## anonymous 5 years ago find the integral S 9z square root 3z^2-7 dz

1. anonymous

2. anonymous

do i divide 9 by du?

3. anonymous

Divide?! Do not pass go. Go back to square one. Are you familiar with integrating using u substitution method.

4. anonymous

this is what i have$u: 3z ^{2}-7 du: 1/3$

5. anonymous

$^9\int\limits_{}^{}\sqrt{u} = 1/3du = 9^{1/2}= 9u ^{3/2}/3/2 \times 1/3du =6u3/2 \times 1/3 =2(3z ^{2}-7)^3/2$

6. 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.

7. anonymous

oooh i see lol