Roboguru

Pertanyaan

begin mathsize 14px style integral subscript 0 superscript straight a straight f open parentheses x minus straight a close parentheses space dx equals... end style

  1. begin mathsize 14px style integral subscript 0 superscript straight a f left parenthesis x right parenthesis space dx end style 

  2. begin mathsize 14px style negative integral subscript 0 superscript straight a f left parenthesis x right parenthesis space dx end style

  3. begin mathsize 14px style integral subscript negative straight a end subscript superscript 0 f left parenthesis x right parenthesis space dx end style

  4. begin mathsize 14px style negative integral subscript negative straight a end subscript superscript 0 f left parenthesis x right parenthesis space dx end style 

  5. begin mathsize 14px style 2 integral subscript negative straight a end subscript superscript 0 f left parenthesis x right parenthesis space dx end style

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.

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Mariyam

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 12 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

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..

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

Hitunglah hasil dari integral tentu berikut ini:

Pembahasan Soal:

Ingatlah bahwa

integral subscript a superscript b f open parentheses x close parentheses d x equals straight F open parentheses b close parentheses minus straight F open parentheses a close parentheses

dengan straight F open parentheses x close parentheses adalah anti turunan dari f open parentheses x close parentheses.

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 2 superscript 4 open parentheses negative x squared plus 6 x minus 8 close parentheses d x end cell equals cell right enclose negative 1 third x cubed plus 3 x squared minus 8 x end enclose subscript 2 superscript 4 end cell row blank equals cell open parentheses negative 1 third open parentheses 4 close parentheses cubed plus 3 open parentheses 4 close parentheses squared minus 8 open parentheses 4 close parentheses close parentheses end cell row blank blank cell negative open parentheses negative 1 third open parentheses 2 close parentheses cubed plus 3 open parentheses 2 close parentheses squared minus 8 open parentheses 2 close parentheses close parentheses end cell row blank equals cell open parentheses negative 1 third open parentheses 64 close parentheses plus 3 open parentheses 16 close parentheses minus 32 close parentheses end cell row blank blank cell negative open parentheses negative 1 third open parentheses 8 close parentheses plus 3 open parentheses 4 close parentheses minus 16 close parentheses end cell row blank equals cell open parentheses negative 64 over 3 plus 48 minus 32 close parentheses minus open parentheses negative 8 over 3 plus 12 minus 16 close parentheses end cell row blank equals cell negative 64 over 3 plus 16 plus 8 over 3 plus 4 end cell row blank equals cell negative 56 over 3 plus 20 end cell row blank equals cell negative 56 over 3 plus 60 over 3 end cell row blank equals cell 4 over 3 end cell end table

Jadi, nilai dari integral subscript 2 superscript 4 open parentheses negative x squared plus 6 x minus 8 close parentheses d x equals 4 over 3.

Roboguru

Jika , maka nilai  adalah ....

Pembahasan Soal:

integral from 1 to 2 of 1 over straight x squared straight f open parentheses 1 plus 2 over straight x close parentheses dx  Substitusi space straight y equals 1 plus 2 over straight x space maka  straight y equals 1 plus 2 straight x to the power of negative 1 end exponent  dy equals negative 2 straight x to the power of negative 2 end exponent straight space dx  dy equals negative 2 over straight x squared dx  box enclose negative 1 half straight space dy equals 1 over straight x squared straight space dx end enclose  Untuk space straight x equals 1 comma space maka space straight y equals 1 plus 2 over 1 equals 1 plus 2 equals box enclose 3  Untuk space straight x equals 2 comma space maka space straight y equals 1 plus 2 over 2 equals 1 plus 1 equals box enclose 2  Sehingga  integral from 1 to 2 of 1 over straight x squared straight f open parentheses 1 plus 2 over straight x close parentheses dx equals integral from 1 to 2 of straight f open parentheses 1 plus 2 over straight x close parentheses.1 over straight x squared dx equals integral from 3 to 2 of straight f open parentheses straight y close parentheses open parentheses negative 1 half straight space dy close parentheses equals negative 1 half integral from 3 to 2 of straight f open parentheses straight y close parentheses dy  equals 1 half integral from 2 to 3 of straight f open parentheses straight y close parentheses dy equals 1 half integral from 2 to 3 of straight f open parentheses straight x close parentheses dx equals box enclose fraction numerator square root of 2 over denominator 2 end fraction end enclose

Roboguru

Nilai adalah ....

Pembahasan Soal:

undefined

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved