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Thread: integration constant

  1. #1
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    integration constant

    why do we put arbitrary constant in integration
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  2. #2
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    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?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    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:
    $\displaystyle F'(x)=f(x)$ $\displaystyle \\\\$ $\displaystyle \forall x \in (a,b)$
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    Quote Originally Posted by prasum View Post
    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).
    Last edited by TheCoffeeMachine; May 25th 2010 at 05:35 PM.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    Writing $\displaystyle \int{x^2}\;{dx} = 2x$
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  6. #6
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    Quote Originally Posted by pickslides View Post
    . Thank you.
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