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Nilai ∫ 0 1 ​ ( x 3 + 2 x − 5 ) d x = ....

Nilai  = ....

  1. begin mathsize 12px style negative 16 over 4 end style

  2. begin mathsize 12px style negative 15 over 4 end style

  3. 0

  4. begin mathsize 12px style 15 over 4 end style

  5. begin mathsize 12px style 16 over 4 end style

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Y. Laksmi

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

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Pembahasan

begin mathsize 12px style integral subscript 0 superscript 1 open parentheses x cubed plus 2 x minus 5 close parentheses d x  equals open square brackets fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent plus 2 times fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent minus 5 x close square brackets blank with 0 below and 1 on top  equals open square brackets 1 fourth x to the power of 4 plus x squared minus 5 x close square brackets blank with 0 below and 1 on top  equals open parentheses 1 fourth left parenthesis 1 right parenthesis to the power of 4 plus open parentheses 1 close parentheses squared minus 5 open parentheses 1 close parentheses close parentheses minus open parentheses 1 fourth open parentheses 0 close parentheses to the power of 4 plus open parentheses 0 close parentheses squared minus 5 open parentheses 0 close parentheses close parentheses  equals open parentheses 1 fourth plus 1 minus 5 close parentheses minus left parenthesis 0 right parenthesis  equals negative 15 over 4 end style

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Lukiskan sketsa grafik y = x 15 ​ untuk domain 3 ≤ x ≤ 5 . Tunjukkan bahwa: 6 < ∫ 0 5 ​ x 15 ​ d x < 10 .

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