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Nilai dari ∫04​2x+1​dx adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style integral subscript 0 superscript 4 square root of 2 x plus 1 end root d x end style adalah ....

  1. begin mathsize 14px style 13 over 3 end style   undefined 

  2. begin mathsize 14px style 16 over 3 end style   undefined 

  3. begin mathsize 14px style 20 over 3 end style   undefined 

  4. begin mathsize 14px style 26 over 3 end style   undefined 

  5. begin mathsize 14px style 28 over 3 end style   undefined 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 0 superscript 4 square root of 2 x plus 1 end root d x end cell equals cell integral subscript 0 superscript 4 open parentheses 2 x plus 1 close parentheses to the power of 1 half end exponent d x end cell row blank equals cell open square brackets fraction numerator begin display style bevelled 1 half end style cross times open parentheses 2 x plus 1 close parentheses over denominator begin display style bevelled 3 over 2 end style end fraction to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 4 end cell row blank equals cell open square brackets 1 third open parentheses 2 x plus 1 close parentheses to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 4 end cell row blank equals cell 1 third open parentheses left parenthesis 2 left parenthesis 4 right parenthesis plus 1 right parenthesis to the power of 3 over 2 end exponent minus left parenthesis 2 left parenthesis 0 right parenthesis plus 1 right parenthesis to the power of 3 over 2 end exponent close parentheses end cell row blank equals cell 1 third open parentheses 27 minus 1 close parentheses end cell row blank equals cell 1 third left parenthesis 26 right parenthesis end cell row blank equals cell 26 over 3 end cell end table end style

Jadi, jawaban yang tepat adalah D

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Nilai dari ∫15​(3x2+4x−5)dx adalah ....

Pembahasan Soal:

Sebelumnya akan dicari terlebih dahulu begin mathsize 14px style integral subscript blank superscript blank left parenthesis 3 x squared plus 4 x minus 5 right parenthesis d x end style.

begin mathsize 14px style integral subscript blank superscript blank left parenthesis 3 x squared plus 4 x minus 5 right parenthesis d x equals fraction numerator 3 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus fraction numerator 4 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus fraction numerator 5 over denominator 0 plus 1 end fraction x to the power of 0 plus 1 end exponent plus C equals x cubed plus 2 x squared minus 5 x plus C end style                  

dengan begin mathsize 14px style C end style konstanta. Selanjutnya 

begin mathsize 14px style integral subscript 1 superscript 5 left parenthesis 3 x squared plus 4 x minus 5 right parenthesis d x equals x cubed plus 2 x squared minus 5 x left enclose blank with 1 below and 5 on top end enclose equals open parentheses 5 cubed plus 2 times 5 squared minus 5 times 5 close parentheses minus open parentheses 1 cubed plus 2 times 1 squared minus 5 times 1 close parentheses equals 150 minus left parenthesis negative 2 right parenthesis right parenthesis equals 152 end style  


Jadi, begin mathsize 14px style integral subscript 1 superscript 5 left parenthesis 3 x squared plus 4 x minus 5 right parenthesis d x equals 152 end style.

0

Roboguru

Hasildari∫01​(3x2−2x+5)dx=....

Pembahasan Soal:

integral subscript 0 superscript 1 3 x squared minus 2 x plus 5 d x equals x cubed minus x squared plus 5 x left enclose table row 1 row 0 end table 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 left parenthesis 1 cubed minus left parenthesis 1 right parenthesis squared plus 5.1 right parenthesis minus 0  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 5

0

Roboguru

Hasil dari ∫01​12x3x2+1​dx adalah ....

Pembahasan Soal:

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Roboguru

Nilai dari ∫12​(x−1)(x2−2x+2)3dx adalah ....

Pembahasan Soal:

Konsep yang digunakan adalah begin mathsize 14px style integral subscript a superscript b f left parenthesis x right parenthesis d x equals F left parenthesis a right parenthesis minus F left parenthesis b right parenthesis end style dengan begin mathsize 14px style F end style adalah antiturunan dari fungsi begin mathsize 14px style f end style.  Menggunakan metode substitusi, misalkan begin mathsize 14px style x squared minus 2 x plus 2 equals p end style, dengan menurunkan kedua ruas diperoleh begin mathsize 14px style left parenthesis 2 x minus 2 right parenthesis d x equals d p end style akibatnya begin mathsize 14px style d x equals fraction numerator 1 over denominator 2 x minus 2 end fraction d p end style. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank open parentheses x minus 1 close parentheses open parentheses x squared minus 2 x plus 2 close parentheses cubed d x end cell equals cell integral subscript blank left parenthesis x minus 1 right parenthesis p cubed times fraction numerator 1 over denominator 2 x minus 2 end fraction d p end cell row blank equals cell 1 half integral subscript blank p cubed d p end cell row blank equals cell 1 half times fraction numerator 1 over denominator 3 plus 1 end fraction p to the power of 3 plus 1 end exponent plus C end cell row blank equals cell 1 over 8 p to the power of 4 plus C end cell end table end style  


Selanjutnya, dengan mensubstitusi kembali nilai undefined, diperoleh

begin mathsize 14px style integral subscript blank left parenthesis x minus 1 right parenthesis left parenthesis x squared minus 2 x plus 2 right parenthesis cubed d x equals fraction numerator 1 over denominator 8 end fraction left parenthesis x squared minus 2 x plus 2 right parenthesis to the power of 4 plus C end style  


Dengan demikian,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 1 superscript 2 left parenthesis x minus 1 right parenthesis left parenthesis x squared minus 2 x plus 2 right parenthesis cubed d x end cell equals cell 1 over 8 left parenthesis x squared minus 2 x plus 2 right parenthesis to the power of 4 left enclose blank with 1 below and 2 on top end enclose space end cell row blank equals cell open square brackets 1 over 8 left parenthesis 2 squared minus 2 times 2 plus 2 right parenthesis to the power of 4 close square brackets minus open square brackets 1 over 8 left parenthesis 1 squared minus 2 times 1 plus 2 right parenthesis to the power of 4 close square brackets space end cell row blank equals cell 16 over 8 minus 1 over 8 end cell row blank equals cell 15 over 8 end cell row blank equals cell 1 7 over 8 end cell end table end style  

 

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Roboguru

Hasil pengintegralan fungsi aljabar ∫−10​(4x−2)(1+x−x2)4dx adalah ....

Pembahasan Soal:

Integral fungsi pada soal dapat dikerjakan dengan metode substitusi. Misalkan begin mathsize 14px style 1 plus x minus x squared equals p end style maka dengan menurunkan kedua ruas diperoleh begin mathsize 14px style left parenthesis 1 minus 2 x right parenthesis d x equals d p end style sehingga begin mathsize 14px style d x equals fraction numerator 1 over denominator 1 minus 2 x end fraction d p end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis p to the power of 4 times fraction numerator 1 over denominator 1 minus 2 x end fraction d p end cell row blank equals cell negative 2 integral subscript blank p to the power of 4 d p end cell row blank equals cell negative fraction numerator 2 over denominator 4 plus 1 end fraction p to the power of 4 plus 1 end exponent plus C end cell row blank equals cell negative 2 over 5 p to the power of 5 plus C end cell end table end style           

dengan begin mathsize 14px style C end style konstanta. Selanjutnya dengan mensubstitusi kembali nilai undefined, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript blank left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell negative fraction numerator 2 over denominator 5 end fraction p to the power of 5 end exponent plus C end cell row blank equals cell negative 2 over 5 left parenthesis 1 plus x minus x squared right parenthesis to the power of 5 plus C end cell end table end style  


sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript negative 1 end subscript superscript 0 left parenthesis 4 x minus 2 right parenthesis left parenthesis 1 plus x minus x squared right parenthesis to the power of 4 d x end cell equals cell negative 2 over 5 left parenthesis 1 plus x minus x squared right parenthesis to the power of 5 left enclose blank with negative 1 below and 0 on top end enclose space end cell row blank equals cell open square brackets negative 2 over 5 left parenthesis 1 plus 0 minus 0 squared right parenthesis to the power of 5 close square brackets minus open square brackets negative 2 over 5 left parenthesis 1 plus left parenthesis negative 1 right parenthesis minus left parenthesis negative 1 right parenthesis squared right parenthesis to the power of 5 close square brackets space end cell row blank equals cell negative 2 over 5 minus 2 over 5 end cell row blank equals cell negative 4 over 5 end cell end table end style    

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