Fundamental Theorem of Calculus with Functions












0












$begingroup$


Please excuse the weird title. I was unsure about how to summarize this.



So lets say we have this integral:



$ int_{kx}^{cx} t^2 dt = f(x)$



If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$



or



$ int_{2x}^{2x} t^2 dt = f(2x)$



Edit:



If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$






calculus integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is wrong with the result? Please explain your problem a little bit more detailed.
    $endgroup$
    – Evan William Chandra
    Jan 29 at 5:26










  • $begingroup$
    What's your question?
    $endgroup$
    – Stefan Lafon
    Jan 29 at 5:26










  • $begingroup$
    I'm just asking to confirm if what I have is right.
    $endgroup$
    – user2793618
    Jan 29 at 5:27










  • $begingroup$
    I've updated for clarity.
    $endgroup$
    – user2793618
    Jan 29 at 5:29










  • $begingroup$
    In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
    $endgroup$
    – Ryan Goulden
    Jan 29 at 7:40
















0












$begingroup$


Please excuse the weird title. I was unsure about how to summarize this.



So lets say we have this integral:



$ int_{kx}^{cx} t^2 dt = f(x)$



If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$



or



$ int_{2x}^{2x} t^2 dt = f(2x)$



Edit:



If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$






calculus integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is wrong with the result? Please explain your problem a little bit more detailed.
    $endgroup$
    – Evan William Chandra
    Jan 29 at 5:26










  • $begingroup$
    What's your question?
    $endgroup$
    – Stefan Lafon
    Jan 29 at 5:26










  • $begingroup$
    I'm just asking to confirm if what I have is right.
    $endgroup$
    – user2793618
    Jan 29 at 5:27










  • $begingroup$
    I've updated for clarity.
    $endgroup$
    – user2793618
    Jan 29 at 5:29










  • $begingroup$
    In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
    $endgroup$
    – Ryan Goulden
    Jan 29 at 7:40














0












0








0





$begingroup$


Please excuse the weird title. I was unsure about how to summarize this.



So lets say we have this integral:



$ int_{kx}^{cx} t^2 dt = f(x)$



If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$



or



$ int_{2x}^{2x} t^2 dt = f(2x)$



Edit:



If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




Please excuse the weird title. I was unsure about how to summarize this.



So lets say we have this integral:



$ int_{kx}^{cx} t^2 dt = f(x)$



If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$



or



$ int_{2x}^{2x} t^2 dt = f(2x)$



Edit:



If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 5:30







user2793618

















asked Jan 29 at 5:17









user2793618user2793618

1047




1047












  • $begingroup$
    what is wrong with the result? Please explain your problem a little bit more detailed.
    $endgroup$
    – Evan William Chandra
    Jan 29 at 5:26










  • $begingroup$
    What's your question?
    $endgroup$
    – Stefan Lafon
    Jan 29 at 5:26










  • $begingroup$
    I'm just asking to confirm if what I have is right.
    $endgroup$
    – user2793618
    Jan 29 at 5:27










  • $begingroup$
    I've updated for clarity.
    $endgroup$
    – user2793618
    Jan 29 at 5:29










  • $begingroup$
    In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
    $endgroup$
    – Ryan Goulden
    Jan 29 at 7:40


















  • $begingroup$
    what is wrong with the result? Please explain your problem a little bit more detailed.
    $endgroup$
    – Evan William Chandra
    Jan 29 at 5:26










  • $begingroup$
    What's your question?
    $endgroup$
    – Stefan Lafon
    Jan 29 at 5:26










  • $begingroup$
    I'm just asking to confirm if what I have is right.
    $endgroup$
    – user2793618
    Jan 29 at 5:27










  • $begingroup$
    I've updated for clarity.
    $endgroup$
    – user2793618
    Jan 29 at 5:29










  • $begingroup$
    In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
    $endgroup$
    – Ryan Goulden
    Jan 29 at 7:40
















$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26




$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26












$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26




$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26












$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27




$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27












$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29




$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29












$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40




$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40










1 Answer
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$begingroup$

$f(2x)=int_{2xk}^{2xc} t^2 dt$



But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.



The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$



But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$






share|cite|improve this answer











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    $begingroup$

    $f(2x)=int_{2xk}^{2xc} t^2 dt$



    But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.



    The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$



    But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $f(2x)=int_{2xk}^{2xc} t^2 dt$



      But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.



      The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$



      But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $f(2x)=int_{2xk}^{2xc} t^2 dt$



        But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.



        The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$



        But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$






        share|cite|improve this answer











        $endgroup$



        $f(2x)=int_{2xk}^{2xc} t^2 dt$



        But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.



        The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$



        But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 1 at 9:17

























        answered Jan 29 at 6:38









        Fareed AFFareed AF

        63212




        63212






























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