why do we put arbitrary constant in integration

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- May 25th 2010, 03:27 AMprasumintegration constant
why do we put arbitrary constant in integration

- May 25th 2010, 03:32 AMpickslides
Differentiate these functions

$\displaystyle y_1 = x^2+10x+7$

$\displaystyle y_2 = x^2+10x-9$

$\displaystyle y_3 = x^2+10x+2$

What do you get? What does this mean? - May 25th 2010, 05:28 AMAlso sprach ZarathustraHey there...
It's...

**The fundamental theorem of calculus:**

$\displaystyle F(x)=\int_a^x{f(t)dt}$

where $\displaystyle f$ is a real-valued function continuous on $\displaystyle [a, b]$. Then, $\displaystyle F$ is continuous on $\displaystyle [a, b]$, differentiable on the open interval $\displaystyle (a, b)$, and:

$\displaystyle F'(x)=f(x)$ $\displaystyle \\\\$ $\displaystyle \forall x \in (a,b)$ - May 25th 2010, 03:58 PMTheCoffeeMachine
Say you want to integrate $\displaystyle f(x) = x^2$. Writing $\displaystyle \int{2x}\;{dx} = x^2$ means to say that the function $\displaystyle f(x) = 2x$ satisfies $\displaystyle F'(x) = x^2$. But is it the only one that satisfies this relation? Clearly, $\displaystyle 2x+1$ satisfies; so do $\displaystyle 2x+2$ and $\displaystyle 2x+3$. In fact, for any number $\displaystyle C$, it's true that $\displaystyle 2x+C$ satisfies the relation. Finding $\displaystyle \int{f(x)}\;{dx} $ means finding the set of all function which satisfy F'(x) = f(x), i.e the set of all anti-derivatives of f(x).

- May 25th 2010, 05:16 PMpickslides
- May 25th 2010, 05:36 PMTheCoffeeMachine