University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E13 Integration : correction of the exercises. 1. (a) Assume that f : [a, b] ? R is a continuous function such that f(x) ≥ 0 for all x ? (a, b), and ∫ b a f(t)dt = 0. Show that f(x) = 0 for all x ? [a, b] ; can you use the fundamental theorem of calculus to prove this result ? (b) Use this to show that if f is continuous on [a, b] and ∫ b a f(t)dt = 0 then there must exist t ? (a, b) such that f(t) = 0. Correction. (a) First,notice that, since f is continuous, proving that f(t) = 0 for all t ? [a, b] is the same as proving that f(t) = 0 for all t ? (a, b). Now, let us prove the contraposite of the result we are interested in ; in other words, let us prove that if f(x) > 0 for some x ? (a, b), f(x) ≥ 0 for all x ? (a, b) and f is continuous on [a, b] then ∫ b a f(t)dt > 0.

• induction using


• x? ?


• additivity theorem


• riemann sum


• then pick


• gives limc?a ∫


• since ti


• there exists

Sujets

• Riemann sum
• Existential quantification

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 25
• Langue: English

Extrait

Rb
1. f: [a,b]→R f(x)≥ 0 x∈ (a,b) f(t)dt = 0
a
f(x) = 0 x∈ [a,b]
Rb
f [a,b] f(t)dt = 0 t∈ (a,b)
a
f(t) = 0
f f(t) = 0 t∈ [a,b]
f(t) = 0 t∈ (a,b)
f(x)> 0 x∈ (a,b) f(x)≥ 0 x∈ (a,b) f
Rb
[a,b] f(t)dt > 0 f x δ > 0a
f(x)
f(y)≥ y∈ [a,b] |y−x|≤δ δ [x−δ,x+δ]⊂ [a,b]
2
Z x+δ f(x)
f(t)dt≥ 2δ =δf(x)> 0 .
2x−δ
R Rx−δ b
f(y)≥ 0 y∈ [a,b] f(t)dt≥ 0 f(t)dt≥ 0
a x+δRb
f(t)dt> 0
a
Rx 0F(x) = f(t)dt F F (x) =f(x)≥ 0 F F(a) = 0
a Rb
f(t)dt = 0 F(b) = 0 F
a
F [a,b] 0 [a,b] f(x) = 0 x∈ [a,b]
f 0 (a,b) f > 0 < 0 [a,b]
Rb
f(t)dt = 0 t∈ (a,b) f(t) = 0a
2
Z b
f: [a,b]→R f(t)dt = (b−a)sup{|f(x)|: x∈ [a,b]}
aZ 1
1
f: [0,1]→R f(t)dt = a∈ (0,1)
20
f(a) =a
R R1 1
f,g [0,1] f(t)dt = g(t)dt c∈ [0,1]
0 0
f(x) =g(c)
g [a,b] g(x) = sup{|f(x)|: x ∈ [a,b]}− f(x)
Rb
g(x) ≥ 0 x ∈ [a,b] f(t)dt = (b− a)sup{|f(x)|: x ∈ [a,b]}
aRb
g(t)dt = 0 g(x) = 0 x ∈ [a,b]
a
f [a,b]
R1
t ∈ (0,1) f(t) > t (f(t)−t)dt > 00R R1 11 1f(t)dt > f(t) < t t∈ (0,1) f(t)dt >
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0 0f(t ) = t x7→ f(x)−x (0,1)
(0,1)
a∈ (0,1) f(a) =a
1 f−g
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n n n n n2 kX X X X X1 n k kπ (k−1)π kπ (−1)
; ; ; sin( −sin( )ln(1+sin( ) ; ; .
2 2 3n+k n +k n 2n 2n 2n k
k=1 k=1 k=1 k=1 k=1
n nX X1 1 1
= kn+k n 1+
nk=1 k=1
1
x7→ [0,1] 1/n
1+xZn 1X 1 dt
lim = = ln(2)
n+k 1+t0k=1
Zn n n 1X X Xn 1 1 n dt
= lim( ) =2 2 2 2 2 2kn +k n n +k 1+t01+k=1 k=1 k=1n
π
(1)− (0) =
4 Zn n n 12 2X X X k 1 k k 12 2
= lim = t dt =
3 3n n n n 30k=1 k=1 k=1
t7→ ln(1+t) [0,1]

(k−1)π kπ kπ 1{ sin ,sin ,sin }2n 2n 2n 2nR1
n→ +∞ ln(1+t)dt
0
Z Z1 1 1
ln(1+t)dt = (t+1)ln(t+1) − 1dt = 2ln(2)−1 .
t=0
0 t=0
n = 2p
p pn n nkX X X X X(−1) 1 1 1 1
u = =− +2 =− +2p
k k 2k k k
k=1 k=1 k=1 k=1 k=1
n 2p pkX X X(−1) 1 1
n = 2p =− =− .
k k p+k
k=1 k=p+1 k=1
1 k−1 k kt7→ [0,1] {[ , ], }1+x p p pR11 dtu − = ln(2) u −u2p 2p+1 2pp 0 1+t
0 lim(u ) =−ln(2) (u )2p+1 n
lim(u ) =−ln(2)n
Zn 1X 1 k k−1
4. f,g: [0,1]→R lim f g = f(t)g(t)dt
n→∞ n n n 0k=1
k k
f g
n n
n n nX X X 1 k k−1 1 k k 1 k k−1 k
f g = f g + f g −g .
n n n n n n n n n n
k=1 k=1 k=1
R1
f(t)g(t)dt 0
0
g [0,1] ε > 0 δ |x−y|≤ δ ⇒ε ε
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k−1 k−1|f(x)−f(y)|≤ε x,y∈ [a,b] n |g −g |≤ε
n n
n nX X 1 k k−1 k 1 k
f g −g ≤ε |f | .
n n n n n n
k=1 k=1
P Rbn1 kf |f| |f | |f(t)|dtn k=1 n a RPn b1 kn |f |≤ |f(t)|dt+1 εk=1n n a RPn b1 k k−1 kK ∈N | f g −g )|≤ ε( |f(t)|dt + 1) n≥ Kk=1n n n n aP n1 k k−1 kf g −g 0 n→ +∞n k=1 n n n
Z 1
k5. f: [0,1]→R f(u)u du = 0 k∈{0,...,n} f
0
n+1 (0,1)
n = 0
Z 1
k1 n f f(u)u du = 0
0Z x
k∈{0,...,n+1} F(x) = f(t)dt f k = 0 F(0) = F(1) = 0
0
k = 1,...,n
Z Z1 1 1k k−1 k−1u du = ku −k u F(u)du
0
0 0
Z 1
k−1u F(u)du = 0 k = 1,...,n
0
F n (0,1) F(0) =F(1) = 0 F n+2
0[0,1] F = f F f n + 1
(0,1)
c ,...,c1 n
c < c < ... < c < c M = max{|f(c )|: i = 1,...,n}1 2 n−1 n i
˙P ={[x ,x ],t} [a,b]i−1 i i i=1,...m
m m mX X X
˙