Evaluating the integral, correct?

[Zack88]

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

 

[rocomath]

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

 

[Zack88]

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.

 

[rocomath]

Yes

$$x^2+\frac b a x + \frac c a=0$$

 

[Zack88]

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

 

[rocomath]

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.

$$x^2-6x+13$$

$$x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13$$

 

[Zack88]

$$x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13$$

wouldnt the 9 - 9 cancel out? so ur still left with the original problem.

 

[rocomath]

would ntthe 9 - 9 cancel out? so ur still left with the original problem.

Yep, and from there it simplifies easiy.

 

[Zack88]

lol so in other words was the square already completed?

 

[rocomath]

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.

 

[Zack88]

k cya thanks again