Integral parsial tanzalin2
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INTEGRAL PARSIALINTEGRAL PARSIALTEKNIK TEKNIK TANZALINTANZALIN
Oleh :
PROGRAM PASCA SARJANA PENDIDIKAN MATEMATIKA UNIVERSITAS SRIWIJAYA
2O12 / 2013
Efuansyah, S.Pd06122502008
Dosen Pengampu : Prof. Dr. Zulkardi, M.Ikom, M.Sc.
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APAYANG AKAN
SAYA PEROLEHDARI BELAJAR
INTEGRALINI ?
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STANDAR KOMPETENSI :
MENGGUNAKAN KONSEP INTEGRAL DALAM PEMECAHAN MASALAH.
KOMPETENSI DASAR
MENGHITUNG INTEGRAL TAK TENTU DAN INTEGRAL TENTU DARI FUNGSI ALJABAR DAN FUNGSI TRIGONOMETRI YANG SEDERHANA.
KEJAR
SAMPAI
DAPAT
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PENGERTIAN INTEGRALPENGERTIAN INTEGRAL
DiferensialDiferensial(f(x))(f(x))
IntegralIntegral(gⁿ (x))(gⁿ (x))
AxAxnn
n.axn.axn-1n-1
(n-1).n.ax(n-1).n.axn-2n-2
......................
00
(ax + b)(ax + b)nn
......................
……………………
PERHATIKAN TABEL BERIKUT :
1)()1(
1 +++
nbaxna
!!)()11(
)1(
1
+++++
+ nbaxna
na +-+
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7121 (3 2)x −
81504 (3 2)x −
DiferensialDiferensial IntegralIntegral
62 (3 2)x x dx− =∫
2x
7 82 1(3 2) (3 2)
21 252x x x C= − − − +
02
6(3 2)x −
62 (3 2)x x dx− =∫
+
-71
212 . (3 2)x x − 812. (3 2)504
x C− − +
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-cos x
diturunkanditurunkan di-integralkandi-integralkan
sinx xdx =∫
x
0
1sin x
sinx xdx =∫
+
-
cosx x− sin x C+ +
sin x−
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-cos x
diturunkanditurunkan di-integralkandi-integralkan
2 sinx xdx =∫
2x
0
2sin x
2 sinx xdx =∫
+
-
2 cosx x− 2sin x C+ +
sin x−
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322
3 ( 1)x −
diturunkanditurunkan di-integralkandi-integralkan
1x x dx− =∫
x
0
1
322
3 ( 1)x x −
( )121x −
+
-524
15( 1)x C− − +
524
15 ( 1)x −
1x x dx− =∫
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32 6 1x x dx− =∫2 sinx xdx =∫
2 1x x dx− =∫3 2 x x dx− =∫
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Di turunkanDi turunkan Di integralkanDi integralkan
13(6 1)x dx− =∫
731
112 (6 1)x dx− =∫
32 6 1x x dx− =∫
431
8 (6 1)x dx− =∫
431
82 . (6 1)x x − 731
1122. (6 1)x C− − +431
4 (6 1)x x= −731
56 (6 1)x C− − +
32 6 1x x dx− =∫
2x20 +
-
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2 sinx xdx =∫
cos x−sin x−
Di turunkanDi turunkan Di integralkanDi integralkan
xx22
2x2x22
sin x
00 cos x
2 cosx x−2 sinx x+
2 sinx x dx =∫2cos x+
2 cos 2 sin 2cosx x x x x C− + + +
+-
+
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diturunkanditurunkan di integralkandi integralkan
XX22
2x2x2200
322 2
3 . ( 1)x x −522
15 -2 . ( 1)x x −
2 1x x dx− =∫12( 1)x −
322
3 ( 1)x −522
15 ( 1)x −724
105 ( 1)x − 724
1052. ( 1)x C+ − +
2 1x x dx− =∫
+
-
+
524
15 .( 1)x x− −728
105 ( 1)x C+ − +3222
3 ( 1)x x= −
+
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Di turunkanDi turunkan Di integralkanDi integralkan
XX
11
00
3 2 x x dx− =∫
12(3 2 )x−
321
3 (3 2 )x− −
521
15 (3 2 )x− −+
-
3 2x xdx− =∫
321
3 (3 2 )x x− −521
15 .(3 2 )x− −
+C
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