I understand that the fundamental theorem of calculus is essentially the statement that the derivative of the anti-derivative [itex]F[/itex] evaluated at [itex]x\in (a,b)[/itex] is equal to the value of the primitive function (integrand) [itex]f[/itex] evaluated at [itex]x\in (a,b)[/itex], i.e. [itex]F'(x)=f(x)[/itex]. However, can one imply directly from this that [itex]F'=f\;\;\forall x\in (a,b)[/itex], such that [tex]F(b)-F(a)=\int_{a}^{b}F'(x)dx\;\;?[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Also, I have attempted to prove the FTC and am interested to know whether my attempt is valid?!

Let [itex]f[/itex] be a continuous function on [itex][a,b][/itex] and let [itex]F: \mathbb{R}\rightarrow\mathbb{R}[/itex] be a function that is continuous on [itex][a,b][/itex] and differentiable on [itex](a,b)[/itex], defined such that [tex]F(x)=\int_{a}^{x}f(t)dt[/tex] where [itex]x\in (a,b)[/itex].

As [itex]f[/itex] is continuous on [itex][a,b][/itex] we therefore know that for any [itex]\varepsilon >0\;\;\exists\delta >0[/itex] such that [tex]0<\vert t-x\vert <\delta\;\;\Rightarrow\;\;\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall x \in [a,b][/tex]

Hence, consider the interval [itex](x,x+h)\subseteq (a,b)[/itex] and let [itex]0<h<\delta[/itex]. Now clearly [itex]x<t<x+h\;\;\Rightarrow\;\;0<t-x<h<\delta[/itex] and so [tex]\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x,x+h)[/tex] which implies that in this interval [tex]f(x)-\varepsilon<f(t)<f(x)+\varepsilon[/tex] Using that [itex]f(x)\leq g(x)\;\;\Rightarrow\;\;\int_{a}^{b}f(x)dx\leq \int_{a}^{b}g(x)dx[/itex] it follows that [tex]\int_{x}^{x+h}\left[f(x)-\varepsilon\right]dt<\int_{x}^{x+h}f(t)dt<\int_{x}^{x+h}\left[f(x)+\varepsilon\right]dt \\ \Rightarrow \left[f(x)-\varepsilon\right]h<\int_{x}^{x+h}f(t)dt<\left[f(x)+\varepsilon\right]h \\ \Rightarrow f(x)-\varepsilon<\frac{1}{h}\int_{x}^{x+h}f(t)dt<f(x)+\varepsilon \\ \Rightarrow\Bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt - f(x)\Bigg\vert <\varepsilon [/tex] and as such the limit [tex]\lim_{h\rightarrow 0^{+}}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x)[/tex] exists.

Next we consider the interval [itex](x-h,x)\subseteq (a,b)[/itex] and again let [itex]0<h<\delta[/itex]. We see that [itex]x-h<t<x\;\;\Rightarrow\;\;-h<t-x<0\;\;\Rightarrow\;\;0<-(t-x)<h[/itex] and so, following the same procedure as before [tex]\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x-h,x) \\ \Rightarrow\int_{x-h}^{x}\left[f(x)-\varepsilon\right]dt<\int_{x-h}^{x}f(t)dt<\int_{x-h}^{x}\left[f(x)+\varepsilon\right]dt \\ \Rightarrow \left[f(x)-\varepsilon\right]h<\int_{x-h}^{x}f(t)dt<\left[f(x)+\varepsilon\right]h\\ \Rightarrow f(x)-\varepsilon<\frac{1}{h}\int_{x-h}^{x}f(t)dt<f(x)+\varepsilon \\ \Rightarrow\Bigg\vert \frac{1}{h}\int_{x-h}^{x}f(t)dt - f(x)\Bigg\vert <\varepsilon [/tex] and as such the limit [tex]\lim_{h\rightarrow 0^{-}}\frac{1}{h}\int_{x-h}^{x}f(t)dt=f(x)[/tex] exists.

Therefore, as the right-handed and left-handed limits exist and are equal, we have that for any [itex]\varepsilon >0\;\;\exists\delta >0[/itex] such that [tex]\left(0<\vert t-x\vert <h<\delta\;\;\Rightarrow\;\; 0<h<\delta\right)\;\;\Rightarrow\;\;\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert <\varepsilon[/tex] i.e. the limit [tex]\lim_{h\rightarrow 0}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x)[/tex] exists. As [tex]\frac{1}{h}\int_{x}^{x+h}f(t)dt= F(x+h)-F(x)[/tex] this implies that [tex]\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x)=f(x)[/tex]

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# Fundamental theorem of calculus - question & proof verifying

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