Integral Tak Tentu

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Transcript Integral Tak Tentu 

By
Fattaku Rohman, S.Pd
Guru Matematika
SMAN Titian Teras Jambi
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
END
LATIHAN SOAL
APLIKASI
INTEGRAL
INTEGRAL LUAS
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
Pengertian
Bila suatu fungsi f(x) mempunyai turunan
fā€™(x), maka bila fā€™(x) diintegralkan akan
menjadi f(x) + C.
š‘Ž
š‘›
Rumus :
š‘Žš‘„ dx =
š‘„ š‘›+1 + š¶
š‘›+1
INTEGRAL TENTU
INTEGRAL LUAS
Contoh soal :
APLIKASI
2
1 0+1
1+1
2š‘„ āˆ’ 1 š‘‘š‘„ =
š‘„
āˆ’ š‘„
+š‘
1+1
1
LATIHAN SOAL
= 2š‘„ 2 āˆ’ š‘„ + š¶
END
Lakukan substitusi
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
Pengertian
Jika U= g(x) dengan g (x) mempunyai turunan
maka f(u)=f(g(x)).
INTEGRAL TENTU
Contoh soal :
INTEGRAL LUAS
4
6 3
(2š‘„ āˆ’ 5) š‘„ š‘‘š‘„
APLIKASI
LATIHAN SOAL
Misal : u = 2š‘„ 4 āˆ’ 5, š‘‘š‘Žš‘›
š‘‘š‘¢ = 8š‘„ 3
š‘‘š‘¢
=š‘„ 3
8
š‘¢6 .
Jawab:
1
š‘‘š‘¢
8
=
(2š‘„ 4 āˆ’ 5)6
8
=
1
1
1
š‘„ (2š‘„ 4 āˆ’ 5)7 =
8
7
56
4
7
(2š‘„ āˆ’ 5)
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
Hubungan Trigonometri :
š‘ š‘–š‘›š‘Žš‘„ š‘‘š‘„ = āˆ’
1. š‘ š‘–š‘›2 x + š‘š‘œš‘  2 š‘„ = 1
2. 1 + š‘”š‘Žš‘›2 š‘„ = š‘ š‘’š‘ 2 š‘„
3. 1 + š‘š‘œš‘” 2 š‘„ = š‘š‘œš‘ š‘’š‘ 2 š‘„
4. Sin2x = 2sinxcosx
1
5. š‘ š‘–š‘›2 š‘„ = 2 1 āˆ’ š‘š‘œš‘ 2š‘„
6. š‘š‘œš‘  2 š‘„ =
1
2
1
cosa š‘„
š‘Ž
+C
1
š‘š‘œš‘ š‘Žš‘„ š‘‘š‘„ = š‘ š‘–š‘› š‘Žš‘„ + C
š‘Ž
š‘ š‘’š‘ 2 š‘„ š‘‘š‘„ = š‘”š‘Žš‘›š‘„ + C
( 1 + cos 2x)
INTEGRAL LUAS
APLIKASI
LATIHAN SOAL
Contoh soal :
š‘ š‘–š‘›2š‘„ āˆ’ 5š‘š‘œš‘ š‘„ š‘‘š‘„ = āˆ’
1
2
cos 2x ā€“ 5 sin 2x +C
END
INTEGRAL TAK
TENTU
Bentuk umum:āˆ« f(x) āˆ™ gāæ (x) dx
INTEGRAL
SUBTITUSI
Rumus:
āˆ« u āˆ™ dv = u āˆ™ v - āˆ« v āˆ™ du
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
APLIKASI
LATIHAN SOAL
Contoh soal :
āˆ« x āˆ™ sin 2x dx
Misal : u = x, v = ā€“½ āˆ™ cos 2x, du = dx
= x (ā€“½ āˆ™ cos 2x) ā€“ āˆ« ā€“½ āˆ™ cos 2x dx
= ā€“½x āˆ™ cos 2x + ¼ āˆ™ sin 2x + c
=-
1
2
š‘„ cos 2š‘„ āˆ’
1
sin 2š‘„
2
+c
Cara mudah dengan menggunakan
Tanzali
END
Deferensial
Integral
+X
sin 2š‘„
-1
1
- cos 2š‘„
2
1
- sin 2š‘„
4
+0
Setelah dikalikan silang,
1
1
maka = - š‘„ cos 2š‘„ + sin 2š‘„ + š¶
2
4
1
1
=š‘„ cos 2š‘„ āˆ’ sin 2š‘„ +
2
2
TERBUKTI, HASILNYA SAMA
BACK
C
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
Pengertian
Bila suatu fungsi f(x) mempunyai turunan fā€™(x),
maka bila fā€™(x) diintegralkan pada selang (a,b)
menjadi :
š‘ ā€²
š‘“
š‘Ž
š‘
š‘„ š‘‘š‘„ = [š‘“ š‘„ ] = f ā€² b āˆ’ f ā€² (a)
š‘Ž
INTEGRAL TENTU
Contoh soal :
INTEGRAL LUAS
APLIKASI
LATIHAN SOAL
5 3
š‘„
0
1
5
+ 2š‘„ š‘‘š‘„ = š‘„ 4 + š‘„ 2 ]
4
0
1 4
= ( 5 + 52 ) āˆ’ (0)
4
625
= + 25
4
725
1
=
= 181
4
4
END
INTEGRAL TAK
TENTU
Pengertian
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
L(R)=
š‘
š‘“
š‘Ž
š‘„ š‘‘š‘„
Contoh soal :
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
APLIKASI
LATIHAN SOAL
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
APLIKASI
:
LATIHAN SOAL
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
Tentukan hasil dari š‘„ š‘„š‘‘š‘„ š‘Žš‘‘š‘Žš‘™š‘Žā„Ž
INTEGRAL PARSIAL
š‘Žš‘„ š‘› dx =
INTEGRAL TENTU
= š‘„. š‘„ š‘‘š‘„
3
2
= š‘„ š‘‘š‘„
=3
LATIHAN SOAL
š‘„ š‘›+1 + š¶
1
2
INTEGRAL LUAS
APLIKASI
š‘Ž
š‘›+1
1
+1
š‘„
3
+1
2
+š¶
2
2 5
= š‘„2 + š¶
5
2 2
= š‘„ š‘„+š¶
5
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
9š‘„ 2
š»š‘Žš‘ š‘–š‘™ š‘‘š‘Žš‘Ÿš‘–
š‘„3 + 8
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
Misalkan: u = š‘„ 3 + 8
du = 3š‘„ 2 š‘‘š‘„
9š‘„ 2
dx =
š‘„ 3 +8
=
=3
APLIKASI
LATIHAN SOAL
š‘‘š‘„ š‘Žš‘‘š‘Žš‘™š‘Žā„Ž ā€¦
=3
3š‘‘š‘¢
š‘¢
1
2
3š‘¢ š‘‘š‘¢
1
1
āˆ’2+1
1
1
2
1
2
š‘¢
1
2
š‘¢
1
āˆ’2+1
+C
+c
= 6š‘¢ +c
= 6 š‘„ 3 + 8 +C
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
Tentukan Integral dari cos 6š‘„ š‘‘š‘„ adalah
INTEGRAL
TRIGONOMETRI
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
Rumus : cos š‘„ š‘‘š‘„ =
1
sin
š‘Ž
š‘Žš‘„ + š¶
1
cos 6š‘„ š‘‘š‘„ = sin 6š‘„ + š¶
6
APLIKASI
LATIHAN SOAL
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
INTEGRAL
TRIGONOMETRI
š»š‘Žš‘ š‘–š‘™ š‘‘š‘Žš‘Ÿš‘–
š‘„ 2 sin š‘„ āˆ’ 4 š‘‘š‘„ š‘Žš‘‘š‘Žš‘™š‘Žā„Ž ā€¦
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
APLIKASI
Misalkan:
u = š‘„2
dv = sin(x-4)dx
du =2xdx v = -cos(x-4)
= uv - š‘£š‘‘š‘¢
= š‘„ 2 āˆ’š‘š‘œš‘  š‘„ āˆ’ 4 - āˆ’š‘š‘œš‘  š‘„ āˆ’ 4
=-š‘„ 2 š‘š‘œš‘  š‘„ āˆ’ 4 + 2š‘„š‘ š‘–š‘›(š‘„ āˆ’ 5)
2š‘„ dx
LATIHAN SOAL
END
INTEGRAL TAK
TENTU
INTEGRAL
SUBTITUSI
Tentukan hasil dari
INTEGRAL
TRIGONOMETRI
3 š‘‘š‘„
1 š‘„2
š‘Žš‘‘š‘Žš‘™š‘Žā„Ž
INTEGRAL PARSIAL
INTEGRAL TENTU
INTEGRAL LUAS
APLIKASI
LATIHAN SOAL
š‘
š‘Ž
š‘
š‘“ ā€² š‘„ š‘‘š‘„ = [š‘“ š‘„ ] = f ā€² b āˆ’ f ā€² (a)
š‘Ž
3 š‘‘š‘„
3 āˆ’2
=
š‘„ š‘‘š‘„ =
1 š‘„2
1
1 3
1
=āˆ’
= āˆ’ āˆ’
š‘„ 1
3
2
=
3
āˆ’š‘„ āˆ’1
3
1
(āˆ’1)
END
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