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anonymous

  • 5 years ago

find the integral S 9z square root 3z^2-7 dz

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  1. anonymous
    • 5 years ago
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    First guess, your u is under the radical. Good indication your guess is right, du is outside radical.

  2. anonymous
    • 5 years ago
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    do i divide 9 by du?

  3. anonymous
    • 5 years ago
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    Divide?! Do not pass go. Go back to square one. Are you familiar with integrating using u substitution method.

  4. anonymous
    • 5 years ago
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    this is what i have\[u: 3z ^{2}-7 du: 1/3 \]

  5. anonymous
    • 5 years ago
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    \[^9\int\limits_{}^{}\sqrt{u} = 1/3du = 9^{1/2}= 9u ^{3/2}/3/2 \times 1/3du =6u3/2 \times 1/3 =2(3z ^{2}-7)^3/2 \]

  6. anonymous
    • 5 years ago
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    du=6z dz du/6=z Don't concern yourself with trying to cancel the 9. It is just a constant and you can bring it front of the integral sign. It seems like you have been distracted by the 9 and did not get the correct derivative of u.

  7. anonymous
    • 5 years ago
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    oooh i see lol

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spraguer (Moderator)
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