jawaban tugas 1 kalkulus 2
DESCRIPTION
Buat peserta kuliah Kalkulus 2 TI.Unpar 2009/2010Jawaban tugas pertama, tugas 2nya menyusulTRANSCRIPT
1. ∫ ∫ +−−+=
−+−+ dx
xxxxdx
xxxx
)12)(1(12
24312
223
221)22)(1(12
22 +−++
−=
+−−+
xxCBx
xA
xxxx
)22)(1()1)(()22(
2
2
+−−−+++−=
xxxxCBxxxA
53
312
220
)1)(()22(12 2
=−=
==−
=+−−=+
−+++−=+
CBA
CACBA
BAxCBxxxAx
∫∫∫ +−+−+
−=
+−−+ dx
xxxdx
xdx
xxxx
2253
13
)12)(1(12
22
Cxxxx
xdxxxx
dxxx
xxx
dxxx
xx
+−++−+−=
+−++−+−=
+−++−+−=
+−
+−−+−=
∫
∫
∫
|1|arctan8|22|ln23|1|ln3
1)1(8|22|ln
23|1|ln3
228|22|ln
23|1|ln3
22
8)22(23
|1|ln3
2
22
22
2
2. ∫ −−+ dxxxx
2)2)(1(23
22 )2(21)2)(1(23
−+
−+
−=
−−+
xC
xB
xA
xxx
2
2
)2)(1()1()2)(1()2(
−−−+−−+−=
xxxCxxBxA
85
5224
3340
)1()23()44(23 22
=−=
==−+
=+−−=+
−++−++−=+
CBA
CBACBA
BAxCxxBxxAx
∫∫∫∫ −+
−−+
−=
−−+ dx
xdx
xdx
xdx
xxx
22 )2(8
25
15
)2)(1(23
Cxx
x +−
−−−=
28
21ln5
3. ∫ −−dx
eee
xx
x
22
tdt
tttdx
eee
tdtdxdxedtet
subtitusi
xx
x
xx
⋅−=
=−−
=⇒=→=
∫∫ 22 22
∫ −−=
22 ttdt
)1)(2(1
21
2 +−=
−− tttt
)1)(2()2()1(
12
+−−++=
++
−=
tttBtA
tB
tA
∫∫ +
−+
−=
−−
−=→−=
=→=
13
1
23
1
2
311
312
2 t
dt
t
dt
ttdt
Bt
At
Cee
Ctt
x
x
++−=
++−=
12ln
31
12ln
31
4. ∫ +++
22
2
)4(32
xxx
∫ ∫∫ +−+
+=
+++
−====
++++=+++++
++=
+++
dxxx
xdxdx
xxx
DCBA
DCxxBAxxxx
DCxx
BAxx
xx
22222
2
222
22222
2
)4(12
4)4(32
1210
)4)((32)4(4)4(
32
Cx
xxdx
xxx
++
+=
+−
++= ∫ ∫
41
2arctan
)4()4(2
2arctan
21
2
2222
5. ∫ +dxxx
x2
21
22 xx
xxx
x
+=
+
41
21
21
2xx
x
+=
ttt
xx
x
xt
subtitusi
222
2
41
21
21
41
+=
+
=
2+
=tt
∫ ∫∫ +−+=
+dt
tdtdt
tt
22
2Ctt ++−= |2|ln2
Cxxdxxx
x ++−=+∫ |2|ln2
241
41