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Say we have two functions $f$ and $g$ such that:

$\int _{a}^{b} [f(x) -g(x) ]dx = N$ where $N$ is any real number

So for the two functions:

$k(x) = f(x) + c$ and $h(x)=g(x) + c$ can we still say:

$\int _{a}^{b} [k(x) -h(x) ]dx = N$ ?

I'm thinking yes but I'm not sure.

Thanks for the help

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$\int _{a}^{b} [k(x) -h(x) ]dx =\int _{a}^{b} [f(x)+c-(g(x)+c) ]dx=\int _{a}^{b} [f(x)+c-g(x)-c ]dx=\int _{a}^{b} [f(x) -g(x) ]dx =N$

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  • $\begingroup$ Thanks! I understand now $\endgroup$ – Dahen Sep 4 '17 at 8:57
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Yes, because $(\forall x\in[a,b]):|k(x)-h(x)|=|f(x)-g(x)|$.

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  • $\begingroup$ Tbh i don't get the notation since I haven't studied it yet, could you maybe write it in terms of integrals? $\endgroup$ – Dahen Sep 4 '17 at 8:55
  • $\begingroup$ @user464154 $\int_a^b|k(x)-h(x)|\,\mathrm dx|=\int_a^b|f(x)+c-g(x)-c|\,\mathrm dx=\int_a^b|f(x)-g(x)|\,\mathrm dx$. $\endgroup$ – José Carlos Santos Sep 4 '17 at 8:57
  • $\begingroup$ Alright thanks! $\endgroup$ – Dahen Sep 4 '17 at 8:58
  • $\begingroup$ @user464154 If my answer was helpful, perhaps that you might mark it as the accepted one. $\endgroup$ – José Carlos Santos Sep 4 '17 at 8:59
  • $\begingroup$ It tells me I cant mark it as accepted until 6 minutes have passed, but i'll be sure to when it lets me $\endgroup$ – Dahen Sep 4 '17 at 9:00
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Think geometrically: if you move both graphs upp by $c$ units - the whole area between them remains unchanged. Area is invariant under translations.

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  • $\begingroup$ Since niether the shape nor size of region will change , the area will remain the same , right? Okay, thanks! $\endgroup$ – Dahen Sep 4 '17 at 9:07
  • $\begingroup$ @user464154 Yes, since you move all the points in the same direction by the same distance. It is the same as to moving the picture as a whole. $\endgroup$ – A.Γ. Sep 4 '17 at 9:10

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