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Diketahui y=2x+1. Tentukan hasil pengintegralan berikut. ∫(y2−y)dx

Pertanyaan

Diketahui begin mathsize 14px style y space equals space 2 x space plus space 1 end style. Tentukan hasil pengintegralan berikut.

begin mathsize 14px style integral left parenthesis y squared minus y right parenthesis space d x end style 

  1. ...space 

  2. ...undefined 

Pembahasan Soal:

Konsep yang digunakan untuk mengerjakan soal di atas adalah

begin mathsize 14px style integral subscript blank straight a x to the power of n d x equals fraction numerator straight a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end style 

dengan begin mathsize 14px style straight a comma n element of straight real numbers end style.

Diketahui begin mathsize 14px style y equals 2 x plus 1 end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank left parenthesis y squared minus y right parenthesis d x end cell equals cell integral subscript blank open square brackets open parentheses 2 x plus 1 close parentheses squared minus open parentheses 2 x plus 1 close parentheses close square brackets d x end cell row blank equals cell integral subscript blank open parentheses 4 x squared plus 4 x plus 1 minus 2 x minus 1 close parentheses d x end cell row blank equals cell integral subscript blank left parenthesis 4 x squared plus 2 x right parenthesis d x end cell row blank equals cell fraction numerator 4 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus fraction numerator 2 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus C end cell row blank equals cell 4 over 3 x cubed plus x squared plus C end cell end table end style 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rizky

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

∫(x​−x​4​)dx= ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses x to the power of 1 half end exponent minus 4 x to the power of negative 1 half end exponent close parentheses d x end cell equals cell 2 over 3 x to the power of 3 over 2 end exponent minus 8 x to the power of 1 half end exponent plus C end cell row blank equals cell 2 over 3 x square root of x minus 8 square root of x plus C end cell end table end style

 

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Mencari hasil pengintegral:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral thin space 4 x squared minus 5 x cubed straight d x end cell equals cell fraction numerator 4 x to the power of 2 plus 1 end exponent over denominator 2 plus 1 end fraction minus fraction numerator 5 x to the power of 3 plus 1 end exponent over denominator 3 plus 1 end fraction plus C end cell row blank equals cell 4 over 3 x cubed minus 5 over 4 x to the power of 4 plus C end cell end table 

Jadi, hasil dari integral open parentheses 4 x squared minus 5 x cubed close parentheses straight d x adalah Error converting from MathML to accessible text..

0

Roboguru

Tentukan integral tak tentu berikut: f. ∫(3x​−xx​+2)dx

Pembahasan Soal:

Rumus dasar integral yaitu:

  • integral k x to the power of n space d x equals fraction numerator k over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus straight C dengan sayarat straight n not equal to negative 1
  • integral k space d x equals k x plus straight C, suatu konstanta

Diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 3 square root of x minus x square root of x plus 2 close parentheses d x end cell row blank equals cell integral 3 square root of x space d x minus integral x square root of x space d x plus integral 2 space d x end cell row blank equals cell integral 3 x to the power of 1 half end exponent space d x minus integral x x to the power of 1 half end exponent space d x plus integral 2 space d x end cell row blank equals cell integral 3 x to the power of 1 half end exponent space d x minus integral x to the power of 1 plus 1 half end exponent space space d x plus integral 2 space d x end cell row blank equals cell integral 3 x to the power of 1 half end exponent space d x minus integral x to the power of 3 over 2 end exponent space space d x plus integral 2 space d x end cell row blank equals cell fraction numerator 3 over denominator begin display style 1 half end style plus 1 end fraction x to the power of 1 half plus 1 end exponent minus fraction numerator 1 over denominator begin display style 3 over 2 end style plus 1 end fraction x to the power of 3 over 2 plus 1 end exponent plus 2 x plus straight C end cell row blank equals cell fraction numerator 3 over denominator begin display style 1 half end style plus begin display style 2 over 2 end style end fraction straight x to the power of 1 half plus 2 over 2 end exponent minus fraction numerator 1 over denominator begin display style 3 over 2 end style plus begin display style 2 over 2 end style end fraction x to the power of 3 over 2 plus 2 over 2 end exponent plus 2 x plus straight C end cell row blank equals cell fraction numerator 3 over denominator begin display style 3 over 2 end style end fraction x to the power of 3 over 2 end exponent minus fraction numerator 1 over denominator begin display style 5 over 2 end style end fraction x to the power of 5 over 2 end exponent plus 2 x plus straight C end cell row blank equals cell down diagonal strike 3 times fraction numerator 2 over denominator down diagonal strike 3 end fraction x to the power of 3 over 2 end exponent minus 1 times 2 over 5 x to the power of 5 over 2 end exponent plus 2 x plus straight C end cell row blank equals cell 2 x to the power of 3 over 2 end exponent minus 2 over 5 x to the power of 5 over 2 end exponent plus 2 x plus straight C end cell row blank equals cell 2 x times x to the power of 1 half end exponent minus fraction numerator 2 x squared times x to the power of begin display style 1 half end style end exponent over denominator 5 end fraction plus 2 x plus straight C end cell row blank equals cell 2 x square root of x minus fraction numerator 2 x squared square root of x over denominator 5 end fraction plus 2 x plus straight C end cell end table

Dengan demikian, intergal tak tentu dari integral open parentheses 3 square root of x minus x square root of x plus 2 close parentheses d x adalah 2 x square root of x minus fraction numerator 2 x squared square root of x over denominator 5 end fraction plus 2 x plus straight C.

0

Roboguru

Diketahui kecepatan suatu benda adalah v(t)=3t2−4t dan posisi benda pada jarak 5 untuk t=2. Rumus fungsi jarak s(t) adalah....

Pembahasan Soal:

Diketahui

table attributes columnalign right center left columnspacing 0px end attributes row cell v open parentheses t close parentheses end cell equals cell 3 t squared minus 4 t end cell row cell s open parentheses 2 close parentheses end cell equals 5 end table

Dengan menggunakan konsep anti turunan atau integral maka fungsi jarak adalah integral dari fungsi kecepatan, sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell s open parentheses t close parentheses end cell equals cell integral v open parentheses t close parentheses space d t end cell row blank equals cell integral space open parentheses 3 t squared minus 4 t close parentheses space d t end cell row blank equals cell fraction numerator 3 over denominator 2 plus 1 end fraction t to the power of 2 plus 1 end exponent minus fraction numerator 4 over denominator 1 plus 1 end fraction t to the power of 1 plus 1 end exponent plus c end cell row blank equals cell t cubed minus 2 t squared plus c end cell end table

Diketahui s open parentheses 2 close parentheses equals 5 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 cubed minus 2 open parentheses 2 close parentheses squared plus c end cell equals 5 row cell 8 minus 8 plus c end cell equals 5 row c equals 5 end table

Dengan demikian fungsi jarak adalah s open parentheses t close parentheses equals t cubed minus 2 t squared plus 5.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Hasil dari ∫(53−3x+2)dx adalah ...

Pembahasan Soal:

Terdapat kesalahan pada soal. Integral yang dimaksud bukan integral open parentheses 5 cubed minus 3 x plus 2 close parentheses space straight d x, melainkan integral open parentheses 5 x cubed minus 3 x plus 2 close parentheses space straight d x.

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell integral a x to the power of n space straight d x end cell equals cell fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent end cell row cell integral open parentheses f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses close parentheses space straight d x end cell equals cell integral f open parentheses x close parentheses space straight d x plus-or-minus integral g open parentheses x close parentheses space straight d x end cell row cell integral k times f open parentheses x close parentheses space straight d x end cell equals cell k integral f open parentheses x close parentheses space straight d x end cell end table  

Maka, hasil dari integral open parentheses 5 x cubed minus 3 x plus 2 close parentheses space straight d x adalah:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral open parentheses 5 x cubed minus 3 x plus 2 close parentheses space straight d x end cell row blank equals cell integral 5 x cubed space straight d x minus integral 3 x space straight d x plus integral 2 space straight d x end cell row blank equals cell 5 integral x cubed space straight d x minus 3 integral x space straight d x plus integral 2 space straight d x end cell row blank equals cell 5 times 1 fourth x to the power of 4 minus 3 times 1 half x squared plus 2 x plus C end cell row blank equals cell 5 over 4 x to the power of 4 minus 3 over 2 x squared plus 2 x plus C end cell end table  

Jadi, Error converting from MathML to accessible text..

0

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