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Hasil dari integral open parentheses 2 x minus 3 close parentheses square root of 2 x squared minus 6 x plus 7 end root straight d x adalah undefined

  1. begin mathsize 14px style 4 over 3 left parenthesis 2 x squared minus 6 x plus 7 right parenthesis square root of 2 x squared minus 6 x plus 7 end root plus c end style

  2. begin mathsize 14px style 2 over 3 left parenthesis 2 x squared minus 6 x plus 7 right parenthesis square root of 2 x squared minus 6 x plus 7 end root plus c end style

  3. begin mathsize 14px style 1 third left parenthesis 2 x squared minus 6 x plus 7 right parenthesis square root of 2 x squared minus 6 x plus 7 end root plus c end style

  4. begin mathsize 14px style negative 1 third left parenthesis 2 x squared minus 6 x plus 7 right parenthesis square root of 2 x squared minus 6 x plus 7 end root plus c end style

  5. begin mathsize 14px style negative 2 over 3 left parenthesis 2 x squared minus 6 x plus 7 right parenthesis square root of 2 x squared minus 6 x plus 7 end root plus c end style

Pembahasan Soal:

Kita akan menyelesaiakan permasalahan integral tersebut menggunakan metode substitusi.

Misalkan begin mathsize 14px style u equals 2 x squared minus 6 x plus 7 end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d u over denominator d x end fraction end cell equals cell 4 x minus 6 end cell row cell d x end cell equals cell fraction numerator d u over denominator 4 x minus 6 end fraction end cell end table end style

Akibatnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 2 x minus 3 close parentheses square root of 2 x squared minus 6 x plus 7 end root d x end cell row blank equals cell integral open parentheses 2 x minus 3 close parentheses square root of u fraction numerator d u over denominator 4 x minus 6 end fraction end cell row blank equals cell integral up diagonal strike open parentheses 2 x minus 3 close parentheses end strike u to the power of 1 half end exponent fraction numerator d u over denominator 2 up diagonal strike open parentheses 2 x minus 3 close parentheses end strike end fraction end cell row blank equals cell 1 half integral u to the power of 1 half end exponent d u end cell row blank equals cell fraction numerator 1 over denominator up diagonal strike 2 end fraction open parentheses fraction numerator up diagonal strike 2 over denominator 3 end fraction u to the power of 3 over 2 end exponent close parentheses plus c end cell row blank equals cell 1 third u square root of u plus c end cell row blank equals cell 1 third open parentheses 2 x squared minus 6 x plus 7 close parentheses square root of 2 x squared minus 6 x plus 7 end root plus c end cell end table end style

Jadi, jawaban yang tepat adalah C.

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Terakhir diupdate 02 Mei 2021

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Pembahasan Soal:

M i s a l space u space equals x squared minus x plus 5 comma space m a k a space d u equals left parenthesis 2 x minus 1 right parenthesis d x. space S e h i n g g a comma  integral subscript space superscript space left parenthesis 2 x minus 1 right parenthesis square root of x squared minus x plus 5 end root space d x equals space integral subscript space superscript space square root of u space d u space  equals fraction numerator 1 over denominator 1 half plus 1 end fraction u to the power of 1 half plus 1 end exponent plus C  equals 2 over 3 u square root of u plus C  equals 2 over 3 left parenthesis x squared minus x plus 5 right parenthesis square root of x squared minus x plus 5 end root plus C

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Hasil dari  adalah ….

Pembahasan Soal:

Dari soal diberikan integral aljabar sebagai berikut.

integral fraction numerator 9 x squared minus 60 x plus 12 over denominator square root of x cubed minus 10 x squared plus 4 x minus 7 end root end fraction blank straight d x

 

Jika x cubed minus 10 x squared plus 4 x minus 7 equals u, maka didapat perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight d u over denominator straight d x end fraction end cell equals cell 3 x squared minus 20 x plus 4 end cell row cell integral open parentheses fraction numerator straight d u over denominator straight d x end fraction close parentheses straight d x end cell equals cell integral open parentheses 3 x squared minus 20 x plus 4 close parentheses straight d x end cell row cell integral straight d u end cell equals cell integral open parentheses 3 x squared minus 20 x plus 4 close parentheses straight d x end cell end table  

Kemudian, perhatikan bahwa

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral fraction numerator 9 x squared minus 60 x plus 12 over denominator square root of x cubed minus 10 x squared plus 4 x minus 7 end root end fraction blank straight d x end cell row blank equals cell integral fraction numerator 3 open parentheses 3 x squared minus 20 x plus 4 close parentheses over denominator square root of x cubed minus 10 x squared plus 4 x minus 7 end root end fraction blank straight d x end cell row blank equals cell 3 integral fraction numerator 1 over denominator square root of u end fraction blank straight d u end cell row blank equals cell 3 integral u to the power of negative 1 half end exponent blank straight d u end cell row blank equals cell 3 times 2 u to the power of 1 half end exponent plus straight C end cell row blank equals cell 6 square root of u plus straight C end cell end table   

Ingat kembali bahwa u equals x cubed minus 10 x squared plus 4 x minus 7 sehingga diperoleh hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 9 x squared minus 60 x plus 12 over denominator square root of x cubed minus 10 x squared plus 4 x minus 7 end root end fraction blank straight d x end cell equals cell 6 square root of u plus C end cell row blank equals cell 6 square root of x cubed minus 10 x squared plus 4 x minus 7 end root plus C end cell end table

Jadi, jawaban yang tepat adalah E.

Roboguru

Pembahasan Soal:

Error converting from MathML to accessible text.

Roboguru

Pembahasan Soal:

Penyelesaian integral tersebut menggunakan metode integral substitusi sebagai berikut.

integral 6 x square root of 6 minus x squared end root space text dx end text

Misal: u equals 6 minus x squared space rightwards arrow space fraction numerator d u over denominator d x end fraction equals negative 2 x space text atau end text space d x equals fraction numerator d u over denominator negative 2 x end fraction

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 6 x square root of 6 minus x squared end root space d x end cell equals cell integral 6 x open parentheses u close parentheses to the power of 1 half end exponent times fraction numerator d u over denominator negative 2 x end fraction end cell row blank equals cell integral negative 3 open parentheses u close parentheses to the power of 1 half end exponent space d u end cell row blank equals cell negative 3 times 2 over 3 open parentheses u close parentheses to the power of 3 over 2 end exponent plus c end cell row blank equals cell negative 2 u square root of u plus c end cell row blank equals cell negative 2 open parentheses 6 minus x squared close parentheses square root of 6 minus x squared end root plus c end cell end table

Dengan demikian, table attributes columnalign right center left columnspacing 0px end attributes row blank blank integral end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 6 minus x squared end root end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank d end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 6 minus x squared close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 6 minus x squared end root end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table 

Roboguru

Pembahasan Soal:

Misalkan u equals x squared minus 1, maka didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight d u end cell equals cell 2 x space straight d x end cell row cell straight d x end cell equals cell fraction numerator straight d u over denominator 2 x end fraction end cell end table 

Substitusikan ke integral x open parentheses x squared minus 1 close parentheses cubed space straight d x dan didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral x open parentheses x squared minus 1 close parentheses cubed space straight d x end cell equals cell integral x times u cubed space fraction numerator straight d u over denominator 2 x end fraction end cell row blank equals cell integral 1 half u cubed space straight d u end cell row blank equals cell 1 half times 1 fourth u to the power of 4 plus c end cell row blank equals cell 1 over 8 u to the power of 4 plus c end cell row blank equals cell 1 over 8 open parentheses x squared minus 1 close parentheses to the power of 4 plus c end cell end table 

Jadi, integral x open parentheses x squared minus 1 close parentheses cubed space straight d x equals 1 over 8 open parentheses x squared minus 1 close parentheses to the power of 4 plus c.

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