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Hasil dari  adalah ....

Pertanyaan

Hasil dari integral open parentheses 3 x plus 8 close parentheses cubed d x adalah ....  

  1. 1 over 12 left parenthesis 3 x plus 8 right parenthesis to the power of 4 plus c 

  2. 1 over 12 left parenthesis x plus 8 right parenthesis to the power of 4 plus c 

  3. negative 1 over 12 left parenthesis x plus 8 right parenthesis to the power of 4 plus c 

  4. 1 over 12 left parenthesis 3 x minus 8 right parenthesis to the power of 4 plus c 

  5. 1 over 12 left parenthesis negative 3 x minus 8 right parenthesis to the power of 4 plus c 

Pembahasan Soal:

Dimisalkan u equals 3 x plus 8 maka

d u equals 3 space d x d x equals 1 third space d u 

Bentuk integral di atas dapat diselesaikan dengan substitusi :

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 3 x plus 8 close parentheses cubed d x end cell equals cell 1 third integral u cubed space d u end cell row blank equals cell 1 third open square brackets 1 fourth times u to the power of 4 plus C close square brackets end cell row blank equals cell 1 over 12 left parenthesis 3 x plus 8 right parenthesis to the power of 4 plus C end cell end table 

Dengan demikian, hasil dari integral open parentheses 3 x plus 8 close parentheses cubed d x adalah 1 over 12 left parenthesis 3 x plus 8 right parenthesis to the power of 4 plus c.

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 07 Juni 2021

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Pertanyaan yang serupa

Pembahasan Soal:

Menggunakan metode subtitusi, akan dicari nilai integral di atas. Misalkan u equals x cubed minus 5. Maka didapat

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell x cubed minus 5 end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell fraction numerator d open parentheses x cubed minus 5 close parentheses over denominator d x end fraction end cell row cell fraction numerator d u over denominator d x end fraction end cell equals cell 3 x squared end cell row cell d u end cell equals cell 3 x squared d x end cell row cell 2 cross times d u end cell equals cell 2 cross times 3 x squared d x end cell row cell 2 d u end cell equals cell 6 x squared d x end cell end table

Dengan demikian diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 6 x squared square root of x cubed minus 5 end root d x end cell equals cell integral open parentheses square root of x cubed minus 5 end root close parentheses 6 x squared d x end cell row blank equals cell integral square root of u 2 d u end cell row blank equals cell 2 integral square root of u d u end cell row blank equals cell 2 integral u to the power of 1 half end exponent d u end cell row blank equals cell 2 cross times fraction numerator 1 over denominator begin display style 1 half end style plus 1 end fraction u to the power of 1 half plus 1 end exponent plus C end cell row blank equals cell fraction numerator 2 over denominator begin display style 3 over 2 end style end fraction u to the power of 3 over 2 end exponent plus C end cell row blank equals cell 4 over 3 u square root of u plus C end cell row blank equals cell 4 over 3 open parentheses x cubed minus 5 close parentheses square root of x cubed minus 5 end root plus C end cell end table

Jadi, integral 6 x squared square root of x cubed minus 5 end root d x equals 4 over 3 open parentheses x cubed minus 5 close parentheses square root of x cubed minus 5 end root plus C 

Roboguru

Integral dari  adalah ...

Pembahasan Soal:

Ingat aturan integral berikut.

integral a x to the power of n space text d end text x equals fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus c

Pada soal tersebut, penyelesaian integral menggunakan metode integral substitusi.

integral 5 x squared open parentheses x cubed minus 2 close parentheses to the power of 7 text  d end text x

Misal: u equals x cubed minus 2 sehingga diperoleh hubungan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator text d end text u over denominator text d end text x end fraction end cell equals cell 3 x squared end cell row cell integral open parentheses fraction numerator text d end text u over denominator text d end text x end fraction close parentheses text d end text x end cell equals cell integral 3 x squared space text d end text x end cell row cell integral text d end text u end cell equals cell integral 3 x squared space text d end text x end cell row cell integral fraction numerator 1 over denominator 3 x squared end fraction text d end text u end cell equals cell integral text d end text x end cell end table

Selanjutnya, perhatikan perhitungan berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral 5 x squared open parentheses x cubed minus 2 close parentheses to the power of 7 space text d end text x end cell row blank equals cell integral 5 x squared times u to the power of 7 times fraction numerator 1 over denominator 3 x squared end fraction text d end text u end cell row blank equals cell integral 5 over 3 u to the power of 7 space text d end text u end cell row blank equals cell 5 over 3 times 1 over 8 u to the power of 8 plus C end cell row blank equals cell 5 over 24 u to the power of 8 plus C end cell end table

Ingat bahwa u equals x cubed minus 2 sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 5 x squared open parentheses x cubed minus 2 close parentheses to the power of 7 space text d end text x end cell equals cell 5 over 24 u to the power of 8 plus C end cell row blank equals cell 5 over 24 open parentheses x cubed minus 2 close parentheses to the power of 8 plus C end cell end table

Dengan demikian, Error converting from MathML to accessible text.

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Hasil dari

Pembahasan Soal:

integral cube root of open parentheses 2 x minus 1 close parentheses squared end root d x equals integral left parenthesis 2 x minus 1 right parenthesis to the power of 2 over 3 end exponent d x  equals 3 over 5 open parentheses 2 x minus 1 close parentheses to the power of 5 over 3 end exponent 1 half plus c  equals 3 over 10 left parenthesis 2 x minus 1 right parenthesis to the power of 5 over 3 end exponent plus c

equals 3 over 10 left parenthesis x minus 1 right parenthesis cube root of left parenthesis 2 x minus 1 right parenthesis squared end root plus c

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

Diberikan bentuk integral tak tentu integral left parenthesis 3 x plus 4 right parenthesis to the power of 7 d x. Dimisalkan

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell 3 x plus 4 end cell row cell fraction numerator d u over denominator d x end fraction end cell equals 3 row cell d u end cell equals cell 3 space d x end cell row cell d x end cell equals cell 1 third space d u end cell end table

Substitusikan permisalan tersebut ke dalam bentuk integral.

table attributes columnalign right center left columnspacing 0px end attributes row cell integral left parenthesis 3 x plus 4 right parenthesis to the power of 7 d x end cell equals cell 1 third integral u to the power of 7 space d u end cell row blank equals cell 1 third open square brackets 1 over 8 u to the power of 8 plus C close square brackets end cell row blank equals cell 1 over 24 left parenthesis 3 x plus 4 right parenthesis to the power of 8 plus C end cell end table 

Dengan demikian, hasil dari integral left parenthesis 3 x plus 4 right parenthesis to the power of 7 d x adalah 1 over 24 left parenthesis 3 x plus 4 right parenthesis to the power of 8 plus C.

Jadi, jawaban yang tepat adalah E.

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

Dapat ditentukan integral dari fungsi yang diminta dengan integral substitusi sebagai berikut.

Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell 3 minus x end cell row cell d u end cell equals cell negative space d x end cell end table 

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 3 minus x close parentheses to the power of 6 space d x end cell equals cell integral u to the power of 6 space open parentheses negative d u close parentheses end cell row blank equals cell negative integral u to the power of 6 space d u end cell row blank equals cell negative open parentheses fraction numerator 1 over denominator 6 plus 1 end fraction u to the power of 6 plus 1 end exponent close parentheses plus C end cell row blank equals cell negative 1 over 7 u to the power of 7 plus C end cell row blank equals cell negative 1 over 7 open parentheses 3 minus x close parentheses to the power of 7 plus C end cell end table 

Jadi, jawaban yang benar adalah D.

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