Evaluating the integral, correct?

  • Thread starter Zack88
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    Integral
In summary: You should have done \int e^{-x}\cos{2x}dxinstead of\int e^{-x}\cos{2x}dxIn summary, the integrand in the homework equation is x^2/m sinmx. Evaluating this integral requires multiple stages of integration by parts. The 'm' is a constant, so that will not be involved in the integration. Making the integral (1/m) · integral[ x sin(mx) ] dx . Now u = x and dv =
  • #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
 
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  • #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.

[tex]ax^2+bx+c=0[/tex]

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

[tex]x^2+\frac b a x + \frac c a=0[/tex]
 
  • #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.

[tex]x^2-6x+13[/tex]

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

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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