Evaluating the integral, correct?

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    Integral
  • #101
Zack88 said:
sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)
It wasn't so much irritating, just disappointing. I just want you to get a real good grade :-] I'm mainly pushing you to check your answer to exericse your differentiating skills. I'm more than happy to help!
 
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  • #102
hey when completing the square when a square roots is involved, do you square both sides?
 
  • #103
Example ...

x^2-\sqrt 2 x+4

This kind of problem?
 
  • #104
more like 1 / [sqrt]9x^2 + 6x -8[/sqrt]
 
  • #105
You don't have to worry about squaring the square root. Just work within your square root and perform your operation as you would.

But don't forget that your coefficient of your highest degree must be 1.
 
  • #106
rocophysics said:
But don't forget that your coefficient of your highest degree must be 1.

I don't know what that means, since i work with the square root do I not pay attention to it until finally getting the final answer. ex.

[sqrt] x^3 + 6x^2 - 4 [/sqrt]

then

x^3 + 6x^2 = 4
x^3 + 6x^2 + 9 = 4 + 9

and so on
 
  • #107
You can't complete the square of a polynomial of degree 3.

I meant that the standard quadratic equation needs to have a coefficient of 1 in front of it's highest term.

ax^2+bx+c=0

a=1
 
  • #108
oh ok and i just made up the example so if i have a number higher than 1 I need to divide that number out so ex. 5x^2 + 6x + 5 and then id divide the whole thing by 5.
 
  • #109
Yes

x^2+\frac b a x + \frac c a=0
 
  • #110
ok came across a bump.

[integral] dx/ [sqrt]x^2 - 6x + 13[/sqrt]

so i have to complete the square, so

x^2 - 6x + 13 = 0
x^2 - 6x = -13

wait before i continue I always thought anything that was negative and squared it become positive but now my calculator is telling me differently.

the i divide -6 by 2 and then square it (using everything squared is positive) and get 9

x^2 - 6x + 9 = -13 + 9
x^2 + 6x +9 + -4
(x-3)^2 = =4

then square both sides but you can't square -4 w/o getting an imaginary number
 
  • #111
What happen here?

x^2 - 6x + 9 = -13 + 9
x^2 + 6x +9 + -4
(x-3)^2 = =4

Don't set it equal to 0.

x^2-6x+13

x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13
 
  • #112
rocophysics said:
x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13

wouldnt the 9 - 9 cancel out? so ur still left with the original problem.
 
  • #113
Zack88 said:
would ntthe 9 - 9 cancel out? so ur still left with the original problem.
Yep, and from there it simplifies easiy.
 
  • #114
lol so in other words was the square already completed?
 
  • #115
Zack88 said:
lol so in other words was the square already completed?
Not completely. I'm going now though, go to this thread ... Find the integral...Please HELP! - https://www.physicsforums.com/showthread.php?t=211212

Post 12, pretty much your problem, just different numbers.
 
  • #116
k cya thanks again
 
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