ritil point, i.e. p C. Corollry 5.21. Let, b R stisy < b, let C([, b]) nd let be dierentible on (, b). Then p, q [, b] suh tht p is mximum point nd q is minimum point or in (, b) (p, q re not unique) nd ll extrem E C, b}. Exmple 5.22. Find ll the extremum points o : [ 1, 3/2] R deined by (x) = x 2 (3 2x). Theorem 5.23 (Rolle). Let, b R stisy < b, let : [, b] R stisy C([, b]), dierentible on (, b) nd () = (b). Then (, b) suh tht () = 0. Theorem 5.24 (Men vlue theorem). Let, b R stisy < b nd let C([, b]) be dierentible on (, b). Then (, b) suh tht (b) () () = b Exmple 5.25. () Let : [ 1, 1] R be deined by (x) = 2 x 2. Then ( 1) = (1) = 1, so by Rolle s theorem ( 1, 1) suh tht () = 0. In t, = 0. (b) Let : [ 1, 0] R be deined by (x) = x 2 (3 2x). Then ( 1) = 5 nd (0) = 0, so by the men vlue theorem ( 1, 0) suh tht () = ( 1) (0) 1 0 = 5 0 1 = 5

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 7 5.3. Consequenes o the Men Vlue Theorem. Unless otherwise stte, ssume tht, b R stisy < b, nd let C([, b]) be dierentible on (, b); these re the hypotheses o the men vlue theorem. Corollry 5.26. I (x) = 0 x (, b), then is onstnt on [, b]. Corollry 5.28. I (x) > 0 x (, b), then is stritly inresing on [, b]. Exmple 5.29. Let : [ 1, 1] R be deined by (x) = x(2 x). Then is polynomil implies tht C([ 1, 1]) nd is dierentible on ( 1, 1). Also (x) = 2 2x = 2(1 x) > 0 or x < 1. Thus by orollry 5.28, is stritly inresing. Deinition 5.30. Let I R be n intervl, let : I R nd p I. Then p is lol mx point i η > 0 suh tht [p η, p + η] I nd x [p η, p + η] = (x) (p). Theorem 5.31 (Neessry onditions or lol mximum). Let p be lol mx point nd be twie dierentible t p. Then (p) = 0 (p). Exmple 5.32. Let : [1/2, 3/2] R be deined by (x) = x 2 (3 2x). Now (1) (x) = 1 x 2 (3 2x) = (x 1) 2 (2x 1) 0 or x [1/2, 3/2] = [p η, p + η] with p = 1, η = 1/2. Hene (x) (1), i.e. 1 is lol mx point. So by theorem 5.31, (1) = 0 (1). This is true sine (x) = 6x(1 x), so (1) = 6(1 1) = 0 nd (x) = 6(1 2x), so (1) = 6(1 2) = 6 < 0. Theorem 5.33 (Suiient onditions or lol mximum). Let p (, b) nd let be twie dierentible t p. Let (p) < 0 = (p), then p is lol mx point. Theorem 5.34 (Tylor s theorem). Let n N, let, b R stisy < b, let I R be n open intervl ontining [, b] nd let : I R be n + 1 times dierentible on I. Then (, b) suh tht (b) = () + (b ) () + where R n is the error term, deined by (b )2 () + + 2! R n = (n+1) ()(b ) n (b ) n! (b )n (n) () + R n n! Exmple 5.36. Apply Tylor s theorem to rigorously pproximte e = exp(1) to n ury o 10 2. 5.4. De l Hospitl s Rule. Suppose we wnt to ind x (x)/g(x) where () = 0 = g() or (x), g(x) 0 s x. De l Hospitl s rule gives us method or inding L without ε δ rguments. Theorem 5.37 (Cuhy MVT). Let, b R stisy < b, let, g C([, b]) be dierentible on (, b). Then (, b) suh tht [(b) ()]g () = [g(b) g()] () Exmple 5.39. Let (x) := x 3 + x 2 nd g(x) := x 2 3x + 2. We im to ind (x) x 1 g(x) = x 3 + x 2 x 1 x 2 3x + 2 Sine (1) = 0 = g(1), lgebr o its n not be used diretly. By Tylor s theorem,, d (1, x) suh tht (x) = (x 1) (1) + ()(x )(x 1) nd g() = (x 1)g (1) + g (d)(x d)( 1). Hene Now s x 1, x 0 nd x d 0, so (x) g(x) = (1) + ()(x ) g (1) + g (d)(x d) (x) x 1 g(x) = (1) g (1)

8 CHARLIE COLLIER UNIVERSITY OF BATH Now (x) = 3x 2 + 1, so (1) = 4 nd g (x) = 2x 3, so g (1) = 1. This rgument suggests tht (x) x 1 g(x) = 4 Theorem 5.40 (De l Hospitls rule). Let, L R, let η > 0 nd let D = ( η, +η)\} nd let, g : D R be dierentible on D. Suppose tht () x (x) = 0 = x g(x); (b) x D, g (x) 0; () x (x)/g (x) = L. Then we n dedue tht x (x)/g(x) = L. Lemm 5.41. Let the hypotheses o theorem 5.40 hold nd deine () = g() = 0. Then, g re ontinuous on ( η, + η). Exmple 5.42. Let = 1, η = 1/2 to give D = (1/2, 3/2) \ 1} with, g : D R deined by (x) = x 3 + x 2 nd g(x) = x 2 3x + 2 s in exmple 5.39. Both, g re polynomils nd so re dierentible on D, lso x 1 (x) = 0 = x 1 g(x). Also g (x) = 2x 3 0 sine 3/2 D nd x 1 (x)/g (x) = 4/( 1) = 4 by lgebr o its. Hene by de L Hospitls rule, x 1 (x)/g(x) = 4, s expeted.

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 9 6. Riemnn Integrtion 6.1. Upper nd Lower Riemnn Sums. Our im is to deine wht the re under grph ormlly mens. We pproh this by sub-dividing [, b] into severl intervls nd pproximte the re with retngles. We either obtin n overestimte (upper sum) or underestimte (lower sum) depending on the height o the retngle. For iner subdivisions, the dierene between these pproximtions beomes smller. b b Deinition 6.1. A subdivision (or prtition/dissetion) P on [, b] is inite set u 0, u 1,..., u n } [, b] stisying = u 0 < u 1 < < u n = b. This divides [, b] into n subintervls. For eh i = 1,..., n we write I i = [u i 1, u i ] or the i th intervl o P nd I i = u i u i 1 or the length o the intervl I i. Deinition 6.2. Let : [, b] R be bounded nd P = u 0,..., u n } be subdivision o [, b]. m i = in(x) u i 1 x u i } nd M i = sup(x) u i 1 x u i }. Then () the lower Riemnn sum o or P is n L(, P ) = m i I i (b) the upper Riemnn sum o or P is n U(, P ) = M i I i Lemm 6.4. Let : [, b] R be bounded nd P be subdivision o [, b]. Then L(, P ) U(, P ). i=1 i=1 Write Lemm 6.5. Let : [, b] R be bounded nd P be subdivision o [, b] nd P = P v} where = u 0 < < u k 1 < v < u k < < u n = b or some k. Then L(, P ) L(, P ) nd U(, P ) U(, P ). Deinition 6.6. Let P = u 0,..., u n } nd R = v 0,..., v m } be subdivisions o [, b]. I or every i = 0,..., n, j 0, 1,... } suh tht u i = v j, then R is lled reinement o P. Lemm 6.7. Let : [, b] R be bounded nd let P be subdivision o [, b] nd R be reinement o P. Then L(, P ) L(, R) nd U(, R) U(, P ). Theorem 6.8. Let : [, b] R be bounded nd let P, Q be subdivisions o [, b]. Then L(, P ) U(, Q). Deinition 6.9. Let : [, b] R be bounded. Then is Riemnn integrble i!σ R suh tht L(, P ) σ U(, P ) or ll subdivisions P o [, b]. Then σ is lled the Riemnn integrl o on [, b], denoted σ = The uniqueness o σ is ruil; i more thn one σ stisies the inequlity, then is not Riemnn integrble. Deinition 6.10. Let : [, b] R be bounded. The upper integrl o is, := inu(, P ) P is subdivision o [, b]}

10 CHARLIE COLLIER UNIVERSITY OF BATH nd the lower integrl o is := supl(, P ) P is subdivision o [, b]} Note tht these re both well-deined sine the upper sums re bounded below by n lower sum nd vie vers. The untion is Riemnn integrble i This deinition is equivlent to deinition 6.9. = Exmple 6.11. Let R nd onsider : [, b] R, (x) :=. Given ny subdivision P = u 0,..., u m } o [, b], or eh i = 1,..., n we hve tht m i = M i =. So L(, P ) = U(, P ) = n (u i u i 1 ) = (u n u 0 ) = (b ) i=1 Thereore L(, P ) (b ) U(, P ) or ny subdivision, so by deinition 6.10 is Riemnn integrble with Exmple 6.12. Let : [, b] R be deined by (x) = = (b ) 0, x R \ Q 1, x Q I P = u 0,..., u n } is ny subdivision o [, b] then m i = 0 sine there is n irrtionl number in the i th intervl I i = [u i, u i 1 ] by the density o the irrtionls in the rels. Similrly M i = 1 sine the rtionls re dense in the rels. Thereore L(, P ) = n 0 I i = 0, U(, P ) = i=1 n (u i u i 1 ) = u n u 0 = b But!σ R suh tht 0 σ b or b, hene is not Riemnn integrble. Alterntively, nd so by deinition is not Riemnn integrble. i=1 = b 0 = Theorem 6.14 (Cuhy riterion or integrbility). Let : [, b] R be bounded. Then is integrble on [, b] i, nd only i, or every ε > 0 there is subdivision P o [, b] suh tht U(, P ) L(, P ) < ε. Exmple 6.15. Let b > 0 nd : [0, b] R be deined by (x) = x. Show tht is integrble on [0, b] nd ind 0. 6.2. Integrbility o Continuous Funtions. Theorem 6.16. I is ontinuous on [, b], then is integrble on [, b].

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES 11 6.3. Properties o the Integrl. Lemm 6.17. Let, g : [, b] R be bounded nd deine + g : [, b] R pointwise. Then or ny subdivision P o [, b], () L( + g, P ) L(, P ) + L(, P ); (b) U( + g, P ) U(, P ) + U(g, P ). Lemm 6.18. Let : [, b] R be bounded nd let R. Then or n subdivision P o [, b], () L(, P ) = L(, P ) nd U(, P ) = U(, P ) or > 0; nd (b) L(, P ) = U(, P ) nd U(, P ) = L(, P ) or < 0. Theorem 6.19 (Additivity). Let < < b, then the untion is integrble on [, b] i, nd only i, is integrble on [, ] nd [, b]. Furthermore, i this is the se then = Theorem 6.20 (Linerity o ddition). I nd g re integrble on [, b], then + g is integrble on [, b] nd ( + g) = Theorem 6.21 (Linerity, homogeneity). I is integrble on [, b] nd R, then is integrble on [, b] nd + = + Deinition 6.22. Let, b R with < b nd let : [, b] R. Then = 0, = g b 6.4. The Fundmentl Theorem o Clulus. Theorem 6.23 (First undmentl theorem o lulus). Let : [, b] R be integrble nd suppose tht = g or some untion g. Then = g(b) g() Deinition 6.24. Let I be n open intervl ontining [, b] nd let : [, b] R nd g : I R stisy = g on [, b]. Then g is lled primitive or ntiderivtive o. Theorem 6.25 (Seond undmentl theorem o lulus). Suppose tht : [, b] R is integrble nd deine F : [, b] R by F (x) = I is ontinuous t (, b), then F is dierentible t nd F () = (). Lemm 6.26. Suppose tht : [, b] R is integrble nd m, M R stisy m (x) M or ll x [, b]. Then m(b ) x M(b ) Lemm 6.27 (Pinhing theorem). Let I be n intervl, let I nd let, g, h : I \ } R stisy (x) g(x) h(x) or ll x I \ }. I x (x) = x h(x) = L, then x g(x) = L.

12 CHARLIE COLLIER UNIVERSITY OF BATH 6.5. Integrtion Tehniques. Theorem 6.28 (Integrtion by prts). Let, g : [, b] R be integrble. Suppose tht F, G : [, b] R re ontinuous nd re ntiderivtives o nd g respetively on (, b). Then G + F g = F (b)g(b) F ()G() Theorem 6.29 (Integrtion by substitution). Let I be n open intervl nd let : I R be ontinuous. Suppose tht u : [, b] I is lso ontinuous on [, b] nd dierentible on (, b) suh tht u hs ontinuous extension on [, b]. Then 6.6. Improper Integrls. u(b) u() = ( u)u Deinition 6.30. Suppose tht : [, b] R or : (, b] R is unbounded, but is integrble on [, b] or ny (, b) nd tht exists. Then we deine + := + Now tht the vlue o () is irrelevnt in this deinition. Similrly, suppose tht : [, b] R or : [, b) R is unbounded but is integrble on [, ] or ny (, b) nd tht exists. Then we deine These re lled improper integrls. b := b Deinition 6.31. Suppose tht : [, ) R is integrble on [, ] or ny > nd tht exists. Then we deine := Similrly, suppose tht : (, b] R is integrble on [, b] or ny < b nd tht exists. Then we deine := Now suppose tht : R R is integrble on [, b] or ny, b R, < b. Then we deine Exmple 6.32. 1 := 0 + 0 1 ( 1 dx = dx = 1 1 ) = 1 x2 1 x2