anonymous
  • anonymous
Evaluate: anti derivative Square root (3x+9x) dx would the answer be square root (3/2x^2+9x) + c
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
sqrt(12x) .... is that correct?
anonymous
  • anonymous
\[\sqrt{3x+9x}\]
anonymous
  • anonymous
Is that what you have?

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anonymous
  • anonymous
\[\sqrt{3x + 9} dx\]
anonymous
  • anonymous
Ah. Have you done u substitutions yet?
amistre64
  • amistre64
(3x + 9)^1/2 you need a 3, so multiply it by 3/3 and draw out the (1/3)
amistre64
  • amistre64
(1/3) [S] 3 (3x+9)^1/2
amistre64
  • amistre64
1/3 [S] u^1/2 du is what it becomes
anonymous
  • anonymous
no I have not I am having trouble with this section. Not sure when I need to substitute
amistre64
  • amistre64
2sqrt(u^3)/9 if I did it right
amistre64
  • amistre64
u = 3x+9 :)
amistre64
  • amistre64
..+C...
amistre64
  • amistre64
after that last "bout" my brain is acting like a neanderthal
radar
  • radar
amistre64 you've the patient of Job!
amistre64
  • amistre64
lol....thanx ;)
anonymous
  • anonymous
As a u substitution you would just say \[u = 3x+9 \implies du = 3dx \implies dx = \frac{du}{3}\] Then \[\int \sqrt{3x+9}\ dx = \int \sqrt{u} * \frac{1}{3}du\] Then take the anti derivative of the u version and when you are done substitute back in 3x+9 for u.
anonymous
  • anonymous
ok but why will I use substitution. What about this problem tells me to substitute?
amistre64
  • amistre64
that "3x+9" under the radical is the clue. in order to derive down to this state, we had to use the chain rule. And that part is now missing from this equation; we essentially need to get it back.
amistre64
  • amistre64
Dx(3x+9) = 3 ..... so we need to account for a "missing" 3
amistre64
  • amistre64
if u = 3x+9 ; then du = 3 dx and dx= du/3
amistre64
  • amistre64
sqrt(u) du -------- is what we can work out easily 3
anonymous
  • anonymous
When you are finding anti-derivatives of a function composed of another function \[\int f(g(x)) dx\] You will want to use a u substitution. In this case \[f(a) = \sqrt{a}\] and \[g(x) = 3x+9\] And you're finding the anti-derivative f(g(x) dx \[f(g(x)) = f(3x+9) = \sqrt{3x+9}\]
anonymous
  • anonymous
is the answer (3x+9)^2/3 / 9 +C
anonymous
  • anonymous
Not quite. What did you get for the anti derivative of \[\frac{1}{3}\sqrt{u}\ du?\]
anonymous
  • anonymous
Actually I see where your mistake was. You need to bring down the reciprocal of the exponent _after_ you add 1 to it.
anonymous
  • anonymous
Actually, it looks like you did that. I think it was just a book keeping mistake.
anonymous
  • anonymous
You're missing a factor of 2 on the front. And your exponent is flipped the wrong way. It should be: \[\int \frac{\sqrt{u}}{3}du \] \[= \int \frac{u^{1/2}}{3} du\] \[= \frac{2}{3}*\frac{u^{3/2}}{3} + C\] \[= \frac{2(3x+9)^{3/2}}{9} + C\]
anonymous
  • anonymous
I see what you are doing but it just isn't clicking for me, but I am working on it. The 2/3 * u^3/2/3 The 2/3 is just the recp of 3/2???
anonymous
  • anonymous
Yes. Recall that the anti-derivative of \[x^a \implies \frac{1}{a+1}*x^{a+1} + C\] We can confirm this is true, by taking the derivative of the right side and verifying that we get the left side.
anonymous
  • anonymous
Where a and C are constants.
anonymous
  • anonymous
I am going to start a new question
anonymous
  • anonymous
Ok!

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