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Hasil dari begin mathsize 14px style 1 half integral e to the power of square root of x end exponent d x equals end style....

  1. begin mathsize 14px style 1 half e to the power of square root of x end exponent open parentheses square root of x minus 1 close parentheses plus C end style

  2. begin mathsize 14px style e to the power of square root of x end exponent open parentheses square root of x minus 1 close parentheses plus C end style

  3. begin mathsize 14px style e to the power of square root of x end exponent open parentheses square root of x plus 1 close parentheses plus C end style

  4. begin mathsize 14px style 2 e to the power of square root of x end exponent open parentheses square root of x minus 1 close parentheses plus C end style

  5. begin mathsize 14px style 2 e to the power of square root of x end exponent open parentheses square root of x plus 1 close parentheses plus C end style 

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

jawaban yang benar adalah B. 

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Pembahasan

Misalkan . Oleh karena itu, didapat Maka, Kemudian kita gunakan integral parsial untuk melanjutkan integral bentuk di atas. Misalkan u = p dan Didapat bahwa dan Sehingga Jadi, jawaban yang benar adalah B.

Misalkan begin mathsize 14px style p equals square root of x equals x to the power of 1 half end exponent end style. Oleh karena itu, didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator d p over denominator d x end fraction end cell equals cell 1 half x to the power of negative 1 half end exponent blank end cell row cell fraction numerator d p over denominator d x end fraction end cell equals cell fraction numerator 1 over denominator 2 x to the power of 1 half end exponent end fraction blank end cell row cell integral fraction numerator d p over denominator d x end fraction d x end cell equals cell integral fraction numerator 1 over denominator 2 x to the power of 1 half end exponent end fraction d x end cell row cell integral d p end cell equals cell integral fraction numerator 1 over denominator 2 x to the power of 1 half end exponent end fraction d x end cell end table end style 

Maka,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 half integral e to the power of square root of x end exponent d x end cell equals cell 1 half integral open parentheses e to the power of square root of x end exponent times fraction numerator 2 x to the power of 1 half end exponent over denominator 2 x to the power of 1 half end exponent end fraction close parentheses d x end cell row blank equals cell 1 half integral open parentheses e to the power of square root of x end exponent times 2 x to the power of 1 half end exponent times fraction numerator 1 over denominator 2 x to the power of 1 half end exponent end fraction close parentheses d x end cell row blank equals cell 1 half integral open parentheses e to the power of p times 2 p close parentheses d p end cell row blank equals cell integral open parentheses e to the power of p times p close parentheses d p end cell end table end style 

Kemudian kita gunakan integral parsial untuk melanjutkan integral bentuk di atas. Misalkan

u = p

dan

begin mathsize 14px style fraction numerator d v over denominator d p end fraction equals e to the power of p end style

Didapat bahwa

begin mathsize 14px style fraction numerator d u over denominator d p end fraction equals 1 end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row v equals cell integral fraction numerator d v over denominator d p end fraction d p end cell row blank equals cell integral e to the power of p d p end cell row blank equals cell e to the power of p 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 e to the power of square root of x end exponent d x end cell equals cell integral e to the power of p times p blank d p end cell row blank equals cell integral open parentheses u times fraction numerator d v over denominator d p end fraction close parentheses d p end cell row blank equals cell open parentheses u v minus integral open parentheses v times fraction numerator d u over denominator d p end fraction close parentheses d p close parentheses end cell row blank equals cell open parentheses p times e to the power of p minus integral open parentheses e to the power of p times 1 close parentheses d p close parentheses end cell row blank equals cell p times e to the power of p minus e to the power of p plus C end cell row blank equals cell e to the power of p open parentheses p minus 1 close parentheses plus C end cell row blank equals cell e to the power of u open parentheses u minus 1 close parentheses plus C end cell row blank equals cell e to the power of square root of x end exponent open parentheses square root of x minus 1 close parentheses plus C blank end cell end table end style 

Jadi, jawaban yang benar adalah B. 

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