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Nilai dari

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Nilai dari integral fraction numerator 2 x over denominator cube root of left parenthesis 8 minus 3 x squared right parenthesis squared end root end fraction space equals space...

Pembahasan Soal:

Dimisalkan 

u equals 8 minus 3 x squared

d u equals negative 6 x space d x d x equals fraction numerator 1 over denominator negative 6 x end fraction d u

sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator 2 x over denominator cube root of left parenthesis 8 minus 3 x squared right parenthesis squared end root end fraction space end cell equals cell space integral space left parenthesis 2 x right parenthesis space left parenthesis u to the power of negative 3 over 2 end exponent right parenthesis space left parenthesis fraction numerator 1 over denominator negative 6 x end fraction right parenthesis space d u end cell row blank equals cell integral space left parenthesis u to the power of negative 3 over 2 end exponent right parenthesis space left parenthesis fraction numerator 1 over denominator negative 3 end fraction right parenthesis space d u end cell row blank equals cell negative 1 third integral space left parenthesis u to the power of negative 3 over 2 end exponent right parenthesis space d u end cell row blank equals cell negative 1 third cross times fraction numerator 1 over denominator left parenthesis negative begin display style 3 over 2 end style plus 1 right parenthesis end fraction space u to the power of negative 3 over 2 plus 1 end exponent end cell row blank equals cell negative 1 third cross times fraction numerator 1 over denominator left parenthesis negative begin display style 1 half end style right parenthesis end fraction space u to the power of negative 1 half end exponent end cell row blank equals cell 2 over 3 u to the power of negative 1 half end exponent end cell row blank equals cell 2 over 3 cross times fraction numerator 1 over denominator square root of u end fraction end cell row blank equals cell fraction numerator 2 over denominator 3 square root of u end fraction end cell end table

Substitusikan nilai u equals 8 minus 3 x squared sehingga:

integral fraction numerator 2 x over denominator cube root of left parenthesis 8 minus 3 x squared right parenthesis squared end root end fraction space equals space fraction numerator 2 over denominator 3 square root of 8 minus 3 x squared end root end fraction

Dengan demikian, nilai dari integral fraction numerator 2 x over denominator cube root of left parenthesis 8 minus 3 x squared right parenthesis squared end root end fraction space equals space fraction numerator 2 over denominator 3 square root of 8 minus 3 x squared end root end fraction.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Indah

Mahasiswa/Alumni Universitas Diponegoro

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Hasil dari  adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator x cubed plus x squared minus 10 x plus 8 over denominator x squared minus 3 x plus 2 end fraction blank d x end cell equals cell integral fraction numerator open parentheses x plus 4 close parentheses open parentheses x minus 1 close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x minus 2 close parentheses end fraction blank d x end cell row blank equals cell integral open parentheses x plus 4 close parentheses blank d x end cell row blank equals cell integral x blank d x plus integral 4 blank d x end cell row blank equals cell open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus C subscript 1 close parentheses plus open parentheses 4 x plus C subscript 2 close parentheses end cell row blank equals cell 1 half x squared plus 4 x plus C subscript 1 plus C subscript 2 end cell end table end style 

Misalkan begin mathsize 14px style C subscript 1 plus C subscript 2 equals C comma end style maka

begin mathsize 14px style 1 half x squared plus 4 x plus C subscript 1 plus C subscript 2 equals 1 half x squared plus 4 x plus C end style 

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Operasikan perkalian dua fungsi terlebih dahulu, lalu lakukan pengintegralan.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator open parentheses x plus 5 close parentheses open parentheses x minus 5 close parentheses over denominator x squared end fraction d x end cell equals cell integral fraction numerator x squared minus 25 over denominator x squared end fraction d x end cell row blank equals cell integral open parentheses 1 minus 25 x to the power of negative 2 end exponent close parentheses d x end cell row blank equals cell x plus 25 x to the power of negative 1 end exponent plus C end cell row blank equals cell x plus 25 over x plus C comma space text dengan end text space C space text konstanta end text end cell end table end style 

Dengan demikian, hasil dari x2(x+5)(x5)dx=x+x25+C.

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Roboguru

Hasil dari

Pembahasan Soal:

integral open parentheses fraction numerator x to the power of 6 minus x cubed plus 3 over denominator x cubed end fraction close parentheses d x equals integral open parentheses x to the power of 6 over x cubed minus x cubed over x cubed plus 3 over x cubed close parentheses d x  equals integral open parentheses x cubed minus 1 plus 3 over 2 x to the power of negative 3 end exponent close parentheses d x  equals 1 fourth x to the power of 4 minus x minus 3 over 2 x to the power of negative 2 end exponent plus c  equals 1 fourth x to the power of 4 minus x minus fraction numerator 3 over denominator 2 x squared end fraction plus c

0

Roboguru

...

Pembahasan Soal:

Uraikan fungsi aljabar tersebut kemudian gunakan konsep integral seperti berikut:

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

Oleh karena itu jawaban yang benar adalah B.

0

Roboguru

Pembahasan Soal:

Integral pada soal merupakan integral tak tentu yang didefinisikan sebagai integral f left parenthesis x right parenthesis d x equals F left parenthesis x right parenthesis plus C. fungsi fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction dapat diubah menjadi fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction equals fraction numerator 2 x cubed over denominator x squared end fraction minus fraction numerator 3 x squared over denominator x squared end fraction plus 1 over x squared equals 2 x minus 3 plus x to the power of negative 2 end exponent comma sehingga diperoleh
 

integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction text d end text x equals integral 2 x minus 3 plus x to the power of negative 2 end exponent text d end text x


Berdasarkan sifat integral tak tentu yaitu integral x to the power of n d x equals fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus C comma n not equal to 1 comma maka
 

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


Oleh karena itu, jawaban yang benar adalah B.

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