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Hasil dari ∫02​6x−2x3dx=...

Pertanyaan

Hasil dari integral subscript 0 superscript 2 6 x minus 2 x cubed space d x equals...

  1. 0

  2. 1 fourth

  3. 1

  4. 2

  5. 4

Pembahasan Soal:

integral subscript 0 superscript 2 6 x minus 2 x cubed d x equals 3 x squared minus 1 half x to the power of 4 right enclose blank end enclose subscript 0 superscript 2 space equals 3 left parenthesis 2 right parenthesis squared minus 1 half left parenthesis 2 right parenthesis to the power of 4  space space space space space space space space space space space space space space space space space space space space space space space space space space equals 12 minus 8 equals 4

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 04 Oktober 2021

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Pertanyaan yang serupa

Hitunglah nilai masing-masing ekspresi integral tertentu berikut. c. ∫12​(k4−k3)dk

Pembahasan Soal:

Diketahui: Integral tentu begin mathsize 14px style integral subscript 1 superscript 2 open parentheses k to the power of 4 minus k cubed close parentheses space dk end style

Ditanya: Nilai dari begin mathsize 14px style integral subscript 1 superscript 2 open parentheses k to the power of 4 minus k cubed close parentheses space dk end style?

Jawab:

Ingat bahwa untuk menentukan suatu integral tentu maka harus mengintegralkan fungsi tersebut kemudian substitusikan nilai batasnya atau perhatikan rumus umum berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript a superscript b f left parenthesis x right parenthesis end cell equals cell left square bracket F left parenthesis x right parenthesis right square bracket subscript a superscript b equals F left parenthesis b right parenthesis minus F left parenthesis a right parenthesis end cell end table

Ingat juga rumus umum dan sifat integral yaitu 

  • integral a x to the power of n space 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 subscript a superscript b f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis space d x end cell equals cell integral subscript a superscript b f left parenthesis x right parenthesis space d x minus integral subscript a superscript b g left parenthesis x right parenthesis space d x end cell end table

Dari rumus-rumus  di atas maka akan didapatkan 

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 1 superscript 2 open parentheses k to the power of 4 minus k cubed close parentheses space dk end cell equals cell integral subscript 1 superscript 2 k to the power of 4 space dk minus integral subscript 1 superscript 2 k cubed space dk end cell row blank equals cell open square brackets fraction numerator 1 over denominator 4 plus 1 end fraction k to the power of 4 plus 1 end exponent close square brackets subscript 1 superscript 2 minus open square brackets fraction numerator 1 over denominator 3 plus 1 end fraction k to the power of 3 plus 1 end exponent close square brackets subscript 1 superscript 2 end cell row blank equals cell open square brackets 1 fifth k to the power of 5 close square brackets subscript 1 superscript 2 minus open square brackets 1 fourth k to the power of 4 close square brackets subscript 1 superscript 2 end cell row blank equals cell open square brackets 1 fifth times 2 to the power of 5 minus 1 fifth times 1 to the power of 5 close square brackets minus open square brackets 1 fourth times 2 to the power of 4 minus 1 fourth times 1 to the power of 4 close square brackets end cell row blank equals cell open square brackets 1 fifth times 32 minus 1 fifth times 1 close square brackets minus open square brackets 1 fourth times 16 minus 1 fourth times 1 close square brackets end cell row blank equals cell open square brackets 32 over 5 minus 1 fifth close square brackets minus open square brackets 16 over 4 minus 1 fourth close square brackets end cell row blank equals cell 31 over 5 minus 15 over 4 end cell row blank equals cell 124 over 20 minus 75 over 20 end cell row blank equals cell 49 over 20 end cell end table

Jadi, dapat disimpulkan bahwa hasil dari begin mathsize 14px style integral subscript 1 superscript 2 open parentheses k to the power of 4 minus k cubed close parentheses space dk end style adalah 49 over 20.

0

Roboguru

Jika hasil ∫−2K​(3x2−6x+2)dx=24, jumlah semua nilai k yang memenuhi adalah....

Pembahasan Soal:

Jawaban B

integral subscript negative 2 end subscript superscript K left parenthesis 3 x squared minus 6 x plus 2 right parenthesis d x equals 24

open square brackets x cubed minus 3 x squared plus 2 x close square brackets stack blank subscript negative 2 end subscript with k on top equals 24

k cubed minus 3 k squared plus 2 k minus left parenthesis negative 8 minus 12 minus 4 right parenthesis equals 24

k cubed minus 3 k squared plus 2 k plus 24 equals 24

k cubed minus 3 k squared plus 2 k equals 0

k left parenthesis k minus 1 right parenthesis left parenthesis k minus 2 right parenthesis equals 0

k=0 atau k=1 atau k=2

Jadi jumlah semua nilai k = 0 + 1 + 2 = 3

0

Roboguru

Nilai integral fungsi  f(x) = 2x - 4 untuk -2  x 1adalah . .

Pembahasan Soal:

integral subscript negative 2 end subscript superscript 1 2 x minus 4 space d x equals x squared minus 4 x right enclose blank end enclose subscript negative 2 end subscript superscript 1 equals space left parenthesis 1 right parenthesis squared minus 4 left parenthesis 1 right parenthesis squared minus left parenthesis negative 2 right parenthesis squared plus 4 left parenthesis negative 2 right parenthesis  space space space space space space space space space space space space space space space space space space space space space space space space equals 1 minus space 16 equals negative 15

0

Roboguru

Nilai dari ∫25​(3x2−6x+1)dx=...

Pembahasan Soal:

Sesuai dengan definisi integral tentu, maka penyelesaian integral tentu tersebut sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 2 superscript 5 open parentheses 3 x squared minus 6 x plus 1 close parentheses d x end cell equals cell open square brackets x cubed minus 3 x squared plus x close square brackets subscript 2 superscript 5 end cell row blank equals cell open square brackets open parentheses 5 cubed minus 3 times 5 squared plus 5 close parentheses minus open parentheses 2 cubed minus 3 times 2 squared plus 2 close parentheses close square brackets end cell row blank equals cell open square brackets open parentheses 125 minus 75 plus 5 close parentheses minus open parentheses 8 minus 12 plus 2 close parentheses close square brackets end cell row blank equals cell open square brackets 55 minus open parentheses negative 2 close parentheses close square brackets end cell row blank equals 57 end table 

Jadi, nilai dari integral subscript 2 superscript 5 open parentheses 3 x squared minus 6 x plus 1 close parentheses d x equals 57

0

Roboguru

Tentukan hasil integral tentu fungsi-fungsi berikut! 5. ∫23​(3x−2)(3x+4)dx

Pembahasan Soal:

begin mathsize 14px style integral subscript 2 superscript 3 open parentheses 3 x minus 2 close parentheses open parentheses 3 x plus 4 close parentheses d x equals integral subscript 2 superscript 3 9 x squared plus 6 x minus 8 space d x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals integral subscript 2 superscript 3 9 x squared space d x plus integral subscript 2 superscript 3 6 x space d x minus integral subscript 2 superscript 3 8 space d x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals right enclose fraction numerator 9 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent end enclose subscript 2 superscript 3 plus right enclose fraction numerator 6 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent end enclose subscript 2 superscript 3 minus right enclose fraction numerator 8 over denominator 0 plus 1 end fraction x to the power of 0 plus 1 end exponent end enclose subscript 2 superscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals right enclose 3 x cubed end enclose subscript 2 superscript 3 plus right enclose 3 x squared end enclose subscript 2 superscript 3 minus right enclose 8 x end enclose subscript 2 superscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open parentheses 3 left parenthesis 3 right parenthesis cubed minus 3 left parenthesis 2 right parenthesis cubed close parentheses plus open parentheses 3 left parenthesis 3 right parenthesis squared minus 3 left parenthesis 2 right parenthesis squared close parentheses minus open parentheses 8 left parenthesis 3 right parenthesis minus 8 left parenthesis 2 right parenthesis close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 57 plus 15 minus 8 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 64 end style

0

Roboguru

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