Looking at the 3rd term
The restricted value here refers to the part of the function as
I was working on a problem (solution shown):
and checking my work with wolfram alpha to find that there is a "restricted x-value" constant that is added in to the end of the integral, in this case 4/9. I don't know how this value was calculated, but Wolfram has the steps it took here: http://www.wolframalpha.com/input/po...e46hi91b8&s=57
Could someone possibly explain the use of this constant and how it was calculated, since it can be eliminated anyway if C=-4/9?
Also, are there other cases when this need be calculated? Could you point me to some places where I can get some practice?
Help is greatly appreciated!
1. Why is 0 chosen to be plugged into the equation?
2. Whats the purpose of the restriction term if it can just be eliminated if C=-4/9?
I meant that a value of x was plugged into the function so that it would satisfy log (0), which would effectively eliminate the term with the log in it, and when that same x value is plugged into the other term (I believe x=-2/3 satisfies this), then one ends up with 4/9. At least thats how I think you get 4/9. How do you really get it? Thats what I asked originally.Where was 0 plugged in? I do not follow this question.
I don't see the point of adding a constant to account for restricted x-values when the constant can be theoretically eliminated if C is the opposite of that constant. C does not have to be -4/9, but if it were then it would just cancel out the restrictive term, so I'm asking what the point of the restrictive term is. I don't really understand what it accounts for either. Could you please explain?How is c = -4/9? Where you given some initial conditions?
I am saying (read above) not
In your case not
This part is the restriction.
It has nothing to do with the
If you like you can say.
Hope this helps some. I understand it can be tricky!