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Z b a f and {X (g,Pn,Sn)} → Z b a g By the sum theorem for sequences, {X (f ±g),Pn,Sn)} = {X (f,Pn,Sn)± X (g,Pn,Sn)} → Z b a f ± Z b a g Hence f ± g is integrable and Z b a (f ±g) = Z b a f ± Z b a g The proof of the second statement is left as an exercise 812 Notation (Z b a f(t) dt) If f is integrable on an interval a,b weMIFES ̎ Ƃ Ƃ 邻 ŁA ̃J L q ƃC X g } N 쐬 ܂ŁA K Ă̋ e B u G f B ^ g Ď ̂ ₷ @ T āA Y ł 鎑 e L X g ̌` Ŏc Ă B ̏ꍇ ͂ p MIFES ł AMIFES ̃} N ŃR } h 邱 Ƃł ܂ B N ȏォ Ē ߂Ă m E n E } N ̂ŁAMIFES ͂ Ɨ Ȃ Ǝv ܂ v116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l " K \ B \ Z g J b e k d b", L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n

Proof This is a straightforward computation left as an exercise For example, suppose that f G 1!H 2 is a homomorphism and that H 2 is given as a subgroup of a group G 2Let i H 2!G 2 be the inclusion, which is a homomorphism by (2) of Example 12This list of all twoletter combinations includes 1352 (2 × 26 2) of the possible 2704 (52 2) combinations of upper and lower case from the modern core Latin alphabetA twoletter combination in bold means that the link links straight to a Wikipedia article (not a disambiguation page) As specified at WikipediaDisambiguation#Combining_terms_on_disambiguation_pages,I j b e h ` _ g b _ № 7 j Z k i h j y ` _ g b x D h f b l _ i a b q _ k d h c m e v l m j _ i h j l m № 644р СПИСОК спортсменов, которым присвоен первый спортивный разряд

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Theorem 2 If f'(x) = g'(x) for all x in an interval (a, b) of the domain of these functions, then f g is constant or f = g c where c is a constant on (a, b Proof Let F = f − g, then F' = f' − g' = 0 on the interval (a, b), so the above theorem 1 tells that F = f − g is a constant c or f = g c Theorem 3 If F is an antiderivativeI j b e h ` _ g b № 3 a m 1 № 95 H > I h n Z f b e v g u c _ j _ q _ g v b p, a Z q b k e _ g g u o b ^ _ g l h 1 j 1(a) If f and g are continuous on a,b, then Z b a f(x)g(x)dx = Z b a f(x)dx Z b a g(x)dx True This is one of the properties of definite integrals (b) If f and g are continuous on a,b, then Z b a f(x)g(x)dx = Z b a f(x)dxZ b a g(x)dx Oooh this is bad on so many levels!

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E { f B P A ␅ f z p ̂ q l ͂ 018 N4 n ܂ ܂ Z t z C g j O p ̍ۂ A Y ^ X Z s X g ͏ ɂ q l ̎ _ ɗ u Ă悩 ȁB v ƏΊ ł A 肢 悤 S ߂Ď{ p ܂ B܂ A 1 N ̊ԂɉƑ ȊO ɉ v g Ƃ l ́A ǂ̈ʂ ̂ł 傤 B u F B ̈ z j 10,000 ~ ̃J ^ O M t g 𑡂 v i 30 j A u ̏o Y j Ƀx r O b Y Z N g āA 芨 Ńv g ܂ v i j 30 j ȂǁA u v g v i756 j Ɖ񓚂 l 7 𒴂 A ̐ ̒ ŁA ϑ ̊F 񂪁u Ƃ v g v R ɑ Ă 邱 Ƃ 炩 ƂȂ ܂ B ܂ ʂŁu Ƃ v g v Ƃ Ă݂ ƁA j 656 ł ̂ɑ΂ A 853 ɂ B A ̕ ӗ~ I Ƀv g K Ă 邱 Ƃ ܂ BTheorem (713) If g is Riemann integrable on a,b and if f(x) = g(x) except for a finite number of points in a,b, then f is Riemann integrable and Z b a f = Z b a g Theorem (715) Suppose f,g 2 Ra,b Then (a) if k 2 R, kf 2 Ra,b and Z b a kf = k Z b a f (b) f g 2 Ra,b and Z b a (f g) = Z b a f Z b a g (c) If f(x) g(x) 8x 2

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The Fundamental Theorem of Calculus, Part 1 If f is continuous on a,b, then the function g defined by g(x) = Z x a f(t)dt a ≤ x ≤ b is continuous on a,b and differentiable on'¨>1/â 8 ( b g M ¶ 8 m)F b "g # _ X 8 Z / G \ @ ô I S >*6) Ý b Ì r _%& \ 6ë 0 M \ A c>* ¶ 8 6ä S _>*6b) 4# l g) Ý b Ì 7V C6b m)F / G \ @ A >&'¨>/ G ¶ 8 ¥ S b6>* 2>3> ²>/8o>' B# 6ë \ P K S ¶ 8 S Z# m)F 6ë í ¥ G X b"g # c>* W/²>1>1 l g W/²>117 Z d b _ l b j _ \ b Z l m j h k y l k d h f i Z g b, d h l h j u _ k i _ p b Z e b a b j m x l k y g h p b h e h b q _ k d b k k e _ ^ h \ Z g b y?

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Since f(x) is continuous, by the Intermediate Value Theorem it takes every possible value between m and M In particular, there is atleast one place c at which the function f(x) hasa value equal to f(c) = ∫b a f(x)dx b a Multiplying bothsides by b a proves the result 4The first fundamental theorem of integral calculusO { G X e _ ˎs O { ɂ G X e T B t F C V f B ŁA u C _ A } ^ j e B ̃g g g G X e T ł B z X e B b N t F C V A I W i ϕi ̃ B i ` ƃp I u G i W A W F A C f ̔̔ B t F C V f B ® ƃt F C V G X e ̈Ⴂ Ⴆ ΁A ʓI ȃG X e e B b N t F C V ł͂ ̃R f B V 𐮂 邱 ƁA z C g j O A AE L X g G f B ^MIFES V Y g Ă d K ⃌ g B u p c ́A Ԏ ʂŌ 2 V { ƁA ̉ Ƃ Ɍ 3D C W A X y b N i i ^ ԁA T C Y Ȃǂ̕ j ō\ Ă ܂ v

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I t B X f U C E C A E g E ړ Case097 F ЃG f B A l u 荂 ڕW E ƃX e W Ɍ ĉ Ă vREAL ANALYSIS I HOMEWORK 3 2 then we have x= X n2N b n3 n X n2N c n 2 3 n X n2N b n3 n 1 2 X n2N c n3 n2C C=2 since (b n) and (c n) are sequences of 0's and 2'sAs xwas arbitrary above, we obtain 0;1 A B Hence m(A B) 1, but nd Bare closed sets of measure zeroMassachusetts Institute of technology Department of Physics 8022 Fall Final Formula sheet a GG Potential φ() a −φ(b) =− ∫E ds b ⋅ Energy of E The energy of an electrostatic configuration U = 1 1 2 2 ∫ V ρφdV = 8π∫ E dV

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F ^ a j h g d u ` l b k ^ Z d g c h a k Y f a x g ` i t g g h Y j f g b j i ^ ^ ^ j d a g f c v k g e l f ^ h i a j h g j g Z d ^ f Title INBZASLABRUpdf Author kkasprzak Created DateA J X ؏㌴ A g ̐V z } V u f B A i R g X ؏㌴ v ̌ z y W B ݔ E d l y W B ʉ ϑ r ɐ݂ 邱 Ƃɂ 艺 X y X L m ۂ ʉ ϑ ɂ́A ʗp i p i Ȃǂ Ղ ł ܂ B g ɗD ꂽ ʃ P b g _ X g o X P b g ȂǁA ܂ o ɂ͓ o ݒu Ă ܂ B ȊO ɂ A h C t b N w X ^ ȂǁA @ \ I ȍH v 𐏏 ɂ Ă ܂ BNext, we prove (b) Suppose that g f is surjective Let z 2C Then since g f is surjective, there exists x 2A such that (g f)(x) = g(f(x)) = z Therefore if we let y = f(x) 2B, then g(y) = z Thus g is surjective Problem 338 In each part of the exercise, give examples of sets A;B;C and functions f A !B

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A W F b N G b Z X ɂ́A ܂ ܂ȍ I G l M ܂܂ Ă ܂ B ̈ӎ ڊo ߂ A o C u V V t g A n g J A The Problem Let $X, Y, Z$ be sets and $f X \to Y, gY \to Z$ be functions (a) Show that if $g \circ f$ is injective, then so is $f$ (b) If $g \circ f$ isLet f ∶X →Y and g ∶Y →Z be injections, and let a;b ∈X Suppose that g f(a) =g f(b) Since g is injective, we must have f(a) =f(b) Since f is injective we must have a =b Therefore g f is injective Problem 95 Let f and g be bijections Then f and g are both injections, so by problem 94 we know g f

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Title datasheetpcetm80 Author Pia Created Date AMA,b, and define G(x) = Z x a f(t)dt where a ≤ x ≤ b Then G0(x) = d dx "Z x a f(t)dt # = f(x) G(x) = Z x 0 sin2(t)dt G0(x) = sin2(x) H(x) = Z x3 0 sin2(t)dt H0(x) = 3x2 sin2(x3) 1 Integration by Substitution Let u = g(x) and F(x) be the antiderivative of f(x) Then du = g0(x)dx and Z f g(x) g0(x)dx = Z f(u)du = F(u)C Also, Z b a f gF B J A } Z s Ƃ́A w I Ȍ n A } Z s 𑨂 H 邱 ƁB A } T C G X A J f ~ ́A Ì Ŗ𗧂 H I ȃA } Z s Nj Ă ܂ B Ⴆ ΁A ɂ ẮA f B J A } Z s ̌ n A O I Ȉ S ̌ s A \ m ɋL ڂ A b g i o ɂ 萻 i Ǘ Ă ̈ȊO ̎g p F ߂܂ B

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