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Pertanyaan

integral open parentheses 1 over open parentheses 5 x minus 2 close parentheses cubed close parentheses space straight d x equals... space 

Pembahasan Soal:

Ingat!

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

Misalkan, u equals 5 x minus 2, maka turunan pertama fungsi u:

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 1 times 5 x to the power of 1 minus 1 end exponent minus 0 end cell row cell fraction numerator straight d u over denominator straight d x end fraction end cell equals 5 end table

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight 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 cell fraction numerator straight d u over denominator 5 end fraction end cell end table.

Maka, integral dari fungsi di atas:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank open parentheses 1 over open parentheses 5 x minus 2 close parentheses cubed close parentheses space d x end cell equals cell integral subscript blank open parentheses 1 over u cubed close parentheses space fraction numerator d u over denominator 5 end fraction end cell row blank equals cell integral subscript blank u to the power of negative 3 end exponent space fraction numerator d u over denominator 5 end fraction end cell row blank equals cell 1 fifth times fraction numerator 1 over denominator negative 3 plus 1 end fraction u to the power of negative 3 plus 1 end exponent plus C end cell row blank equals cell negative 1 over 10 u to the power of negative 2 end exponent plus C end cell row blank equals cell negative fraction numerator 1 over denominator 10 u squared end fraction plus C end cell row blank equals cell negative fraction numerator 1 over denominator 10 open parentheses 5 x minus 2 close parentheses squared end fraction plus C end cell end table 

Jadi, Error converting from MathML to accessible text..

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 01 Mei 2021

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Pertanyaan yang serupa

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.

 

Roboguru

Jika nilai , hasil dari adalah....

Pembahasan Soal:

Misalkan begin mathsize 14px style u equals 5 minus x space maka space fraction numerator d u over denominator d x end fraction equals negative 1 end style

Perhatikan 

begin mathsize 14px style u equals 5 minus x open curly brackets table attributes columnalign left end attributes row cell x equals 2 comma space u equals 3 end cell row cell x equals 3 comma space u equals 2 end cell end table close end style

sehingga 

23f(5x)dx===32f(u)(du)(23f(u)du)23f(u)du.() 

Persamaan (*) ekuivalen dengan begin mathsize 14px style integral subscript 2 superscript 3 f left parenthesis x right parenthesis d x equals 12 end style.

Dengan demikian, hasil dari begin mathsize 14px style integral subscript 2 superscript 3 f left parenthesis 5 minus x right parenthesis d x end style adalah begin mathsize 14px style 12 end style.

Jadi, jawaban yang tepat adalah A.

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

Roboguru

Pembahasan Soal:

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

Roboguru

Hasil dari  adalah

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.

Roboguru

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