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Nilai adalah

Nilai integral subscript 0 superscript 2 left parenthesis 3 x plus 9 right parenthesis square root of x squared plus 6 x end root space d x adalah

  1. 4

  2. 8

  3. 16

  4. 32

  5. 64

Jawaban:

i)     Gunakan integral substitusi

integral subscript 0 superscript 2 open parentheses 3 x plus 9 close parentheses square root of x squared plus 6 x end root d x

Substitusi y space equals space x squared plus 6 x, maka

d y space equals space 2 x plus 6 space d x  1 half d y space equals space x plus 3 space d x  3 over 2 d y space equals space 3 x plus 9 space d x

Untuk x = 0, maka y space equals space 0 squared plus 6 open parentheses 0 close parentheses space equals space 0.

Untuk x = 2, maka y space equals space 2 squared plus 6 open parentheses 2 close parentheses space equals space 16 .

Sehingga

integral subscript 0 superscript 2 open parentheses 3 x plus 9 close parentheses square root of x squared plus 6 x end root d x space equals space integral subscript 0 superscript 2 square root of x squared plus 6 x end root open parentheses 3 x plus 9 close parentheses d x space equals space integral subscript 0 superscript 16 square root of y open parentheses 3 over 2 d y close parentheses space equals space 3 over 2 space integral subscript 0 superscript 16 y to the power of 1 half end exponent d y  equals right enclose fraction numerator begin display style 3 over 2 end style y to the power of begin display style 1 half end style plus 1 end exponent over denominator begin display style 1 half end style plus 1 end fraction end enclose subscript 0 superscript 16 space equals space right enclose fraction numerator begin display style 3 over 2 y to the power of 3 over 2 end exponent end style over denominator begin display style 3 over 2 end style end fraction end enclose subscript 0 superscript 16 space equals space right enclose y to the power of begin inline style 3 over 2 end style end exponent end enclose subscript 0 superscript 16 space equals space 16 to the power of begin inline style 3 over 2 end style end exponent space minus space 0 to the power of begin inline style 3 over 2 end style end exponent space equals space 16 square root of 16 minus 0 space equals space 64

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