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Jika f ( x ) = a x + b , ∫ 0 2 ​ f ( x ) d x = 4 dan ∫ 2 4 ​ f ( x ) d x = 6 ,maka selisih a dan b adalah ....

Jika  dan , maka selisih  dan  adalah .... 

  1. begin mathsize 14px style 5 end style 

  2. begin mathsize 14px style 4 end style 

  3. begin mathsize 14px style 3 end style 

  4. begin mathsize 14px style 2 end style 

  5. begin mathsize 14px style 1 end style 

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S. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E. 

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Pembahasan

Cari persamaan dari dengan substitusi seperti berikut: Cari persamaan dari dengan substitusi seperti berikut: Kemudian cari nilai dan dengan metode elimasi substitusi persamaan dan persamaan seperti berikut: Kemudian subtitusi kepersamaan seperti berikut: Selisih dan adalah . Jadi, jawaban yang tepat adalah E.

Cari persamaan dari begin mathsize 14px style integral subscript 0 superscript 2 f open parentheses x close parentheses d x equals 4 end style dengan substitusi begin mathsize 14px style f open parentheses x close parentheses equals a x plus b end style seperti berikut:

begin mathsize 14px style space space space space space space space space space space space space space space space space space space space space space space space space space space space space integral subscript 0 superscript 2 open parentheses a x plus b close parentheses d x equals 4 space space space space space space space space space space space space space space space space space space space space integral subscript 0 superscript 2 a x d x plus integral subscript 0 superscript 2 b d x equals 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space right enclose fraction numerator a x squared over denominator 2 end fraction plus b x end enclose subscript space 0 end subscript superscript space 2 end superscript equals 4 open parentheses fraction numerator a open parentheses 2 close parentheses squared over denominator 2 end fraction plus b open parentheses 2 close parentheses close parentheses minus open parentheses fraction numerator a open parentheses 0 close parentheses squared over denominator 2 end fraction plus b open parentheses 0 close parentheses close parentheses equals 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses fraction numerator 4 a over denominator 2 end fraction plus 2 b close parentheses minus 0 equals 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 a plus 2 b equals 4 space space space space space... left parenthesis 1 right parenthesis end style   

Cari persamaan dari begin mathsize 14px style integral subscript 2 superscript 4 f open parentheses x close parentheses d x equals 6 end style dengan substitusi begin mathsize 14px style f open parentheses x close parentheses equals a x plus b end style seperti berikut:

begin mathsize 14px style space space space space space space space space space space space space space space space space space space space space space space space space space space space integral subscript 2 superscript 4 open parentheses a x plus b close parentheses d x equals 6 space space space space space space space space space space space space space space space space space space integral subscript 2 superscript 4 a x d x plus integral subscript 2 superscript 4 b x d x equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space right enclose fraction numerator a x squared over denominator 2 end fraction plus b x end enclose subscript space 2 end subscript superscript space 4 end superscript equals 6 open parentheses fraction numerator a open parentheses 4 close parentheses squared over denominator 2 end fraction plus b open parentheses 4 close parentheses close parentheses minus open parentheses fraction numerator a open parentheses 2 close parentheses squared over denominator 2 end fraction plus b open parentheses 2 close parentheses close parentheses equals 6 space space space space space space space space space space space open parentheses fraction numerator 16 a over denominator 2 end fraction plus 4 b close parentheses minus open parentheses fraction numerator 4 a over denominator 2 end fraction plus 2 b close parentheses equals 6 space space space space space space space space space space space space space space space space space space space space space space space 8 a plus 4 b minus 2 a minus 2 b equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 6 a plus 2 b equals 6 space space space space space... left parenthesis 2 right parenthesis end style 

Kemudian cari nilai begin mathsize 14px style a end style dan begin mathsize 14px style b end style dengan metode elimasi substitusi persamaan begin mathsize 14px style open parentheses 1 close parentheses end style dan persamaan begin mathsize 14px style open parentheses 2 close parentheses end style seperti berikut:

begin mathsize 14px style 2 a plus down diagonal strike 2 b end strike equals 4 bottom enclose 6 a plus down diagonal strike 2 b end strike equals 6 space space minus end enclose space space space space space space space minus 4 a equals negative 2 space space space space space space space space space space space a equals fraction numerator negative 2 over denominator negative 4 end fraction space space space space space space space space space space space a equals 1 half end style 

Kemudian subtitusi begin mathsize 14px style a equals 1 half end style ke persamaan begin mathsize 14px style open parentheses 1 close parentheses end style seperti berikut:

begin mathsize 14px style space space space space space 2 a plus 2 b equals 4 space 2 open parentheses 1 half close parentheses plus 2 b equals 4 space space space space space space space space 1 plus 2 b equals 4 left parenthesis 1 minus 1 right parenthesis plus 2 b equals 4 minus 1 space space space space space space space space space space space space space space 2 b equals 3 space space space space space space space space space space space space space space space space b equals 3 over 2 end style 

Selisih begin mathsize 14px style a end style dan begin mathsize 14px style b end style adalah begin mathsize 14px style 3 over 2 minus 1 half equals 2 over 2 equals 1 end style.

Jadi, jawaban yang tepat adalah E. 

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Hiutnglah nilai integral berikut: b. ∫ 1 3 ​ ( 3 x + 2 ) ( x − 1 ) d x − ∫ 4 3 ​ ( 3 x + 2 ) ( x − 1 ) d x

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