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Hasil dari integral x left parenthesis x plus 4 right parenthesis to the power of 7 straight d x equals horizontal ellipsis.

  1. 1 over 9 open parentheses 2 left parenthesis x plus 4 right parenthesis left parenthesis x minus 2 right parenthesis to the power of 8 plus c close

  2. 1 over 9 left parenthesis 2 x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis to the power of 8 plus c

  3. 1 over 18 left parenthesis 2 x minus 1 right parenthesis left parenthesis x plus 2 right parenthesis to the power of 8 plus c

  4. 1 over 18 left parenthesis 2 x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis to the power of 8 plus c

  5. 1 over 18 left parenthesis 2 x plus 1 right parenthesis left parenthesis x minus 4 right parenthesis to the power of 8 plus c

Pembahasan Soal:

Diberikan bentuk integral integral x left parenthesis x plus 4 right parenthesis to the power of 7 straight d x. Karena terdiri dari perkalian dua fungsi maka dapat dimisalkan

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell x plus 4 end cell row x equals cell u minus 4 end cell row cell straight d x end cell equals cell straight d u end cell end table

Maka dengan metode substitusi, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral x left parenthesis x plus 4 right parenthesis to the power of 7 straight d x end cell equals cell integral open parentheses u minus 4 close parentheses u to the power of 7 d u end cell row blank equals cell integral open parentheses u times u to the power of 7 minus 4 u to the power of 7 close parentheses d u end cell row blank equals cell integral open parentheses u to the power of 8 minus 4 u to the power of 7 close parentheses d u end cell row blank equals cell 1 over 9 u to the power of 9 minus 4 over 8 u to the power of 8 plus C end cell row blank equals cell 1 over 9 open parentheses x plus 4 close parentheses to the power of 9 minus 1 half open parentheses x plus 4 close parentheses to the power of 8 plus C end cell row blank equals cell open parentheses x plus 4 close parentheses to the power of 8 open parentheses 1 over 9 open parentheses x plus 4 close parentheses minus 1 half close parentheses plus C end cell row blank equals cell open parentheses x plus 4 close parentheses to the power of 8 open parentheses 1 over 9 x plus 4 over 9 minus 1 half close parentheses plus C end cell row blank equals cell open parentheses x plus 4 close parentheses to the power of 8 open parentheses 1 over 9 x plus 8 over 18 minus 9 over 18 close parentheses plus C end cell row blank equals cell open parentheses x plus 4 close parentheses to the power of 8 open parentheses 1 over 9 x minus 1 over 18 close parentheses plus C end cell row blank equals cell 1 over 9 open parentheses x plus 4 close parentheses to the power of 8 open parentheses x minus 1 half close parentheses plus C end cell row blank equals cell 1 over 9 open parentheses x plus 4 close parentheses to the power of 8 open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses plus C end cell row blank equals cell 1 over 18 left parenthesis 2 x minus 1 right parenthesis open parentheses x plus 4 close parentheses to the power of 8 plus C end cell end table

Jadi,  jawaban soal ini terdapat pada pilihan (D)

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Pembahasan Soal:

Gunakan konsep integral tak tentu.

begin mathsize 14px style 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 plus c end style 

Perhatikan perhitungan berikut.

begin mathsize 14px style integral open parentheses 2 x minus 3 close parentheses space straight d x equals fraction numerator 2 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus 3 x plus c equals x squared minus 3 x plus c end style  

Jadi, diperoleh hasilnya adalah begin mathsize 14px style x squared minus 3 x plus c end style.

Roboguru

Tentukan hasil dari integral berikut: b.

Pembahasan Soal:

Dengan menggunakan konsep integral fungsi aljabar diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell integral x squared open parentheses square root of x minus 1 over x close parentheses squared d x end cell equals cell integral x squared open parentheses x to the power of 1 half end exponent minus x to the power of negative 1 end exponent close parentheses d x end cell row blank equals cell integral open parentheses x squared times x to the power of 1 half end exponent minus x squared space times x to the power of negative 1 end exponent to the power of times close parentheses d x end cell row blank equals cell integral open parentheses x to the power of 2 plus 1 half end exponent minus x to the power of 2 minus 1 end exponent close parentheses d x end cell row blank equals cell integral open parentheses x to the power of 5 over 2 end exponent minus x to the power of 1 close parentheses d x end cell row blank equals cell fraction numerator 1 over denominator begin display style 5 over 2 end style plus 1 end fraction x to the power of 5 over 2 plus 1 end exponent minus fraction numerator 1 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 fraction numerator 1 over denominator begin display style 7 over 2 end style end fraction x to the power of 7 over 2 end exponent minus 1 half x squared plus C end cell row blank equals cell 2 over 7 x cubed square root of x minus 1 half x squared plus C end cell end table

Dengan demikian hasil dari Error converting from MathML to accessible text..

Roboguru

= ...

Pembahasan Soal:

Ingat sifat-sifat integral sebagai berikut :

table attributes columnalign right center left columnspacing 2px end attributes row cell integral subscript blank superscript blank x to the power of n d x end cell equals cell fraction numerator 1 over denominator n plus 1 end fraction x to the power of n plus 1 end exponent plus C end cell row cell integral subscript blank superscript blank open parentheses f open parentheses x close parentheses plus g open parentheses x close parentheses close parentheses d x end cell equals cell integral subscript blank superscript blank f open parentheses x close parentheses d x plus integral subscript blank superscript blank g open parentheses x close parentheses d x end cell row cell integral subscript blank superscript blank open parentheses f open parentheses x close parentheses minus g open parentheses x close parentheses close parentheses d x end cell equals cell integral subscript blank superscript blank f open parentheses x close parentheses d x minus integral subscript blank superscript blank g open parentheses x close parentheses d x end cell end table 

Sehingga penyelesaian dari integral tak tentu pada integral subscript blank open parentheses 2 x cubed plus 3 x squared minus 2 x minus 1 close parentheses d x sebagai berikut ;

integral subscript blank open parentheses 2 x cubed plus 3 x squared minus 2 x minus 1 close parentheses d x equals integral subscript blank 2 x cubed d x plus integral subscript blank 3 x squared d x minus integral subscript blank 2 x d x minus integral subscript blank 1 d x equals 1 fourth 2 x to the power of 4 plus 1 third 3 x cubed minus 1 half 2 x squared minus x plus C equals 1 half x to the power of 4 plus x cubed minus x squared minus x plus C 

Jadi, integral subscript blank open parentheses 2 x cubed plus 3 x squared minus 2 x minus 1 close parentheses d x = 1 half x to the power of 4 plus x cubed minus x squared minus x plus C 

 

Roboguru

Pembahasan Soal:

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

Dengan menggunakan rumus tersebut, diperoleh 

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

Dengan demikian, jawaban yang tepat adalah A.

Roboguru

Selesaikanlah integral berikut! b.

Pembahasan Soal:

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell integral space 2 open parentheses 1 over x minus x close parentheses space straight d x end cell equals cell 2 integral 1 over x minus x space straight d x end cell row blank equals cell 2 open parentheses ln open vertical bar x close vertical bar minus 1 half x squared close parentheses plus C end cell row blank equals cell 2 ln open vertical bar x close vertical bar minus x squared plus C end cell end table  

Jadi, Error converting from MathML to accessible text..

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