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Hasil dari  adalah...

Pertanyaan

Hasil dari integral 12 square root of straight x dx adalah...

  1. 4 x square root of x space plus C

  2. 6 x square root of x space plus C

  3. 8 x square root of x space plus C

  4. 12 x square root of x space plus C

  5. 18 x square root of x space plus C

Pembahasan Soal:

Hasil dari integral 12 square root of straight x dx adalah...

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 12 square root of straight x dx end cell equals cell 12 integral square root of straight x space dx end cell row blank equals cell 12 integral straight x to the power of 1 half end exponent dx space end cell row blank equals cell 12 x open parentheses fraction numerator 1 over denominator begin display style 1 half end style plus 1 end fraction close parentheses straight x to the power of 1 half plus 1 end exponent plus straight C end cell row blank equals cell 12 x open parentheses fraction numerator 1 over denominator begin display style 3 over 2 end style end fraction close parentheses straight x to the power of 3 over 2 end exponent plus straight C end cell row blank equals cell 12 x open parentheses 2 over 3 close parentheses square root of straight x cubed end root plus straight C end cell row blank equals cell 8 x square root of straight x cubed end root plus C end cell end table

Jadi, Hasil dari integral 12 square root of straight x dx adalah 8 x square root of x space plus C

Oleh karena itu, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rahmawati

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

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Pembahasan Soal:

Integral tak tentu

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator x squared open parentheses 2 x minus 3 close parentheses over denominator square root of x end fraction straight d x end cell equals cell integral fraction numerator x squared open parentheses 2 x minus 3 close parentheses over denominator x to the power of begin display style 1 half end style end exponent end fraction straight d x end cell row blank equals cell integral x squared x to the power of negative 1 half end exponent open parentheses 2 x minus 3 close parentheses straight d x end cell row blank equals cell integral x to the power of 2 minus 1 half end exponent open parentheses 2 x minus 3 close parentheses straight d x end cell row blank equals cell integral x to the power of 3 over 2 end exponent open parentheses 2 x minus 3 close parentheses straight d x end cell row blank equals cell integral 2 x to the power of 3 over 2 end exponent x minus 3 x to the power of 3 over 2 end exponent space straight d x end cell row blank equals cell integral 2 x to the power of 5 over 2 end exponent minus 3 x to the power of 3 over 2 end exponent space straight d x end cell row blank equals cell fraction numerator 2 over denominator 5 over 2 plus 1 end fraction x to the power of 5 over 2 plus 1 end exponent minus 3 over blank to the power of 3 over 2 plus 1 end exponent x to the power of 3 over 2 plus 1 end exponent plus C end cell row blank equals cell fraction numerator 2 over denominator 7 over 2 end fraction x to the power of 7 over 2 end exponent minus 3 over blank to the power of 5 over 2 end exponent x to the power of 5 over 2 end exponent plus C end cell row blank equals cell 2 times 2 over 7 times x cubed times square root of x minus 3 times 2 over 5 times x squared times square root of x plus C end cell row blank equals cell 4 over 7 x cubed square root of x minus 6 over 5 x squared square root of x plus C end cell end table

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator x squared open parentheses 2 x minus 3 close parentheses over denominator square root of x end fraction straight d x end cell equals cell 4 over 7 x cubed square root of x minus 6 over 5 x squared square root of x plus C end cell end table.

Roboguru

Jika  dan  maka ....

Pembahasan Soal:

Diketahui:

  • turunan pertama fungsi f open parentheses x close parenthesesf apostrophe open parentheses x close parentheses equals 3 x squared minus 6 x 
  • f open parentheses 2 close parentheses equals 3  

Ditanya: fungsi f open parentheses x close parentheses  

Jawab:

Sebuah intergral tak tentu dapat diselesaikan dengan menggunakan rumus

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell integral f apostrophe open parentheses x close parentheses d x end cell row blank equals cell integral x to the power of n d x end cell row blank 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 end table

dengan n not equal to negative 1 dan C adalah konstanta. 

Dari soal, diketahui bahwa fungsi yang akan diintegralkan adalah f apostrophe open parentheses x close parentheses equals 3 x squared minus 6 x.

Oleh karena dalam menyelesaikan suatu permasalahan integral, berlaku sifat perkalian konstanta fungsi dan sifat pengurangan fungsi, maka diperoleh nilai integral dari fungsi yang diketahui adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell integral f apostrophe open parentheses x close parentheses d x end cell row blank equals cell integral 3 x squared minus 6 x d x end cell row blank equals cell 3 integral x squared d x minus 6 integral x to the power of 1 d x end cell row blank equals cell 3 open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent close parentheses minus 6 open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent close parentheses plus C end cell row blank equals cell 3 open parentheses 1 third x cubed close parentheses minus 6 open parentheses 1 half x squared close parentheses plus C end cell row blank equals cell x cubed minus 3 x squared plus C end cell end table

Pada fungsi f open parentheses x close parentheses yang diperoleh di atas terdapat nilai C yang belum diketahui. Oleh karena f open parentheses 2 close parentheses equals 3, maka dengan menyubtitusi x equals 2 ke fungsi f open parentheses x close parentheses di atas, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses 2 close parentheses end cell equals 3 row cell 2 cubed minus 3 open parentheses 2 close parentheses squared plus C end cell equals 3 row cell 8 minus 12 plus C end cell equals 3 row cell negative 4 plus C end cell equals 3 row C equals cell 3 plus space 4 end cell row C equals 7 end table  

Jadi, diperoleh fungsi f open parentheses x close parentheses equals x cubed minus 3 x squared plus 7. space 

Roboguru

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 

Untuk menyelesaikan soal di atas terlebih dahulu selesaikan perpangkatan.

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell integral open parentheses x squared plus 1 close parentheses squared space straight d x end cell equals cell integral open parentheses x squared plus 1 close parentheses open parentheses x squared plus 1 close parentheses space straight d x end cell row blank equals cell integral open parentheses x to the power of 4 plus 2 x squared plus 1 close parentheses space straight d x end cell row blank equals cell fraction numerator 1 over denominator 4 plus 1 end fraction x to the power of 4 plus 1 end exponent plus fraction numerator 2 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus x plus c end cell row cell integral open parentheses x squared plus 1 close parentheses squared space straight d x end cell equals cell 1 fifth x to the power of 5 plus 2 over 3 x cubed plus x plus c end cell end table end style  

Jadi, diperoleh hasilnya adalah begin mathsize 14px style 1 fifth x to the power of 5 plus 2 over 3 x cubed plus x plus c end style.

Roboguru

Pembahasan Soal:

Ingat bagaimana cara mengintegralkan suatu fungsi dengan integral tak tentu yaitu integral a x to the power of n 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.

table attributes columnalign right center left columnspacing 0px end attributes row cell integral 9 x squared d x end cell equals cell fraction numerator 9 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C end cell row blank equals cell 9 over 3 x cubed plus C end cell row blank equals cell 3 x cubed plus C end cell end table

Jadi, dapat disimpulkan bahwa integral 9 x squared d x adalah 3 x cubed plus C.

Oleh karena itu, jawaban yang benar adalah A.

Roboguru

Diberikan  yang merupakan turunan kedua dari . Untuk  fungsi bernilai 14 dan untuk , fungsi  bernilai 6. Tentukan fungsi .

Pembahasan Soal:

Terlebih dahulu kita cari anti-turunan (integral) dari f apostrophe apostrophe open parentheses x close parentheses untuk memperoleh f apostrophe open parentheses x close parentheses, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe open parentheses x close parentheses end cell equals cell integral f apostrophe apostrophe open parentheses x close parentheses d x space end cell row blank equals cell integral open parentheses 120 x squared minus 30 x close parentheses d x end cell row blank equals cell 40 x cubed plus 15 x squared plus C comma space text dengan end text space C text  konstanta end text end cell end table 

Selanjutnya kita integralkan f apostrophe open parentheses x close parentheses untuk memperoleh f open parentheses x close parentheses, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell integral open parentheses 40 x cubed plus 15 x squared plus C close parentheses d x end cell row blank equals cell 10 x to the power of 4 plus 5 x cubed plus C x plus D comma space text dengan end text space C space text dan  end text D space text konstanta end text end cell end table 

Diketahui bahwa untuk begin mathsize 14px style x equals negative 1 end style fungsi begin mathsize 14px style f left parenthesis x right parenthesis space end stylebernilai begin mathsize 14px style 14 end style.

table row cell f open parentheses x close parentheses equals 10 x to the power of 4 plus 5 x cubed plus C x plus D end cell row cell f open parentheses negative 1 close parentheses equals 10 open parentheses negative 1 close parentheses to the power of 4 plus 5 open parentheses negative 1 close parentheses cubed plus C open parentheses negative 1 close parentheses plus D end cell row cell 14 equals 10 minus 5 minus C plus D end cell row cell 14 equals 5 minus C plus D end cell row cell negative C plus D equals 9 space horizontal ellipsis space left parenthesis 1 right parenthesis end cell end table 

Dan untuk begin mathsize 14px style x equals 1 end style, fungsi undefined bernilai begin mathsize 14px style 6 end style.

table row cell f open parentheses x close parentheses equals 10 x to the power of 4 plus 5 x cubed plus C x plus D end cell row cell f open parentheses 1 close parentheses equals 10 open parentheses 1 close parentheses to the power of 4 plus 5 open parentheses 1 close parentheses cubed plus C open parentheses 1 close parentheses plus D end cell row cell 14 equals 10 plus 5 plus C plus D end cell row cell 14 equals 15 plus C plus D end cell row cell C plus D equals negative 1 space horizontal ellipsis space open parentheses 2 close parentheses end cell end table 

Eliminasi persamaan (1) dan persamaan (2), diperoleh:

table row cell up diagonal strike negative C end strike plus D equals 9 end cell row cell up diagonal strike C plus D equals negative 1 end cell row cell 2 D equals 8 end cell row cell D equals 8 over 2 end cell row cell D equals 4 end cell end table plus 

Substitusi nilai D ke salah satu persamaan, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell C plus D end cell equals cell negative 1 end cell row cell C plus 4 end cell equals cell negative 1 end cell row C equals cell negative 5 end cell end table 

Kembalikan nilai C dan D pada fungsi f open parentheses x close parentheses maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell 10 x to the power of 4 plus 5 x cubed plus C x plus D end cell row blank equals cell 10 x to the power of 4 plus 5 x cubed minus 5 x plus 4 end cell end table

Dengan demikian, fungsi undefined adalah Error converting from MathML to accessible text..

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