Θέματα Ιούνιος 2012- Π2 ΖΑΦΕΙΡΙΔΟΥ

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I (α) x +3 x 2 - x +5 dx (β ) 1 e 1 x dx x 3 (γ ) 1 e 0 + dx x ln 2 x (δ ) 2 0 x 2 4 - x 2 dx x = 2 sin t (α) n=1 ( n+1 n ) n 2 3 n , (β ) n=1 2 · 5 · ... · (3n - 1) 1 · 5 · ... · (4n - 3) , (γ ) n=1 (-1) n 2n 2 3n 2 +5 , (δ ) 2 x sin x x 4 + x 2 +1 dx, (ε) 3 0 + cos x x dx α f :[a, b] R f Riemann [a, b] β f :[a, b] R Riemann [a, b] F (x)= x a f (t)dt, x [a, b] γ n=1 a n k 1 n=k+1 a n n=1 a n = a 1 + ... + a k + n=k+1 a n n=1 (-1) n+1 3 n-1 nx n . (α) f n (x)= x n n =1, 2, ... Δ=[ 1 5 , 3 5 ] (β ) n=1 x n (1 - x) [0, 1]

Transcript of Θέματα Ιούνιος 2012- Π2 ΖΑΦΕΙΡΙΔΟΥ

Page 1: Θέματα Ιούνιος 2012- Π2 ΖΑΦΕΙΡΙΔΟΥ

PRAGMATIKH ANALUSH II

1. Na brejoÔn ta oloklhr¸mata: (32)

(α)

∫x + 3

x2 − x + 5dx, (β)

∫ ∞

1

e1x dx

x3, (γ)

∫ 1e

0+

dx

x ln2 x,

(δ)

∫ 2

0

x2√

4− x2dx ( Upìdeixh: qrhsimopoi ste thn antikat�stash x = 2 sin t.)

2. Na melethjoÔn wc proc th sÔgklish oi seirèc: (30)

(α)∞∑

n=1

(n+1

n

)n2

3n, (β)

∞∑n=1

2 · 5 · . . . · (3n− 1)

1 · 5 · . . . · (4n− 3), (γ)

∞∑n=1

(−1)n 2n2

3n2 + 5,

kai ta oloklhr¸mata:

(δ)

∫ ∞

2

x sin x

x4 + x2 + 1dx, (ε)

∫ 3

0+

cos x√x

dx

Na diatupwjoÔn ta krit ria sÔgklishc (   apìklishc) pou ja qrhsimopoihjoÔn.

3. Na apodeiqjoÔn oi prot�seic: (30)

(α) An h sun�rthsh f : [a, b] → R eÐnai suneq c, tìte h f eÐnai oloklhr¸simh kat�Riemann sto [a, b].

(β) An h sun�rthsh f : [a, b] → R eÐnai oloklhr¸simh kat� Riemann sto di�sthma

[a, b], tìte h sun�rthsh F (x) =

∫ x

a

f(t)dt, x ∈ [a, b], eÐnai suneq c.

(γ) An h seir�∞∑

n=1

an sugklÐnei, tìte gia k�je fusikì arijmì k ≥ 1 h seir�∞∑

n=k+1

an

sugklÐnei kai∞∑

n=1

an = a1 + ... + ak +∞∑

n=k+1

an .

4. Na brejeÐ h aktÐna sÔgklishc kai to di�sthma sÔgklishc thc dunamoseir�c

∞∑n=1

(−1)n+13n−1n xn.

Sth sunèqeia na brejeÐ èna di�sthma sto opoÐo h seir� sugklÐnei omal� kai èna di�sthmasto opoÐo h seir� sugklÐnei kat� shmeÐo all� ìqi omal� (dikaiolog ste thn ap�nthshsac). (15)

5. Na apodeiqjeÐ ìti (15)

(α) H akoloujÐa sunart sewn fn(x) = xn, n = 1, 2, ..., sugklÐnei omal� sto di�sthma∆ = [1

5, 3

5].

(β) H seir�∞∑

n=1

xn(1− x) sugklÐnei kat� shmeÐo all� ìqi omal� sto di�sthma [0, 1].