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∫ 0 1 ​ x ( 2 x − 4 ) 4 dx = ....

   

  1. begin mathsize 14px style 152 over 5 end style   

  2. begin mathsize 14px style 184 over 5 end style  

  3. begin mathsize 14px style negative 216 over 5 end style  

  4. begin mathsize 14px style negative 184 over 5 end style   

  5. begin mathsize 14px style negative 152 over 5 end style   

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

jawaban yang tepat adalah A.

Pembahasan

Kita cari hasil integral di atas dengan menggunakan integral parsial. Misalkan u = x dan , maka dan Sehingga Jadi, jawaban yang tepat adalah A.

Kita cari hasil integral di atas dengan menggunakan integral parsial.
Misalkan u = x dan begin mathsize 14px style dv over dx equals left parenthesis 2 straight x minus 4 right parenthesis to the power of 4 end style, maka

begin mathsize 14px style du over dx equals 1 end style  

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight v equals cell integral dv over dx dx end cell row blank equals cell integral left parenthesis 2 straight x minus 4 right parenthesis to the power of 4 space dx semicolon space gunakan space metode space substitusi end cell row blank equals cell 1 over 10 left parenthesis 2 straight x minus 4 right parenthesis to the power of 5 plus straight C end cell end table end style    

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 1 straight x left parenthesis 2 straight x minus 4 right parenthesis to the power of 4 space dx end cell equals cell straight x times 1 over 10 left parenthesis 2 straight x minus 4 right parenthesis to the power of 5 vertical line table row 1 row 0 end table minus integral subscript 0 superscript 1 1 over 10 left parenthesis 2 straight x minus 4 right parenthesis to the power of 5 space dx end cell row blank equals cell open square brackets 1 over 10 left parenthesis 2 straight x minus 4 right parenthesis to the power of 5 close square brackets vertical line table row 1 row 0 end table minus 1 over 120 left parenthesis 2 straight x minus 4 right parenthesis to the power of 6 vertical line table row 1 row 0 end table end cell row blank equals cell open square brackets 1 over 10 times 1 left parenthesis 2 times 1 minus 4 right parenthesis to the power of 5 minus fraction numerator 1 over denominator 10 end fraction times 0 left parenthesis 2 times 0 minus 4 right parenthesis to the power of 5 end exponent close square brackets minus open square brackets 1 over 120 left parenthesis 2 times 1 minus 4 right parenthesis to the power of 6 minus fraction numerator 1 over denominator 120 end fraction left parenthesis 2 times 0 minus 4 right parenthesis to the power of 6 close square brackets end cell row blank equals cell open square brackets 1 over 10 left parenthesis negative 2 right parenthesis to the power of 5 minus 0 close square brackets minus open square brackets 1 over 120 left parenthesis negative 2 right parenthesis to the power of 6 minus 1 over 120 left parenthesis negative 4 right parenthesis to the power of 6 close square brackets end cell row blank equals cell negative 32 over 10 minus 64 over 120 plus fraction numerator 4.096 over denominator 120 end fraction end cell row blank equals cell fraction numerator negative 384 minus 64 plus 4.096 over denominator 120 end fraction end cell row blank equals cell fraction numerator 3.648 over denominator 120 end fraction end cell row blank equals cell 152 over 5 end cell end table end style    

Jadi, jawaban yang tepat adalah A.

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