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Diketahui f(x)=4x3−5xdang(x)=2x2−2 Nilai dari ∫04​[f(x)−g(x)]dxadalah...

Pertanyaan

Diketahui f left parenthesis x right parenthesis equals space 4 x cubed minus 5 x space d a n space g left parenthesis x right parenthesis equals space 2 x squared minus 2

Nilai dari integral subscript 0 superscript 4 open square brackets f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis close square brackets d x spaceadalah...

  1. 180 1 third

  2. 180 2 over 3

  3. 181

  4. 181 1 third

  5. 181 2 over 3

Pembahasan Soal:

f left parenthesis x right parenthesis space equals space 4 x cubed minus 5 x space d a n space g left parenthesis x right parenthesis equals space 2 x squared minus x  integral subscript 0 superscript 4 open square brackets f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis close square brackets d x space equals integral subscript 0 superscript 4 open square brackets 4 x cubed minus 5 x minus 2 x squared plus x close square brackets 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 equals integral subscript 0 superscript 4 open square brackets 4 x cubed minus 2 x squared minus 4 x close square brackets 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 equals space x to the power of 4 minus 2 over 3 x cubed minus 2 x squared left enclose blank with 0 below and 4 on top end enclose  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 space open parentheses 4 to the power of 4 minus 2 over 3 left parenthesis 4 right parenthesis cubed minus 2 left parenthesis 4 right parenthesis squared close parentheses minus space left parenthesis 0 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 space space space space space space space equals 256 minus 128 over 3 minus 32 space equals 544 over 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 equals 181 1 third

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 04 Oktober 2021

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Pembahasan Soal:

Ingat sudut rangkap berikut ini:

cos squared 2 x minus sin squared 2 x equals space sin space 4 x

Jika diketahui:

f left parenthesis x right parenthesis space equals space cos squared 2 x  g left parenthesis x right parenthesis space equals space sin squared 2 x    m a k a colon  integral subscript 0 superscript fraction numerator 2 straight pi over denominator 3 end fraction end superscript f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis space d x equals integral subscript 0 superscript fraction numerator 2 straight pi over denominator 3 end fraction end superscript cos space squared space 2 x minus space sin squared space 2 x space d x  equals integral subscript 0 superscript fraction numerator 2 straight pi over denominator 3 end fraction end superscript cos space 4 x space d x  equals 1 fourth sin space 4 x left enclose blank with 0 below and begin inline style fraction numerator 2 straight pi over denominator 3 end fraction end style on top end enclose  equals 1 fourth sin space 4 space open parentheses fraction numerator 2 straight pi over denominator 3 end fraction close parentheses plus 1 fourth sin space 4 space left parenthesis 0 right parenthesis  equals 1 fourth open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses plus 0  equals fraction numerator square root of 3 over denominator 8 end fraction

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∫02​(x3−8x2+6x−3)dx = ....

Pembahasan Soal:

Ingat rumus integral fungsi aljabar:

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 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 2 open parentheses x cubed minus 8 x squared plus 6 x minus 3 close parentheses d x end cell equals cell right enclose 1 fourth x to the power of 4 minus 8 over 3 x cubed plus 3 x squared minus 3 x end enclose subscript 0 superscript 2 end cell row blank equals cell open parentheses 1 fourth times 2 to the power of 4 minus 8 over 3 times 2 cubed plus 3 times 2 squared minus 3 times 2 close parentheses end cell row blank blank cell negative open parentheses 1 fourth times 0 to the power of 4 minus 8 over 3 times 0 cubed plus 3 times 0 squared minus 3 times 0 close parentheses end cell row blank equals cell open parentheses 1 fourth times 16 minus 8 over 3 times 8 plus 3 times 4 minus 6 close parentheses end cell row blank blank cell negative open parentheses 1 fourth times 0 minus 8 over 3 times 0 plus 3 times 0 minus 0 close parentheses end cell row blank equals cell open parentheses 4 minus 64 over 3 plus 12 minus 6 close parentheses end cell row blank blank cell negative open parentheses 0 minus 0 plus 0 minus 0 close parentheses end cell row blank equals cell open parentheses 10 minus 64 over 3 close parentheses minus 0 end cell row blank equals cell 30 over 3 minus 64 over 3 end cell row blank equals cell negative 34 over 3 end cell end table 

Jadi, jawaban yang benar adalah B.

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Hasil dari ∫01​(3x2−2x+5)dx=...

Pembahasan Soal:

integral subscript 0 superscript 1 left parenthesis 3 x squared minus 2 x plus 5 right parenthesis d x  rightwards double arrow left square bracket x cubed minus x squared plus 5 x right square bracket subscript 0 superscript 1  equals left parenthesis 1 cubed minus 1 squared plus 5 left parenthesis 1 right parenthesis right parenthesis minus left parenthesis 0 right parenthesis  equals 1 minus 1 plus 5  equals 5

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Hasil∫−33​(x2+6x−2)dx=...

Pembahasan Soal:

integral subscript negative 3 end subscript superscript 3 left parenthesis x squared plus 6 x minus 2 right parenthesis d x equals left square bracket 1 third x cubed plus 6 left parenthesis 1 half x squared right parenthesis minus 2 x right square bracket subscript negative 3 end subscript superscript 3  equals left square bracket 1 third x cubed plus 3 x squared minus 2 x right square bracket subscript negative 3 end subscript superscript 3  equals left parenthesis 1 third.3 cubed plus 3 left parenthesis 3 squared right parenthesis minus 2 left parenthesis 3 right parenthesis right parenthesis minus left parenthesis 1 third. left parenthesis negative 3 right parenthesis cubed plus 3 left parenthesis negative 3 right parenthesis squared minus 2 left parenthesis negative 3 right parenthesis right parenthesis  equals 30 minus 24  equals 6

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Roboguru

Lukiskan sketsa kurva y=4x−x2. Hitunglah luas daerah tertutup yang dibatasi kurva tersebut dan sumbu X.

Pembahasan Soal:

Ingat kembali rumus luas daerah fungsi f open parentheses x close parentheses dengan batas atas a dan batas bawah b berikut:

L=ab(x)dx 

Diketahui kurva undefined. Kurva tersebut dapat digambarkan sebagai berikut:
 


 

Dari gambar di atas, dapat dilihat bahwa kurva memotong sumbu x di titik x equals 0 dan x equals 4. Maka luas kurva tersebut dapat dihitung dengan cara:

table attributes columnalign right center left columnspacing 0px end attributes row straight L equals cell integral subscript b superscript a f open parentheses x close parentheses space d x end cell row blank equals cell integral subscript b superscript a y space d x end cell row blank equals cell integral subscript 0 superscript 4 open parentheses 4 x minus x squared close parentheses space d x end cell row blank equals cell right enclose fraction numerator 4 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent end enclose subscript 0 superscript 4 end cell row blank equals cell right enclose 2 x squared minus 1 third x cubed end enclose subscript 0 superscript 4 end cell row blank equals cell open parentheses 2 times 4 squared minus 1 third times 4 cubed close parentheses minus open parentheses 2 times 0 squared minus 1 third times 0 cubed close parentheses end cell row blank equals cell open parentheses 2 times 16 minus 1 third times 64 close parentheses minus open parentheses 2 times 0 minus 1 third times 0 close parentheses end cell row blank equals cell open parentheses 32 minus 64 over 3 close parentheses minus open parentheses 0 minus 0 close parentheses end cell row blank equals cell open parentheses 96 over 3 minus 64 over 3 close parentheses minus 0 end cell row blank equals cell 32 over 3 end cell row blank equals cell 10 2 over 3 end cell end table 

Jadi,  luas daerah tertutup yang dibatasi kurva tersebut dan sumbu undefined adalah 10 2 over 3 satuan luas.

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