# integration constant

• May 25th 2010, 03:27 AM
prasum
integration constant
why do we put arbitrary constant in integration
• May 25th 2010, 03:32 AM
pickslides
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 AM
Also sprach Zarathustra
Hey 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 PM
TheCoffeeMachine
Quote:

Originally Posted by prasum
why do we put arbitrary constant in integration

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 PM
pickslides
Quote:

Originally Posted by TheCoffeeMachine
Writing $\displaystyle \int{x^2}\;{dx} = 2x$

(Surprised)
• May 25th 2010, 05:36 PM
TheCoffeeMachine
Quote:

Originally Posted by pickslides
(Surprised)

(Swear). Thank you. (Itwasntme)