Θέματα Ιούνιος 2012- Π2 ΖΑΦΕΙΡΙΔΟΥ
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Transcript of 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].