- #1
Natasha1
- 494
- 9
Am I right in saying that the integral of 1/(x-6)^2 dx
= (x-6)^-1 / -1
= - 1 / (x-6)
= (x-6)^-1 / -1
= - 1 / (x-6)
Simplifying the Integral of 1/(x-6)^2 and Checking for Accuracy
- Thread starter Natasha1
- Start date
-
- Tags
- Integral
Click For Summary
Homework Help Overview
The discussion revolves around the integral of the function 1/(x-6)^2, with participants exploring the simplification of the integral and the importance of the integration constant.
Discussion Character
- Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the simplification of the integral and the necessity of including an integration constant. Questions arise regarding the nature and significance of the integration constant in the context of indefinite integrals.
Discussion Status
Some participants have provided explanations regarding the integration constant and its role in representing a family of functions. The conversation includes attempts to clarify the concept without reaching a definitive conclusion on the broader implications.
Contextual Notes
Participants are engaging with the concept of integration in a homework context, with an emphasis on understanding rather than solving the integral completely. There is a focus on the implications of the integration constant and its necessity in representing multiple functions.
- #2
siddharth
Homework Helper
Gold Member
- 1,145
- 0
Add the Integration constant, and you are correct.
- #3
Natasha1
- 494
- 9
siddharth said:
I will thanks very much :-)... oh could someone explain exactly what this integral constant is? And why is it needed?
- #4
TD
Homework Helper
- 1,016
- 0
When we write [itex]\int x dx[/itex] we're looking for the function(s) which give x again if we take the derivative. Now, you know this integral is x²/2 since (x²/2)' = x but when you add a constant, it's still valid: (x²/2 + c)' = x. This is because the derivative of a constant is zero.
- #5
Natasha1
- 494
- 9
TD said:
thanks TD for your explanation
- #6
TD
Homework Helper
- 1,016
- 0
You're welcome 
- #7
rea
- 17
- 0
If you have a function
f(x) = x^2 there are other functions of the same shape that difer only in the position respect "y"...
if f(x)=x; is like if you have f(x)=x+0=x+C, the +C let yo translate up or down the graph of the function.
Is necesary when integrating a function, because you don't know which of this infinite posible values to C is the one, thus, instead of write a infinite set of functions that only difer in the a real number "you select" one specific function of those many posibles that represent such infinite funtions and call it F(x)+C, thus ;), you have one function that can represent any function with the same shape and that only difer in the y position instead of have a infinite number of functions to analise.
f(x) = x^2 there are other functions of the same shape that difer only in the position respect "y"...
if f(x)=x; is like if you have f(x)=x+0=x+C, the +C let yo translate up or down the graph of the function.
Is necesary when integrating a function, because you don't know which of this infinite posible values to C is the one, thus, instead of write a infinite set of functions that only difer in the a real number "you select" one specific function of those many posibles that represent such infinite funtions and call it F(x)+C, thus ;), you have one function that can represent any function with the same shape and that only difer in the y position instead of have a infinite number of functions to analise.