anonymous
  • anonymous
find the integral S 9z square root 3z^2-7 dz
Mathematics
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anonymous
  • anonymous
find the integral S 9z square root 3z^2-7 dz
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
First guess, your u is under the radical. Good indication your guess is right, du is outside radical.
anonymous
  • anonymous
do i divide 9 by du?
anonymous
  • anonymous
Divide?! Do not pass go. Go back to square one. Are you familiar with integrating using u substitution method.

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anonymous
  • anonymous
this is what i have\[u: 3z ^{2}-7 du: 1/3 \]
anonymous
  • anonymous
\[^9\int\limits_{}^{}\sqrt{u} = 1/3du = 9^{1/2}= 9u ^{3/2}/3/2 \times 1/3du =6u3/2 \times 1/3 =2(3z ^{2}-7)^3/2 \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
oooh i see lol

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