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Pertanyaan

Hasil integral open parentheses x plus 5 close parentheses square root of 4 minus 3 x end root d x equals... 

Pembahasan Soal:

Misalnya 

u equals 4 minus 3 x

Maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell d u end cell equals cell d open parentheses 4 minus 3 x close parentheses end cell row cell d u end cell equals cell negative 3 d x end cell row cell d x end cell equals cell negative 1 third d u end cell row blank blank blank row u equals cell 4 minus 3 x end cell row cell 3 x end cell equals cell 4 minus u end cell row x equals cell 4 over 3 minus u over 3 end cell end table 

Sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses x plus 5 close parentheses square root of 4 minus 3 x end root d x end cell equals cell integral open parentheses open parentheses 4 over 3 minus u over 3 close parentheses plus 5 close parentheses square root of u open parentheses negative 1 third d u close parentheses end cell row blank equals cell negative 1 third integral open parentheses 19 over 3 minus u over 3 close parentheses square root of u d u end cell row blank equals cell negative 1 over 9 integral open parentheses 19 minus u close parentheses square root of u d u end cell row blank equals cell negative 1 over 9 integral open parentheses 19 square root of u minus u square root of u close parentheses d u end cell row blank equals cell negative 1 over 9 integral open parentheses 19 u to the power of 1 half end exponent minus u to the power of 3 over 2 end exponent close parentheses d u end cell row blank equals cell negative 1 over 9 open parentheses 19 times 2 over 3 u to the power of 3 over 2 end exponent minus 2 over 5 u to the power of 5 over 2 end exponent close parentheses plus C end cell row blank equals cell negative 38 over 27 u to the power of 3 over 2 end exponent plus 2 over 45 u to the power of 5 over 2 end exponent plus C end cell row blank equals cell negative 38 over 27 open parentheses 4 minus 3 x close parentheses to the power of 3 over 2 end exponent plus 2 over 45 open parentheses 4 minus 3 x close parentheses to the power of 5 over 2 end exponent plus C end cell end table 

Jadi, integral open parentheses x plus 5 close parentheses square root of 4 minus 3 x end root d x equals negative 38 over 27 open parentheses 4 minus 3 x close parentheses to the power of 3 over 2 end exponent plus 2 over 45 open parentheses 4 minus 3 x close parentheses to the power of 5 over 2 end exponent equals C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Hasil dari = ...

Pembahasan Soal:

Ingat konsep turunan dan integral substitusi:

table row cell u equals a x to the power of n end cell rightwards arrow cell fraction numerator straight d u over denominator straight d x end fraction equals a times n x to the power of n minus 1 end exponent end cell row cell integral a x to the power of n end cell rightwards arrow cell fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end cell end table 

Misalkan u equals x squared minus 4 x plus 1 maka:

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 2 x minus 4 end cell row dx equals cell fraction numerator 1 over denominator 2 x minus 4 end fraction straight d u end cell end table 

kemudian substitusikan:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses x minus 2 close parentheses open parentheses x squared minus 4 x plus 1 close parentheses cubed space straight d x end cell equals cell integral open parentheses x minus 2 close parentheses u cubed space fraction numerator 1 over denominator 2 x minus 4 end fraction space straight d u end cell row blank equals cell integral fraction numerator x minus 2 over denominator 2 open parentheses x minus 2 close parentheses end fraction u cubed space straight d u end cell row blank equals cell integral 1 half u cubed space straight d u end cell row blank equals cell 1 half integral u squared space straight d u end cell row blank equals cell 1 half open square brackets 1 third u to the power of 4 plus C close square brackets end cell row blank equals cell 1 over 6 open parentheses x squared minus 4 x plus 1 close parentheses to the power of 4 plus C end cell end table 

Dengan demikian hasil dari integral open parentheses x minus 2 close parentheses open parentheses x squared minus 4 x plus 1 close parentheses cubed space straight d x adalah Error converting from MathML to accessible text..

0

Roboguru

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 8 square root of open parentheses x squared minus 2 x to the power of 4 close parentheses end root d x end cell equals cell integral 8 square root of x squared open parentheses 1 minus 2 x squared close parentheses end root d x end cell row blank equals cell integral 8 x square root of open parentheses 1 minus 2 x squared close parentheses end root d x end cell row blank equals cell integral 8 x times open parentheses 1 minus 2 x squared close parentheses to the power of 1 half end exponent d x end cell end table

Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell open parentheses 1 minus 2 x squared close parentheses end cell row cell d u end cell equals cell negative 4 x space d x end cell row cell d x end cell equals cell fraction numerator d u over denominator negative 4 x end fraction end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 8 x times open parentheses 1 minus 2 x squared close parentheses to the power of 1 half end exponent d x end cell equals cell integral 8 x times u blank to the power of 1 half end exponent fraction numerator d u over denominator negative 4 x end fraction end cell row blank equals cell integral negative 2 times u to the power of 1 half end exponent d u end cell row blank equals cell negative 4 over 3 u to the power of 3 over 2 end exponent plus C end cell row blank equals cell negative 4 over 3 open parentheses 1 minus 2 x squared close parentheses to the power of 3 over 2 end exponent plus C end cell end table

Jadi hasil dari integral 8 square root of open parentheses x squared minus 2 x to the power of 4 close parentheses end root d x adalah Error converting from MathML to accessible text.

 

0

Roboguru

Selesaikan

Pembahasan Soal:

Integral tersebut dapat diselesaikan dengan metode substitusi seperti di bawah ini 

Misalnya 

u equals 3 x minus 5 space rightwards double arrow straight d u equals 3 straight d x space space space space space space space space space space space space space space space space space rightwards double arrow straight d x equals fraction numerator straight d u over denominator 3 end fraction  

Maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses 3 x minus 5 close parentheses to the power of 12 d x end cell equals cell integral u to the power of 12 fraction numerator straight d u over denominator 3 end fraction end cell row blank equals cell 1 third integral u to the power of 12 straight d u end cell row blank equals cell 1 third times fraction numerator 1 over denominator 12 plus 1 end fraction u to the power of 12 plus 1 end exponent plus C end cell row blank equals cell 1 third times 1 over 13 times u to the power of 13 plus C end cell row blank equals cell 1 over 39 u to the power of 13 plus C end cell row blank equals cell 1 over 39 open parentheses 3 x minus 5 close parentheses to the power of 13 plus C end cell end table 

Jadi integral open parentheses 3 x minus 5 close parentheses to the power of 12 d x equals 1 over 39 open parentheses 3 x minus 5 close parentheses to the power of 13 plus C

0

Roboguru

Pembahasan Soal:

Misalkan

 u=xadxdu=1du=dx , batas nilai u=xax=0u=ax=au=0

sehingga

0af(xa)dx==a0f(u)dua0f(x)dx

Jadi, pilihan jawaban yang tepat adalah C.

 

0

Roboguru

Pembahasan Soal:

Misalkan:

  begin mathsize 14px style u equals x plus 4 rightwards arrow d u equals d x x equals u minus 4 end style 

Sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral x open parentheses x plus 4 close parentheses to the power of 5 dx end cell equals cell integral open parentheses straight u minus 4 close parentheses straight u to the power of 5 space du end cell row blank equals cell integral straight u to the power of 6 minus 4 straight u to the power of 5 space du end cell row blank equals cell 1 over 7 straight u to the power of 7 minus 4 over 6 straight u to the power of 6 plus C end cell row blank equals cell 1 over 7 open parentheses straight x plus 4 close parentheses to the power of 7 minus 2 over 3 open parentheses straight x plus 4 close parentheses to the power of 6 plus C end cell row blank equals cell 1 over 21 open parentheses x plus 4 close parentheses to the power of 6 open square brackets 3 open parentheses x plus 4 close parentheses minus 14 close square brackets plus C end cell row blank equals cell 1 over 21 open parentheses x plus 4 close parentheses to the power of 6 open parentheses 3 x plus 12 minus 14 close parentheses plus C end cell row blank equals cell 1 over 21 open parentheses 3 x minus 2 close parentheses open parentheses x plus 4 close parentheses to the power of 6 plus C end cell row blank blank blank end table end style

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Roboguru

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