Pertanyaan

Hasil dari ....

Hasil dari $begin mathsize 14px style integral open parentheses fraction numerator x cubed minus square root of x sin invisible function application x over denominator square root of x end fraction close parentheses d x equals end style$ ....

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

Sederhanakan bentuk persamaan tersebut terlebih dahulu kemudian integralkan.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses fraction numerator x cubed minus square root of x sin invisible function application x over denominator square root of x end fraction close parentheses d x end cell equals cell integral open parentheses fraction numerator x cubed minus x to the power of 1 half end exponent times sin invisible function application x over denominator x to the power of 1 half end exponent end fraction close parentheses d x end cell row blank equals cell integral open parentheses x to the power of 5 over 2 end exponent minus sin invisible function application x close parentheses d x end cell row blank equals cell fraction numerator 1 over denominator 1 plus 5 over 2 end fraction x to the power of 1 plus 5 over 2 end exponent plus cos invisible function application x plus C end cell row blank equals cell 2 over 7 x to the power of 7 over 2 end exponent plus cos invisible function application x plus C end cell row blank equals cell 2 over 7 x cubed square root of x plus cos invisible function application x plus C end cell end table end style$