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Pertanyaan

integral x open parentheses 3 x minus 1 close parentheses cos open parentheses 2 x cubed minus x squared plus 12 close parentheses d x equals horizontal ellipsis

  1. 1 half x squared open parentheses 3 x minus 1 close parentheses cos open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

  2. x open parentheses 3 x minus 1 close parentheses sin open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

  3. 2 sin open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

  4. 1 half sin open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

  5. 1 half cos open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

Pembahasan Soal:

Gunakan integral subtitusi :

u equals 2 x cubed minus x squared plus 12  fraction numerator d u over denominator d x end fraction equals 6 x squared minus 2 x  d x equals fraction numerator d u over denominator 2 x open parentheses 3 x minus 1 close parentheses end fraction

Jadi,

integral x open parentheses 3 x minus 1 close parentheses cos open parentheses 2 x cubed minus x squared plus 12 close parentheses d x  equals integral x open parentheses 3 x minus 1 close parentheses cos u fraction numerator d u over denominator 2 x open parentheses 3 x minus 1 close parentheses end fraction  equals integral 1 half cos u d u  equals 1 half sin u plus C  equals 1 half sin open parentheses 2 x cubed minus x squared plus 12 close parentheses plus C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Ayu

Mahasiswa/Alumni Universitas Negeri Padang

Terakhir diupdate 16 Desember 2020

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Pertanyaan yang serupa

Nilai ∫136​x​(3+x​23​3​dx adalah...

Pembahasan Soal:

i right parenthesis G u n a k a n space i n t e g r a l space s u b s t i t u s i  integral from 1 to 36 of fraction numerator 3 over denominator square root of x open parentheses 3 plus square root of x close parentheses to the power of 3 over 2 end exponent end fraction d x  S u b s t i t u s i space y equals 3 plus square root of x comma space m a k a  y equals 3 plus x to the power of 1 half end exponent  d y equals 1 half x to the power of negative 1 half end exponent space d x  d y equals fraction numerator 1 over denominator 2 square root of x end fraction space d x  2 space d y equals fraction numerator 1 over denominator square root of x end fraction space d x  6 space d y equals fraction numerator 3 over denominator square root of x end fraction space d x  U n t u k space x equals 1 comma space m a k a space y equals 3 plus square root of 1 equals 4. space  U n t u k space x equals 36 comma space m a k a space y equals 3 plus square root of 36 equals 9. space space    S e h i n g g a  integral from 1 to 36 of fraction numerator 3 over denominator square root of x open parentheses 3 plus square root of x close parentheses to the power of 3 over 2 end exponent end fraction d x equals integral from 1 to 36 of 1 over open parentheses 3 plus square root of x close parentheses to the power of 3 over 2 end exponent bullet fraction numerator 3 over denominator square root of x end fraction d x equals integral from 4 to 9 of 1 over y to the power of 3 over 2 end exponent open parentheses 6 space d y close parentheses equals 6 integral from 4 to 9 of y to the power of negative 3 over 2 end exponent d y  equals open fraction numerator 6 y to the power of negative 3 over 2 plus 1 end exponent over denominator negative 3 over 2 plus 1 end fraction close square brackets subscript 4 superscript 9 equals open fraction numerator 6 y to the power of negative 1 half end exponent over denominator negative 1 half end fraction close square brackets subscript 4 superscript 9 equals negative 12 open y to the power of negative 1 half end exponent close square brackets subscript 4 superscript 9 equals open negative fraction numerator 12 over denominator square root of y end fraction close square brackets subscript 4 superscript 9 equals negative fraction numerator 12 over denominator square root of 9 end fraction minus open parentheses negative fraction numerator 12 over denominator square root of 4 end fraction close parentheses equals negative 12 over 3 plus 12 over 2  equals negative 4 plus 6 equals 2

0

Roboguru

Hasil dari  adalah ….

Pembahasan Soal:

Akan dicari hasil dari

undefined    

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell d over dx open square brackets open parentheses 4 x plus 1 close parentheses to the power of 6 close square brackets end cell equals cell 6 ⋅ open parentheses 4 x plus 1 close parentheses to the power of 5 ⋅ open parentheses 4 x plus 1 close parentheses to the power of apostrophe end cell row blank equals cell 6 ⋅ open parentheses 4 x plus 1 close parentheses to the power of 5 ⋅ 4 end cell row blank equals cell 24 open parentheses 4 x plus 1 close parentheses to the power of 5 end cell end table end style    

Akibatnya, kita peroleh   

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral 24 left parenthesis 4 x plus 1 right parenthesis to the power of 5 space dx end cell equals cell integral fraction numerator straight d over denominator dx end fraction left square bracket left parenthesis 4 straight x plus 1 right parenthesis to the power of 6 right square bracket space dx end cell row cell integral 24 left parenthesis 4 straight x plus 1 right parenthesis to the power of 5 space dx end cell equals cell left parenthesis 4 straight x plus 1 right parenthesis to the power of 6 plus straight C end cell end table end style

Sehingga, perhatikan bahwa

 Error converting from MathML to accessible text. 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Tentukan hasil dari ∫02​(2x+4)xdx=…

Pembahasan Soal:

Perhatikan perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 2 open parentheses 2 x plus 4 close parentheses x space d x end cell equals cell integral subscript 0 superscript 2 open parentheses 2 x squared plus 4 x close parentheses space d x end cell row blank equals cell right enclose open parentheses 2 over 3 x to the power of 2 plus 1 end exponent plus 4 over 2 x to the power of 1 plus 1 end exponent close parentheses end enclose subscript 0 superscript 2 end cell row blank equals cell right enclose open parentheses 2 over 3 x cubed plus 2 x squared close parentheses end enclose subscript 0 superscript 2 end cell row blank equals cell open parentheses 2 over 3 open parentheses 2 cubed close parentheses plus 2 open parentheses 2 squared close parentheses close parentheses minus 0 end cell row blank equals cell 16 over 3 plus 8 end cell row blank equals cell 40 over 3 end cell end table

Dengan demikian, hasil dari integral subscript 0 superscript 2 open parentheses 2 x plus 4 close parentheses x space d x equals 40 over 3.

0

Roboguru

Hasil dari  adalah ….

Pembahasan Soal:

Dari soal diberikan

 

Error converting from MathML to accessible text.  

 

Jika dimisalkan begin mathsize 14px style x cubed plus 6 x squared plus 9 equals u end style, maka didapat

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell du over dx end cell equals cell 3 x squared plus 12 x end cell row cell integral open parentheses du over dx close parentheses dx end cell equals cell integral open parentheses 3 x squared plus 12 x close parentheses dx end cell row cell integral du end cell equals cell integral open parentheses 3 x squared plus 12 x close parentheses dx end cell end table end style  

 

Kemudian, perhatikan bahwa

 

Error converting from MathML to accessible text. 

 

Ingat kembali bahwa begin mathsize 14px style u equals x cubed plus 6 x squared plus 9 end style. Maka diperoleh

 

Error converting from MathML to accessible text. 

 

Jadi, jawaban yang tepat adalah C.

0

Roboguru

∫6x2x3−5​dx=....

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 

0

Roboguru

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