arctgx = -
■/3
\D\ =
<=» x = s/3 — arc tg x) dx
1.0 pt.
0.5 pt.
arc tg x dx = x arc tg x — ^ ln (a:2 4-1) + C
• |£>j = ^ - xarctgx)% + ^ln (x2 + l)j = ln 2.
0.5 pt.
1.5 pt.
1.5 pt.
x2 dx
x4 -t- x2 + 1
x2 dx
x2 dx
x4 + x2 + 1 / x4 + x2 + 1
0.5 pt.
0£
dx
- < —-- — |
——- dla x |
1 X4 + X2 |
X2 + 1 |
0 |
0 |
/ dx |
7T f |
lim / 0 | |
pętcoj x2 + i |
2 ^ J |
s |
—oo |
x2 dx
dx 7T
1.5 pt.
dx
— lim
x4 + x2 + 1 x2 dx
zbieżna z kryterium porównawego
1.5 pt.
X2 + 1 T—*oo J x2 + 1 2
o
O
r x2 dx.
X4 + X2 + 1
zbieżna z kryterium porównawczego
1.0 pt.
x4 + x2 + 1
zbieżna
0.5 pt.
% = |
ln(y + 1) - e, | |
(e,°),,+ |H |
[e, 0)u?, = —evx + evy | |
( Vx ~ Vy = 0 l Mą-1 | ||
V- f-L -L' |
) " *=-{ |
J_ jJ |
lv2’ V2. |
.y/2' y/2) |
1.5 pt. 0.5 pt.
1.0 pt.
1.0 pt.
1.0 pt.
4. • |^=4y + 4x3, ~ = 4x + 4y3
ax aj/
( x = 0, f |
X = 1, f |
•S ' V < | |
\ y = 0 I |
y = —i \ |
x = —1, 2/ -1
0 4 4 0 12 4
4 12
12 4
4 12
—16 < 0 ==> (0,0) nie ma ekstremum d2f
= 128 > 0, ^-(1, -1) = 12 > 0
1.0 pt.
1.0 pt.
1.0 pt.
0.5 pt.
(1, —1) min. lok. wł. (—1,1) min. lok. wł.
1.5 pt.
1