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SzeregiTayloraica“ki
Zadanie1.Dok“adno–¢uzyskanegowynikubƒdziezale»a“aodilo–ciwyraz ó w,
kt ó rezsumujemy.Wrozwi¡zaniuwypisujƒjedyniekilkapierwszychwyraz ó wrozwi-
niƒcia.
a) e 0 , 2 1+0 , 2+ (0 , 2) 2
2 + (0 , 2) 3
3! ++ (0 , 2) 4
4! + ...
b)cos(0 , 1) 1 (0 , 1) 2
2! + (0 , 1) 4
4! (0 , 1) 6
6! + (0 , 1) 8
8! + ...
c)sin(0 , 1) 0 , 1 (0 , 1) 3
3! + (0 , 1) 5
5! (0 , 1) 7
7! + (0 , 1) 9
9! + ...
p e = e 0 , 5 = 1+0 , 5+ (0 , 5) 2
2 + (0 , 5) 3
3! + (0 , 5) 4
d)
4! + ...
e)ln(1+ x )= x x
2 + x 2
3 x 3
4 + ··· +( 1) n +1 x n
n + ...
2 + (0 , 1) 2
3 (0 , 1) 3
4 + (0 , 1) 4
zatemln(1 , 1) 1 0 , 1
5 + ...
f)sin(1) 1 1
3! + 1
5! 1
7! + 1
9! + ...
g)cos(1) 1 1
2! + 1
4! 1
6! + 1
8! + ...
h)ln(2) 0 , 1 1
2 + 1
3 1
4 + 1
5 1
6 + 1
7 1
8 + 1
9 + ...
Zadanie2.
a) 2
3 x 3 + 3
2 x 2 x + C b)5ln( | x +4 | ) 1
3 x 3 + C
7 q
c) cos( x 1) sin( x +1)+ C d) 8
7
( x +1) 8 + 1
15 x 5 + C
e) 4
5 x 5 2
3 x 3 + e x + C f)3 3 p
e x +2
p
x 7+ C
1
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Zadanie3.
5 Z
( x (5 x )+1) dx =( 5
2 x 1
5
= 25
2 125
a) P =
3 x 3 + x )
3 +5
0
0
3 Z
p
q
3
p
64 p
b) P =
x +1 dx =
( x +1) 3
=
1=7
Z
0
0
c) P =
sin xdx =cos x
=2
0
e x dx = e x
0
ln5
ln5
= 1
5 + 1
2 = 3
d) P =
10
ln2
ln2
0 Z
0
1
3 x 3 1
= 1
3 + 1
2 = 1
e) P =
( x ( x +1)) dx =
2 x
6
1
1
b Z
2
2
b
= b 2
2 + b = b +( b +
f) P =
( x + ) dx =
+ x
2
0
0
Zadanie4.
b Z
4
b
4
a)
(2 x +1) 2 dx =
3 x 3 +2 x 2 + x
=
3 ( b 3 a 3 )+2( b 2 a 2 )+( b a )
a
a
2 Z
9
2
9
5 · 32+16+2
b)
(3 x 2 +1) 2 dx =
5 x 5 +2 x 3 + x
=
0
0
1
!
Z Z
Z
1 cos(2 x )
2 =
2 x sin(2 x )
= 2
2
c)
sin 2 x
4
0
0
0
xdx = x 2
a Z
a
d)
= a 2
0
0
2 Z
Z
2
e)
sin xdx = ( cos x )
= (0 ( 1))=
0
0
Z
ln3
e 2 x dx =
ln3
=
2
1
9 1
= 4
9
f)
2 e x
0
0
2
Z
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Zgłoś jeśli naruszono regulamin