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Anti - turunan
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Anti - turunan Teorema A Teorema C Teorema B Teorema D
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Anti - turunan Teorema A
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Anti - turunan Anti - π»ππππππ π»ππππππ π¦β²=πβ² π₯ π¦=π π₯ π¦=π π₯ π¦β²=πβ² π₯ 4π₯β5
= π₯ 1+1 β5π₯+πΆ = 2 π₯ 2 β5π₯+πΆ = π₯ 6+1 β π₯ 4+1 +πΆ = π₯ 7 β6 π₯ 5 +πΆ = π₯ πΆ = π₯ πΆ π»ππππππ π¦β²=πβ² π₯ π¦=π π₯ π¦=π π₯ π¦β²=πβ² π₯ 4π₯β5 = π₯ 1+1 β5π₯+πΆ = 2 π₯ 2 β5π₯+πΆ 2 π₯ 2 β5π₯+3 4π₯β5 π₯ 7 β6 π₯ 5 β4 7 π₯ 6 β30 π₯ 4 7 π₯ 6 β30 π₯ 4 = π₯ 6+1 β π₯ 4+1 +πΆ = π₯ 7 β6 π₯ 5 +πΆ π₯ 4 3 π₯ 1 3 4 3 π₯ 1 3 = π₯ πΆ = π₯ πΆ
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Anti - turunan Teorema B
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Teorema C 5 π₯ 2 ππ₯ =5 π₯ 2 ππ₯ Anti - turunan (i) ππ π₯ ππ₯=π π π₯ ππ₯
Teorema C memiliki 3 sifat : (i) ππ π₯ ππ₯=π π π₯ ππ₯ Contoh : 5 π₯ 2 ππ₯ =5 π₯ 2 ππ₯
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Teorema C Anti - turunan (4π₯+5)ππ₯ = 4π₯ππ₯ + 5ππ₯
(ii) π π₯ +π π₯ ππ₯ = π π₯ ππ₯ + π π₯ ππ₯ Contoh : (4π₯+5)ππ₯ = 4π₯ππ₯ + 5ππ₯
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Teorema C Anti - turunan (4π₯β5)ππ₯ = 4π₯ππ₯ β 5ππ₯
π π₯ βπ π₯ ππ₯ = π π₯ ππ₯ β π π₯ ππ₯ (iii) Contoh : (4π₯β5)ππ₯ = 4π₯ππ₯ β 5ππ₯
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Anti - turunan Teorema D π π₯ π .π π₯ ππ₯ = 1 π+1 π π₯ π+1 +πΆ
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Teorema D Anti - turunan Contoh soal : Misal π₯ 3 +6π₯ 5 . 6 π₯ 2 +12 ππ₯
π₯ 3 +6π₯ π₯ ππ₯ Misal π 5 2ππ’ π= π₯ 3 +6π₯ ππ π₯ ππ₯=π π π₯ ππ₯ =2 π 5 ππ’ ππ’ ππ₯ =3 π₯ 2 +6 = π 6 +πΆ ππ’=(3 π₯ 2 +6)ππ₯ 3 = π₯ 3 +6π₯ 6 +πΆ 2ππ’=2 3 π₯ 2 +6 ππ₯
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Latihan Soal dan Soal Bonus
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Latihan Soal : Anti - turunan =3. 1 β2+1 π₯ β2+1 β2. 1 β3+1 π₯ β3+1 +C
π π₯ = 3 π₯ 2 β 2 π₯ 3 =3. 1 β2+1 π₯ β2+1 β2. 1 β3+1 π₯ β3+1 +C 3 π₯ 2 β 2 π₯ 3 ππ₯ =β3 π₯ β1 + π₯ β2 +C = (3 π₯ β2 β2 π₯ β3 )ππ₯ = β3 π₯ + 1 π₯ 2 +πΆ
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Soal Bonus: Anti - turunan π π₯ = 2π₯ π₯ + 3 π₯ 5
π π₯ = 2π₯ π₯ + 3 π₯ 5 = β π₯ β β5+1 π₯ β5+1 +πΆ π π₯ = 2 π₯ π₯ + 3 π₯ 5 = π₯ β 3 4 π₯ β4 +πΆ π π₯ = 2 π₯ β π₯ β5 =2 2 π₯ β 3 4 π₯ 4 +πΆ 2 π₯ β π₯ β5 ππ₯
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Latihan Soal : Anti - turunan π π₯ = 4 π₯ 6 +3 π₯ 4 π₯ 3
π π₯ = 4 π₯ 6 +3 π₯ 4 π₯ 3 = π₯ π₯ 1+1 +πΆ π π₯ = π₯ 3 4 π₯ 3 +3π₯ π₯ 3 = π₯ π₯ 2 +πΆ = π₯ π₯ 2 +πΆ π π₯ =4 π₯ 3 +3π₯ = (4 π₯ 3 +3π₯)ππ₯
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Soal Bonus: Anti - turunan π π₯ = π₯ 6 βπ₯ π₯ 3
π π₯ = π₯ 6 βπ₯ π₯ 3 = π₯ 3+1 β 1 β2+1 π₯ β2+1 +πΆ π π₯ = π₯ 6 π₯ 3 β π₯ π₯ 3 = 1 4 π₯ 4 + π₯ β1 +πΆ = 1 4 π₯ π₯ +πΆ π π₯ = π₯ 3 β π₯ β2 = ( π₯ 3 β π₯ β2 )ππ₯
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Latihan Soal : Anti - turunan sinπ₯ 1+cosπ₯ 4 ππ₯ sinπ₯ . π 4 .ππ₯
Misal : = sinπ₯ . π ππ’ βsinπ₯ π=1+cosπ₯ ππ’ ππ₯ =βsinπ₯ = β π 4 ππ’ ππ₯= ππ’ βsinπ₯ =β 1 5 π 5 +πΆ
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Soal Bonus: Anti - turunan sinπ₯cosπ₯ . π 1 2 . ππ₯ Misal :
sinπ₯cosπ₯ 1+ sin 2 π₯ ππ₯ = π πΆ sinπ₯cosπ₯ . π ππ₯ Misal : = π πΆ sinπ₯cosπ₯ . π ππ’ 2sinπ₯cosπ₯ π=1+ sin 2 π₯ ππ’ ππ₯ =2sinπ₯cosπ₯ = π ππ’ = 1 3 π πΆ ππ₯= ππ’ 2sinπ₯cosπ₯ = π πΆ = sin 2 π₯ 3 +πΆ
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Latihan Soal : Anti - turunan πβ²β² π₯ = π₯ 4 +1 π₯ 3
πβ²β² π₯ = π₯ π₯ 3 πβ² π₯ = π₯ β3+1 π₯ β3+1 + πΆ 1 π π₯ = π₯ π₯ + πΆ 1 π₯+ πΆ 2 πβ²β² π₯ = π₯ 4 π₯ π₯ 3 πβ² π₯ = 1 2 π₯ 2 β 1 2 π₯ β2 + πΆ 1 π π₯ = π₯ 2 β 1 2 π₯ β2 + πΆ 1 ππ₯ πβ²β² π₯ =π₯+ π₯ β3 πβ² π₯ = (π₯+ π₯ β3 )ππ₯ π π₯ = π₯ 2+1 β β2+1 π₯ β2+1 + πΆ 1 π₯+ πΆ 2
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Soal Bonus: Anti - turunan πβ²β² π₯ =2 3 π₯+1 π π₯ = 3 2 π₯+1 4 3 + πΆ 1 ππ₯
π π₯ = π₯ πΆ 1 ππ₯ πβ²β² π₯ =2 π₯ π π₯ = π₯ πΆ 1 π₯+ πΆ 2 πβ² π₯ = π₯ ππ₯ π π₯ = π₯ πΆ 1 π₯+ πΆ 2 πβ² π₯ = π₯ πΆ 1 πβ² π₯ = π₯ πΆ 1
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