Proof.

continuity
If \(f\in R[a,b]\), then \(f\) is bounded on \([a,b]\) and so is \(\begin {vmatrix} f\\ \end {vmatrix}\). Let \(M\) be the upper bound of \(\begin {vmatrix} f\\ \end {vmatrix}\) on \([a,b]\).Then given \(\varepsilon > 0\, , \, \exists \, \delta > 0\, \ni \, 0 < M\,\delta < \varepsilon \,\) and if \(x\in [a,b],\hspace {0.2cm}x + h\in [a,b]\) and \(\begin {vmatrix} h\\ \end {vmatrix} < \delta \), then \begin {align*} \begin {vmatrix} F(x + h) - F(x)\\ \end {vmatrix} & = \begin {vmatrix} \displaystyle {\int ^{x+h}_af(t)\, dt - \int ^x_af(t)\,dt}\\ \end {vmatrix} = \begin {vmatrix} \displaystyle {\int ^{x+h}_af(t)\,dt + \int ^a_cf(t)\,dt}\\ \end {vmatrix}\\\\ & = \begin {vmatrix} \displaystyle {\int ^{x + h}_xf(t)\, dt}\\ \end {vmatrix}\\\\ & \leq \int ^{x + h}_x\begin {vmatrix} f(t)\\ \end {vmatrix} \, dt \\\\ & \leq M \, \begin {vmatrix} h\\ \end {vmatrix} < M\, \delta < \varepsilon \end {align*}

Hence \(F\) is continuous on \([a,b]\).

Differentiability at \(c\in [a,b]\)
Since \(f\) is continuous at \(c\in [a,b]\) given \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(\begin {vmatrix} x - c\\ \end {vmatrix} < \delta \\ \, \implies \, \begin {vmatrix} f(x) - f(c)\\ \end {vmatrix} < \varepsilon \). Then for \(\, s,\, t \in [a,b]\) with \(s\neq t\) , \(\hspace {0.3cm} c-\delta < s\leq c \leq t < c + \delta \) \begin {align*} \begin {vmatrix} \dfrac {F(t) - F(s)}{t - s} - F(c)\\ \end {vmatrix} & = \begin {vmatrix} \dfrac {1}{t - s}\, \displaystyle {\int ^t_s\Big [f(x) - f(c)\Big ]\, dx}\\ \end {vmatrix}\\\\ & \leq \frac {1}{t - s}\, \int ^t_s\begin {vmatrix} f(x) - f(c)\\ \end {vmatrix}\,dx < \varepsilon \end {align*}

Since \(\varepsilon > 0\) is arbitrary, then \(F'(c) = f(c)\)

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Proof. For any fixed \(x\in [a,b]\) chose \(h\neq 0\) and \(x + h \in [a,b]\). Then we have \begin {align*} F(x + h) - F(x) & = \int ^{x+h}_af(t)\,dt - \int ^x_af(t)\, dt\\\\ & = \int ^{x + h}_af(t)\, dt + \int ^a_xf(t)\, dt\\\\ & = \int ^{x + h}_xf(t)\, dt \end {align*}

Since \(f\) is continuous on the closed and bounded interval \([a,b]\, [x, x+h]\). Let \(M\) be the maximum value of \(f\) and \(m\) its minimum value on \([x,x+h]\), then the intermediate value theorem implies that there exist points \(t_1,\,t_2\in [x,x+h]\ni f(t_1) = m,\, f(t_2)= M\) and \(m \leq f(t) \leq M\). Thus \[\int ^{x + h}_xm\, dt \leq \int ^{x + h}_xf(t)\, dt \leq \int ^{x + h}_xM\,dt\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (1)\] So that \(\hspace {0.3cm}\displaystyle {m\, h \leq \int ^{x + h}_x f(t)\, dt \leq M\, h}.\hspace {0.3cm}\) Thus, there is \(\theta \) with \(m < \theta < M\ni \theta = \displaystyle {\frac {1}{h}\, \int ^{x + h}_x f(t)\, dt}.\hspace {0.2cm}\)
Again by the intermediate value theorem of continuous functions, there exist a point \(C(h)\) in \([x,x+h]\ni f\big (C(h)\big ) = \theta \). Thus if \(h> 0\, \exists \, C(h) \in [x,x+h]\ni \displaystyle {f\big [C(h)\big ] = \dfrac {1}{h}\,\int ^{x + h}_xf(t)\, dt}\hspace {0.2cm}\) so that \[\dfrac {F(x + h) - F(x)}{h} = f\big (C(h)\big )\hspace {0.3cm} \cdots \cdots \cdots \hspace {0.3cm}(2)\] Since \(x\leq C(h)\leq x + h\, \) we have \(\, \lim \limits _{h\rightarrow 0}\, C(h) = x.\,\) Since \(f\) is continuous at \(x\), the right side of (2) has the limit \(f(x)\) and the left side of (2) approaches \(F'(x)\) as \(h\rightarrow 0\). So we get \[F'(x) = \lim _{h\rightarrow 0} \, \dfrac {F(x + h) - F(x)}{h} = f(x)\] if \(h<0\), we take \([x + h, x]\) instead of \([x, x+h]\) and make necessary modification in the argument.


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Proof. Let \(\displaystyle {F(x) = \int ^{x}_a f(t)\, dt}\). Since \(f\) is continuous the first fundamental theorem implies that
\(F'(x) = f(x)\) for \(a\leq x\leq b\). By the hypothesis \(\Phi '(x) = f(x)\). Hence, we have \(F(x) = \Phi (x) + c\) for \(a\leq x \leq b\) and for a constant \(c\). Therefore \begin {align*} F(b) - F(a) & = \big [\Phi (b) + c\big ] - \big [\Phi (a) + c\big ]\\ & = \Phi (b) - \Phi (a) \end {align*}

But \(\hspace {0.2cm}\displaystyle {F(a) = \int ^a_af(t)\,dt = 0}\hspace {0.1cm}\) by definition. Thus \(\hspace {0.1cm} F(b) = \Phi (b) - \Phi (a)\). Since \(\hspace {0.2cm}\displaystyle {F(b) = \int ^b_af(t)\, dt}\hspace {0.1cm}\) we have \(\hspace {0.1cm}\displaystyle {\int ^b_af(t)\, dt = \Phi (b) - \Phi (a)}\hspace {0.1cm}\) as required.

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