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hasildari∫02​3(x+1)(x−6)dx=...

Pertanyaan

h a s i l space d a r i space integral subscript 0 superscript 2 3 left parenthesis x plus 1 right parenthesis left parenthesis x minus 6 right parenthesis d x equals...

  1. -58

  2. -56

  3. -28

  4. -16

  5. -14

Pembahasan Soal:

integral subscript 0 superscript 2 3 left parenthesis x plus 1 right parenthesis left parenthesis x minus 6 right parenthesis d x equals integral subscript 0 superscript 2 left parenthesis 3 x squared minus 15 x minus 18 right parenthesis d x  equals x cubed minus 15 over 2 x squared minus 18 x space right square bracket subscript 0 superscript 2  equals 2 cubed minus 15 over 2 2 squared minus 18.2 – left parenthesis 0 cubed minus 15 over 2 0 squared minus 18.0 right parenthesis  equals 8 – 30 – 36 – 0  equals negative 58

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Mustikowati

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 04 Oktober 2021

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Pertanyaan yang serupa

∫24​(4x3+9x−2)dx=

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 2 superscript 4 left parenthesis 4 x cubed plus 9 x minus 2 right parenthesis space d x end cell equals cell open square brackets x to the power of 4 plus 9 over 2 x squared minus 2 x close square brackets subscript 2 superscript 4 end cell row blank equals cell open parentheses 4 close parentheses to the power of 4 plus 9 over 2 open parentheses 4 close parentheses squared minus 2 open parentheses 4 close parentheses minus open parentheses open parentheses 2 close parentheses to the power of 4 plus 9 over 2 open parentheses 2 close parentheses squared minus 2 open parentheses 2 close parentheses close parentheses end cell row blank equals cell 256 plus 9 over 2 open parentheses 16 close parentheses minus 8 minus open parentheses 16 plus 9 over 2 open parentheses 4 close parentheses minus 4 close parentheses end cell row blank equals cell 256 plus 72 minus 8 minus open parentheses 16 plus 18 minus 4 close parentheses end cell row blank equals cell 320 minus 30 end cell row blank equals 290 end table end style

Jadi,  integral subscript 2 superscript 4 left parenthesis size 14px 4 size 14px x to the power of size 14px 3 size 14px plus size 14px 9 size 14px x size 14px minus size 14px 2 size 14px right parenthesis size 14px dx size 14px equals size 14px 290

 

0

Roboguru

Hasil dari ∫02​3(x+1)(x−8)=...

Pembahasan Soal:

Perhatikan rumus integral tentu berikut:

integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx equals open square brackets straight F left parenthesis straight x right parenthesis close square brackets subscript straight x equals straight a end subscript superscript straight x equals straight b end superscript equals straight F left parenthesis straight b right parenthesis minus straight F left parenthesis straight a right parenthesis.

Penyelesaian integral tentu pada soal adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 2 3 open parentheses straight x plus 1 close parentheses open parentheses straight x minus 8 close parentheses end cell equals cell integral subscript 0 superscript 2 open parentheses 3 straight x plus 3 close parentheses open parentheses straight x minus 8 close parentheses end cell row blank equals cell integral subscript 0 superscript 2 3 straight x squared minus 24 straight x plus 3 straight x minus 24 end cell row blank equals cell integral subscript 0 superscript 2 3 straight x squared minus 21 straight x minus 24 end cell row blank equals cell open square brackets 3 over 3 straight x cubed minus 21 over 2 straight x squared minus 24 straight x close square brackets subscript straight x equals 0 end subscript superscript straight x equals 2 end superscript end cell row blank equals cell open square brackets straight x cubed minus 21 over 2 straight x squared minus 24 straight x close square brackets subscript straight x equals 0 end subscript superscript straight x equals 2 end superscript end cell row blank equals cell open parentheses 2 cubed minus 21 over 2 open parentheses 2 squared close parentheses minus 24 open parentheses 2 close parentheses close parentheses minus open parentheses 0 cubed minus 21 over 2 open parentheses 0 squared close parentheses minus 24 open parentheses 0 close parentheses close parentheses end cell row blank equals cell 8 minus 21 over 2 open parentheses 4 close parentheses minus 48 end cell row blank equals cell 8 minus 42 minus 48 end cell row blank equals cell negative 82 end cell end table 

Dengan demikian, hasil dari integral subscript 0 superscript 2 3 open parentheses x plus 1 close parentheses open parentheses x minus 8 close parentheses adalah negative 82.

0

Roboguru

∫−21​(2x−4)2dx​======​∫−21​(2x−4)(2x−4)dx∫−21​(4x2−16x+16)dx[3....​x3−....x2+....x]−21​{(34​....3−8....2+16....)−(34​(....)3−8(....)2+16(....))}(....−....+....)−(....−....+....)....​

Pembahasan Soal:

Integral Tentu

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 2 end subscript superscript 1 open parentheses 2 x minus 4 close parentheses squared text dx end text end cell equals cell integral subscript negative 2 end subscript superscript 1 open parentheses 2 x minus 4 close parentheses open parentheses 2 x minus 4 close parentheses text dx end text end cell row blank equals cell integral subscript negative 2 end subscript superscript 1 open parentheses 4 x squared minus 16 x plus 16 close parentheses d text x end text end cell row blank equals cell open square brackets 4 over 3 x cubed minus 8 x squared plus 16 x close square brackets subscript negative 2 end subscript superscript 1 end cell row blank equals cell open curly brackets open parentheses 4 over 3 open parentheses 1 close parentheses cubed minus 8 open parentheses 1 close parentheses squared plus 16 open parentheses 1 close parentheses close parentheses minus open parentheses 4 over 3 open parentheses negative 2 close parentheses cubed minus 8 open parentheses negative 2 close parentheses squared plus 16 open parentheses negative 2 close parentheses close parentheses close curly brackets end cell row blank equals cell open parentheses 4 over 3 minus 8 plus 16 close parentheses minus open parentheses 32 over 3 minus open parentheses negative 32 close parentheses plus open parentheses negative 32 close parentheses close parentheses end cell row blank equals cell 4 over 3 minus 32 over 3 plus 8 end cell row blank equals cell negative 28 over 3 plus 24 over 3 end cell row blank equals cell negative 4 over 3 end cell end table end style

Jadi, begin mathsize 14px style integral subscript negative 2 end subscript superscript 1 open parentheses 2 x minus 4 close parentheses squared d x equals negative 4 over 3 end style.

0

Roboguru

Diketahui . Nilai p yang memenuhi adalah......

Pembahasan Soal:

integral subscript negative 1 end subscript superscript 2 open parentheses 3 x squared plus 2 x minus p close parentheses d x equals 12  open x cubed plus x squared minus p x close square brackets subscript negative 1 end subscript superscript 2 equals 12  open parentheses 2 cubed plus 2 squared minus p open parentheses 2 close parentheses close parentheses minus open parentheses open parentheses negative 1 close parentheses cubed plus open parentheses negative 1 close parentheses squared minus p open parentheses negative 1 close parentheses close parentheses equals 12  open parentheses 2 cubed plus 2 squared minus p open parentheses 2 close parentheses close parentheses minus open parentheses open parentheses negative 1 close parentheses cubed plus open parentheses negative 1 close parentheses squared minus p open parentheses negative 1 close parentheses close parentheses equals 12  open parentheses 12 minus 2 p close parentheses minus open parentheses negative 1 plus 1 minus p close parentheses equals 12  12 minus 2 p plus p equals 12  p equals 0

0

Roboguru

Hasil dari ∫−13​(6x2+5)dx adalah ....

Pembahasan Soal:

begin mathsize 14px style integral subscript negative 1 end subscript superscript 3 open parentheses 6 x squared plus 5 close parentheses d x end style

begin mathsize 14px style equals open parentheses fraction numerator 6 x cubed over denominator 3 end fraction plus 5 x close parentheses table row cell x equals 3 end cell row blank row cell x equals negative 1 end cell end table end style 

 begin mathsize 14px style equals open parentheses 2 x cubed plus 5 x close parentheses table row cell x equals 3 end cell row cell x equals negative 1 end cell end table end style 

masukkan batas atas, masukkan batas bawah, lalu kurangkan.

begin mathsize 14px style equals open parentheses 2 open parentheses 27 close parentheses plus 5 open parentheses 3 close parentheses close parentheses minus open parentheses 2 open parentheses negative 1 close parentheses plus 5 open parentheses negative 1 close parentheses close parentheses end style 

begin mathsize 14px style equals 54 plus 15 plus 2 plus 5 end style 

begin mathsize 14px style equals 76 end style.

Jadi, 

begin mathsize 14px style integral subscript negative 1 end subscript superscript 3 open parentheses 6 x squared plus 5 close parentheses d x end style begin mathsize 14px style equals 76 end style 

0

Roboguru

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