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Nilai dari ∫01​x(x2−1)5dx adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style integral subscript 0 superscript 1 x left parenthesis x squared minus 1 right parenthesis to the power of 5 d x end style adalah ....

  1. undefined     undefined 

  2. undefined     undefined 

  3. begin mathsize 14px style negative 1 half end style     undefined 

  4. begin mathsize 14px style negative 1 over 6 end style     undefined 

  5. begin mathsize 14px style negative 1 over 12 end style     undefined 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 1 x left parenthesis x squared minus 1 right parenthesis to the power of 5 d x end cell equals cell open square brackets 1 half open parentheses x squared minus 1 close parentheses to the power of 6 over 6 close square brackets subscript 0 superscript 1 end cell row blank equals cell open square brackets blank over 12 close square brackets subscript 0 superscript 1 end cell row blank equals cell 1 over 12 open square brackets open parentheses 1 squared minus 1 close parentheses to the power of 6 minus open parentheses 0 squared minus 1 close parentheses to the power of 6 close square brackets end cell row blank equals cell 1 over 12 left parenthesis 0 minus 1 right parenthesis end cell row blank equals cell negative 1 over 12 end cell end table end style 

Jadi, jawaban yang tepat adalah E

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Hasil dari ∫01​12x3x2+1​dx adalah ....

Pembahasan Soal:

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Nilai dari ∫02​(2x−1)(x2−x−2)4dx adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 2 left parenthesis 2 x minus 1 right parenthesis left parenthesis x squared minus x minus 2 right parenthesis to the power of 4 d x end cell equals cell open square brackets left parenthesis x squared minus x minus 2 right parenthesis to the power of 5 over 5 close square brackets subscript 0 superscript 2 end cell row blank equals cell 1 fifth open square brackets left parenthesis 2 squared minus 2 minus 2 right parenthesis to the power of 5 minus left parenthesis 0 squared minus 0 minus 2 right parenthesis to the power of 5 close square brackets end cell row blank equals cell 1 fifth left parenthesis 0 minus 32 right parenthesis end cell row blank equals cell negative 32 over 5 end cell end table end style 

Jadi, jawaban yang tepat adalah E

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Hasil pengintegralan fungsi aljabar ∫−10​(4x−2)(1+x−x2)4dx adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style 1 plus x minus x squared equals p end style maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style left parenthesis 1 minus 2 x right parenthesis d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 1 minus 2 x end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis p to the power of 4 times fraction numerator 1 over denominator 1 minus 2 x end fraction d p end cell row blank equals cell negative 2 integral subscript blank p to the power of 4 d p end cell row blank equals cell negative fraction numerator 2 over denominator 4 plus 1 end fraction p to the power of 4 plus 1 end exponent plus C end cell row blank equals cell negative 2 over 5 p to the power of 5 plus C end cell end table end style           

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell negative fraction numerator 2 over denominator 5 end fraction p to the power of 5 end exponent plus C end cell row blank equals cell negative 2 over 5 left parenthesis 1 plus x minus x squared right parenthesis to the power of 5 plus 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 negative 1 end subscript superscript 0 left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell negative 2 over 5 left parenthesis 1 plus x minus x squared right parenthesis to the power of 5 left enclose blank with negative 1 below and 0 on top end enclose space end cell row blank equals cell open square brackets negative 2 over 5 left parenthesis 1 plus 0 minus 0 squared right parenthesis to the power of 5 close square brackets minus open square brackets negative 2 over 5 left parenthesis 1 plus left parenthesis negative 1 right parenthesis minus left parenthesis negative 1 right parenthesis squared right parenthesis to the power of 5 close square brackets space end cell row blank equals cell negative 2 over 5 minus 2 over 5 end cell row blank equals cell negative 4 over 5 end cell end table end style    

   .

0

Roboguru

Hasil pengintegralan fungsi aljabar ∫01​x(2−x2)5dx adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style 2 minus x squared equals p end style maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style negative 2 x d x equals d p end style sehingga begin mathsize 14px style d x equals negative fraction numerator 1 over denominator 2 x end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank x left parenthesis 2 minus x squared right parenthesis to the power of 5 d x end cell equals cell integral subscript blank x p to the power of 5 times open parentheses negative fraction numerator 1 over denominator 2 x end fraction d p close parentheses end cell row blank equals cell negative 1 half integral subscript blank p to the power of 5 d p end cell row blank equals cell negative 1 half times fraction numerator 1 over denominator 5 plus 1 end fraction p to the power of 5 plus 1 end exponent plus C end cell row blank equals cell negative 1 over 12 p to the power of 6 plus C end cell end table end style          

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank x left parenthesis 2 minus x squared right parenthesis to the power of 5 d x end cell equals cell negative fraction numerator 1 over denominator 12 end fraction p to the power of 6 plus C end cell row blank equals cell negative 1 over 12 left parenthesis 2 minus x squared right parenthesis to the power of 6 plus 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 x left parenthesis 2 minus x squared right parenthesis to the power of 5 d x end cell equals cell negative 1 over 12 left parenthesis 2 minus x squared right parenthesis to the power of 6 left enclose blank with 0 below and 1 on top end enclose space end cell row blank equals cell open square brackets negative 1 over 12 left parenthesis 2 minus 1 squared right parenthesis to the power of 6 close square brackets minus open square brackets negative 1 over 12 left parenthesis 2 minus 0 squared right parenthesis to the power of 6 close square brackets end cell row blank equals cell negative 1 over 12 minus open parentheses negative 64 over 12 close parentheses end cell row blank equals cell negative 63 over 12 end cell row blank equals cell negative 21 over 4 end cell end table end style  

   .

0

Roboguru

Hasil dari ∫01​12x3x2+1​dx adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style 3 x squared plus 1 equals p end style maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style 6 x d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 6 x end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank 12 x square root of left parenthesis 3 x squared plus 1 right parenthesis end root d x end cell equals cell integral subscript blank 12 x square root of p times fraction numerator 1 over denominator 6 x end fraction d p end cell row blank equals cell 2 integral subscript blank square root of p d p end cell row blank equals cell fraction numerator 2 over denominator begin display style 1 half end style plus 1 end fraction p to the power of 1 half plus 1 end exponent plus C end cell row blank equals cell 4 over 3 p square root of p plus C end cell end table end style                  

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank 12 x square root of 3 x squared plus 1 end root d x end cell equals cell 4 over 3 p square root of p plus C end cell row blank equals cell 4 over 3 left parenthesis 3 x squared plus 1 right parenthesis square root of 3 x squared plus 1 end root plus 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 12 x square root of 3 x squared plus 1 end root d x end cell equals cell 4 over 3 left parenthesis 3 x squared plus 1 right parenthesis square root of 3 x squared plus 1 end root left enclose blank with 0 below and 1 on top end enclose end cell row blank equals cell open parentheses 4 over 3 left parenthesis 3.1 squared plus 1 right parenthesis square root of 3.1 squared plus 1 end root close parentheses minus open parentheses 4 over 3 left parenthesis 3.0 squared plus 1 right parenthesis square root of 3.0 squared plus 1 end root close parentheses end cell row blank equals cell 32 over 3 minus 4 over 3 end cell row blank equals cell 28 over 3 end cell end table end style 


Jadi, jawaban yang benar adalah E.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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