integral_definido._exercicios_resolvidos
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integral_definido._exercicios_resolvidos


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\u222b \u22c5+ \u2212
1
0
1 dx
ee xx
. 
 
 Fazemos a substituição te x = . Então temos: 
t
dtdxtnlxte x =\u21d2=\u21d2= . 
Determinamos os limites de integração para a variável t : 
 
10 0 ==\u21d2= etx nfinfi , eetx upsups ==\u21d2=
11 . 
 
Portanto 
( ) ==\u22c5
+
=\u22c5\u22c5
+
=\u22c5
+
=\u22c5
+ \u222b\u222b\u222b\u222b \u2212
e
ee
x
x
xx
tarctgdt
t
dt
t
t
t
dx
e
e
dx
ee 11
2
1
1
0
1
0 1
11
1
1
1
11
 
4
1 pi\u2212=\u2212= earctgarctgearctg . 
Matemática 1 Anatolie Sochirca ACM DEETC ISEL 
 
 6 
Exercício 13. \u222b \u22c5+\u22c5
4
0
2 9 dxxx . 
 
 Fazemos a substituição tx =+ 92 . Então temos: 
9
999
2
2222
\u2212
\u22c5
=\u21d2\u2212=\u21d2=+\u21d2=+
t
dttdxtxtxtx . 
Determinamos os limites de integração para a variável t : 
 
390 ==\u21d2= nfinfi tx , 5944 2 =+=\u21d2= upsups tx . 
 
Portanto 
3
98
3
3
3
5
39
99
335
3
35
3
2
5
3
2
2
4
0
2
=\u2212=\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
=\u22c5=\u22c5
\u2212
\u22c5\u22c5\u2212=\u22c5+\u22c5 \u222b\u222b\u222b
tdttdt
t
t
ttdxxx . 
 
Exercício 14. \u222b
+++
2
0
3)1(1 xx
dx
. 
 
 Fazemos a substituição tx =+1 . Então temos: 
dttdxtxtxtx \u22c5=\u21d2\u2212=\u21d2=+\u21d2=+ 2111 22 . 
 
Determinamos os limites de integração para a variável t : 
 
110 ==\u21d2= nfinfi tx , 3122 =+=\u21d2= upsups tx . 
 
Portanto 
 
=
+
\u22c5
\u22c5=
+
\u22c5
=
+++
=
+++
\u222b\u222b\u222b\u222b
3
1
2
3
1
3
2
0
3
2
0
3 )1(2
2
)1(1)1(1 tt
dtt
tt
dtt
xx
dx
xx
dx
 
 
( ) ( )
643
21322
1
2 3
1
3
1
2
pipipi
=\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
\u2212\u22c5=\u2212\u22c5=\u22c5=
+
\u22c5= \u222b arctgarctgtarctgt
dt
. 
 
 
Exercício 15. \u222b ++
2
0 1
pi
xoscxsen
dx
. 
 
 Fazemos a substituição txtg =\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
2
. Então temos: 
 
2
2
2 1
1
,
1
2
t
t
xosc
t
t
senx
+
\u2212
=
+
= . 
Matemática 1 Anatolie Sochirca ACM DEETC ISEL 
 
 7 
21
22
22 t
dtdxtarctgxtarctgxtxtg
+
\u22c5
=\u21d2\u22c5=\u21d2=\u21d2=\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
. 
 
Determinamos os limites de integração para a variável t : 
 
0)0(0 ==\u21d2= tgtx nfinfi , 142 =\uf8f7\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
=\u21d2=
pipi
tgtx upsups . 
 
Portanto 
=
+
=
\u2212+++
\u22c5
=
+
\u22c5
\u22c5
+
\u2212
+
+
+
=
++ \u222b\u222b\u222b\u222b
1
0
1
0
22
1
0
2
2
2
2
2
0 1121
2
1
2
1
1
1
21
1
1 t
dt
ttt
dt
t
dt
t
t
t
txoscxsen
dx
pi
 
 
( ) ( ) 201111)1(
1
1 1
0
1
0
nlnlnltnltd
t
=+\u2212+=+=+\u22c5
+
= \u222b . 
 
Exercício 16. \u222b +
4
0 1 x
dx
. 
 
 Fazemos a substituição tx = . Então temos: 
dttdxtxtx \u22c5=\u21d2=\u21d2= 22 . 
 
Determinamos os limites de integração para a variável t : 
 
000 ==\u21d2= nfinfi tx , 244 ==\u21d2= upsups tx . 
 
Portanto 
 
=\u22c5\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
+
\u2212
+
+
\u22c5=\u22c5
+
\u2212+
\u22c5=
+
\u22c5
\u22c5=
+
\u22c5
=
+ \u222b\u222b\u222b\u222b\u222b
2
0
2
0
2
0
2
0
4
0 1
1
1
12
1
112
1
2
1
2
1
dt
tt
tdt
t
t
t
dtt
t
dtt
x
dx
 
 
( ) ( ) =+\u22c5\u2212\u22c5=+\u22c5
+
\u22c5\u2212\u22c5= \u222b\u222b
2
0
2
0
2
0
2
0
122)1(
1
122 tnlttd
t
dt 
 
( ) 32401212)02(2 nlnlnl \u22c5\u2212=+\u2212+\u22c5\u2212\u2212\u22c5= . 
 
Exercício 17. \u222b
\u2212
++
0
1
3 11 x
dx
. 
 
 Fazemos a substituição tx =+3 1 . Então temos: 
dttdxtxtxtx \u22c5=\u21d2\u2212=\u21d2=+\u21d2=+ 2333 3111 . 
 
Matemática 1 Anatolie Sochirca ACM DEETC ISEL 
 
 8 
Determinamos os limites de integração para a variável t : 
 
0111 3 =+\u2212=\u21d2\u2212= nfinfi tx , 1100 3 =+=\u21d2= upsups tx . 
 
Portanto 
 
=\u22c5\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
+
\u2212
+
+
\u22c5=\u22c5
+
\u2212+
\u22c5=
+
\u22c5=
+
=
++ \u222b\u222b\u222b\u222b\u222b
\u2212
dt
t
t
t
dt
t
t
t
dtt
t
dtt
x
dx 1
0
21
0
21
0
21
0
20
1
3 1
1
1
13
1
113
1
3
1
3
11
 
 
=\u22c5\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
\u2212+
+
\u22c5=\u22c5\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
+
+\u2212
+
+
\u22c5= \u222b\u222b dttt
dt
t
tt
t
1
0
1
0
1
1
13
1
)1)(1(
1
13 
 
( ) ( ) =\u22c5\u2212\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\u22c5++\u22c5=\u22c5\u2212\u22c5\u22c5+\u22c5
+
\u22c5= \u222b\u222b\u222b
1
0
1
0
21
0
1
0
1
0
1
0
3
2
31333
1
13 tttnldtdttdt
t
 
 
( )
2
323)01(3
2
0
2
1301113
22
\u2212\u22c5=\u2212\u22c5\u2212\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\u2212\u22c5++\u2212+\u22c5= nlnlnl . 
 
Exercício 18. \u222b \u22c5
\u2212
3
0 6
dx
x
x
. 
 
 Fazemos a substituição t
x
x
=
\u22126
. Então temos: 
\u21d2
+
=\u21d2\u22c5\u2212=\u21d2=
\u2212
\u21d2=
\u2212
2
2
222
1
66
66 t
t
xtxtxt
x
x
t
x
x
 
 
( ) ( ) dtt
tdt
t
ttttdt
t
tdx \u22c5
+
=\u22c5
+
\u22c5\u2212+
=\u22c5
\u2032
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
+
= 2222
22
2
2
1
12
1
26)1(12
1
6
. 
 
Determinamos os limites de integração para a variável t : 
 
00 =\u21d2= nfinfi tx , 13 =\u21d2= upsups tx . 
 
Portanto 
 
( ) ( ) ( ) ( ) =+
\u22c5
\u22c5\u22c5=
+
\u22c5\u22c5
\u22c5=\u22c5
+
\u22c5=\u22c5
+
\u22c5=\u22c5
\u2212
\u222b\u222b\u222b\u222b\u222b
1
0
22
1
0
22
1
0
22
21
0
22
3
0 1
26
1
26
1
12
1
12
6 t
dtt
t
t
dtttdt
t
tdt
t
t
tdx
x
x
 
 
Na continuação integramos por partes: 
 
dtdUtU == , ; 
Matemática 1 Anatolie Sochirca ACM DEETC ISEL 
 
 9 
( ) ( ) ( ) ( )
( )
2
122
22
2
22
2
2222 1
1
12
1
1
)1(
1
)(
1
2
,
1
2
t
t
t
td
t
td
t
dttV
t
dttdV
+
\u2212=
+\u2212
+
=
+
+
=
+
=
+
\u22c5
=
+
\u22c5
=
+\u2212
\u222b\u222b\u222b . 
Obtemos 
 
( ) =\uf8fa
\uf8fb
\uf8f9
\uf8ef
\uf8f0
\uf8ee
+\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
+
\u2212
+
\u2212\u22c5=
\uf8fa
\uf8fa
\uf8fb
\uf8f9
\uf8ef
\uf8ef
\uf8f0
\uf8ee
+
+\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
+
\u2212\u22c5= \u222b
1
022
1
0
2
1
0
2 01
0
11
16
11
6 tarctg
t
dt
t
t
 
 
2
)2(3
42
16 \u2212\u22c5=\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
+\u2212\u22c5=
pipi
. 
 
 
Exercício 19. \u222b +
1
0 1
xe
dx
. 
 
 Fazemos a substituição te x = . Então temos: 
t
dtdxtnlxte x =\u21d2=\u21d2= . 
 
Determinamos os limites de integração para a variável t : 
 
10 0 ==\u21d2= etx nfinfi , eetx upsups ==\u21d2=
11 . 
 
Portanto 
=\u22c5\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\u22c5+
\u2212
\u22c5+
+
=\u22c5
\u22c5+
\u2212+
=
\u22c5+
=
+ \u222b\u222b\u222b\u222b
eee
x
dt
tt
t
tt
tdt
tt
tt
tt
dt
e
dx
111
1
0 )1()1(
1
)1(
1
)1(1 
 
( ) ( ) =+\u2212=+\u22c5
+
\u2212\u22c5=\u22c5\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
+
\u2212= \u222b\u222b\u222b
ee
eee
tnltnltd
t
dt
t
dt
tt 11111
1)1(
1
11
)1(
11
 
 
( ) ( )
1
2)1()2()1(22)1(1
+
=+\u2212=+\u2212+=\u2212+\u2212\u2212=
e
e
nlenlenlenlnlnelnlenlnlnel . 
 
 
Exercício 20. \u222b
+
3
1
32 )1( x
dx
. 
 
 Fazemos a substituição ttgx = . Então temos: 
tosc
dtdx 2= , ( ) tt
tsent
t
tsen
t
tsen
ttgx 22
22
2
22
22
cos
1
cos
cos
cos
1
cos
111 =+=+=\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
+=+=+ . 
 
Porque xarctgtttgx =\u21d2= , determinamos os limites de integração para a 
variável t : 
Matemática 1 Anatolie Sochirca ACM DEETC ISEL 
 
 10 
4
)1(1 pi==\u21d2= arctgtx nfinfi , 3)3(3
pi
==\u21d2= arctgtx upsups . 
 
Portanto 
( ) =
\u22c5
==
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
=
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
+
=
+
\u222b\u222b\u222b\u222b\u222b\u222b
3
4
2
33
4 3
23
4
3
23
4
3
2
23
1
3
2
3
1
32 1111)1(
pi
pi
pi
pi
pi
pi
pi
pi tosc
dttosc
tosc
tosc
dt
tosc
tosc
dt
tosc
tosc
dt
x
dx
x
dx
 
( )
2
23
2
2
2
3
43
3
3
3
4
\u2212
=\u2212=\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
\u2212\uf8f7
\uf8f8
\uf8f6
\uf8ec
\uf8ed
\uf8eb
==\u22c5= \u222b
pipipi
pi
pi
pi
sensensentdttosc