Iklan

Iklan

Pertanyaan

Hasil dari ∫ x sin 2 x d x adalah ...

Hasil dari   adalah ...

  1. begin mathsize 14px style 1 over 8 open parentheses 2 x squared minus 4 x sin invisible function application open parentheses 2 x close parentheses minus cos invisible function application open parentheses 2 x close parentheses close parentheses plus C end style

  2. begin mathsize 14px style 1 over 8 open parentheses 2 x squared minus 2 x sin invisible function application open parentheses 2 x close parentheses minus cos invisible function application open parentheses 2 x close parentheses close parentheses plus C end style

  3. begin mathsize 14px style 1 over 8 open parentheses 2 x squared minus 4 x cos invisible function application open parentheses 2 x close parentheses minus sin invisible function application open parentheses 2 x close parentheses close parentheses plus C end style

  4. begin mathsize 14px style 1 over 8 open parentheses 2 x squared minus 2 x cos invisible function application open parentheses 2 x close parentheses minus sin invisible function application open parentheses 2 x close parentheses close parentheses plus C end style

  5. begin mathsize 14px style 1 over 8 open parentheses 2 x squared minus 4 x sin invisible function application open parentheses 2 x close parentheses plus cos invisible function application open parentheses 2 x close parentheses close parentheses plus C end style 

Iklan

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

jawaban yang tepat adalah B.

Iklan

Pembahasan

Perhatikan perhitungan berikut, Perhatikan integral berikut. ∫x cos⁡2x dx Akan digunakan metode integral parsial untuk menyelesaikan integral tersebut. Misalkan u = x dan . Didapat dan Kemudian, perhatikan perhitungan berikut. Jadi, jawaban yang tepat adalah B.

Perhatikan perhitungan berikut,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral x blank sin squared x blank d x end cell equals cell integral x open parentheses 1 half minus 1 half cos invisible function application 2 x close parentheses blank d x end cell row blank equals cell 1 half integral open parentheses x minus x cos invisible function application 2 x close parentheses d x end cell row blank equals cell 1 half open parentheses integral x blank d x minus integral x cos invisible function application 2 x d x close parentheses end cell end table end style

Perhatikan integral berikut.

∫x cos⁡2x dx

Akan digunakan metode integral parsial untuk menyelesaikan integral tersebut. Misalkan u = x dan begin mathsize 14px style fraction numerator d v over denominator d x end fraction equals cos invisible function application 2 x end style . Didapat

undefined 

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell integral fraction numerator d v over denominator d x end fraction d x end cell row blank equals cell integral cos invisible function application 2 x d x end cell row blank equals cell fraction numerator sin invisible function application 2 x over denominator 2 end fraction plus C end cell end table end style

Kemudian, perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral x blank sin squared x blank d x end cell equals cell 1 half open parentheses integral x blank d x minus integral x cos invisible function application 2 x blank d x close parentheses end cell row blank equals cell 1 half open parentheses integral x blank d x minus open parentheses x times fraction numerator sin invisible function application 2 x over denominator 2 end fraction minus integral fraction numerator sin invisible function application 2 x over denominator 2 end fraction d x close parentheses close parentheses end cell row blank equals cell 1 half open parentheses x squared over 2 minus open parentheses fraction numerator x sin invisible function application 2 x over denominator 2 end fraction minus 1 half times open parentheses negative fraction numerator cos invisible function application 2 x over denominator 2 end fraction close parentheses close parentheses close parentheses end cell row blank equals cell 1 half open parentheses x squared over 2 minus open parentheses fraction numerator x sin invisible function application 2 x over denominator 2 end fraction plus fraction numerator cos invisible function application 2 x over denominator 4 end fraction close parentheses close parentheses end cell row blank equals cell 1 half open parentheses x squared over 2 minus fraction numerator x sin invisible function application 2 x over denominator 2 end fraction minus fraction numerator cos invisible function application 2 x over denominator 4 end fraction close parentheses end cell row blank equals cell 1 over 8 open parentheses 2 x squared minus 2 x sin invisible function application 2 x minus cos invisible function application 2 x close parentheses plus C end cell end table end style  

Jadi, jawaban yang tepat adalah B.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Iklan

Iklan

Pertanyaan serupa

Hasil dari ∫ x 2 e 2 x d x adalah ...

1

5.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia