THEOREM

If a function \(f\) is continuous on an interval containing the numbers \(a\), \(b\), and \(c\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \displaystyle\int_{a}^{b}f(x)\,dx = \displaystyle\int_{a}^{c}f(x)\,dx + \displaystyle\int_{c}^{b}f(x)\,dx}} \]

Proof

Since \(f\) is continuous on an interval containing \(a\), \(b\), and \(c\), the three integrals above exist.

Part 1 Assume \(a < b < c.\) Since \(f\) is continuous on \([ a,b] \) and on \([ b,c] ,\) given any \(\varepsilon > 0,\) there is a number \(\delta_{1} > 0\), so that \begin{equation} \left\vert \sum\limits_{i=1}^{k}f( u_{i}) \Delta x_{i}-\displaystyle\int_{a}^{b}f( x) \,dx\right\vert < \dfrac{\varepsilon }{2}\tag{1} \end{equation}

for every Riemann sum \(\sum\limits_{i=1}^{k}f(u_{i}) \Delta x_{i}\) for \(f\) on \([a,b] ,\) where \(x_{i-1}\leq u_{i}\leq x_{i}\), \(i = 1,2, \ldots ,k,\) and whose partition \(P_{1}\) of \([a,b] \) has norm \( \Vert P_{1}\Vert < \delta _{1}.\) There is also a number \(\delta_{2} > 0\) for which \begin{equation} \left\vert \sum\limits_{i=k+1}^{n}f( u_{i}) \Delta x_{i}-\int_{b}^{c}f( x) \,dx\right\vert <\dfrac{\varepsilon }{2}\tag{2} \end{equation}

for every Riemann sum \(\sum\limits_{i=k+1}^{n}f( u_{i}) \Delta x_{i}\) for \(f\) on \([ b,c] ,\) where \(x_{i-1}\leq u_{i}\leq x_{i}\), \(i = k + 1, k + 2,\ldots , n,\) and whose partition \(P_{2}\) of \([b,c] \) has norm \(\Vert P_{2}\Vert < \delta_{2}.\)

Let \(\delta \) be the smaller of \(\delta_{1}\) and \(\delta_{2}.\) Then (1) and (2) hold, with \(\delta \) replacing \(\delta _{1}\) and \(\delta _{2}.\) If (1) and (2) are added and if \(\Vert P_{1}\Vert <\delta \) and \( \Vert P_{2}\Vert < \delta,\) then \begin{equation*} \left\vert \sum\limits_{i=1}^{k}f( u_{i}) \Delta x_{i}-\int_{a}^{b}f( x) \,dx\right\vert + \left\vert \sum\limits_{i=k+1}^{n}f( u_{i}) \Delta x_{i}-\int_{b}^{c}f( x) \,dx \right\vert < \dfrac{\varepsilon }{2} + \dfrac{\varepsilon }{2} = \varepsilon \end{equation*}

Using the Triangle Inequality, this result implies that for \(\Vert P_{1}\Vert < \delta \) and \(\Vert P_{2}\Vert < \delta\) , \begin{equation} \left\vert \sum\limits_{i=1}^{k}f( u_{i}) \Delta x_{i}-\int_{a}^{b}f( x) \,dx+\sum\limits_{i = k + 1}^{n}f( u_{i}) \Delta x_{i}-\int_{b}^{c}f( x) \,dx\right\vert < \varepsilon\tag{3} \end{equation}

Denote \(P_{1}\cup P_{2}\) by \(P\)*. Then \(P\)* is a partition of \([ a,c]\) having the number \(b = x_{k}\) as an endpoint of the \(k\)th subinterval. So, \begin{equation*} \sum\limits_{i = 1}^{k}f( u_{i}) \Delta x_{i} + \sum\limits_{i = k + 1}^{n}f( u_{i}) \Delta x_{i} = \sum\limits_{i = 1}^{n}f( u_{i}) \Delta x_{i} \end{equation*}

are Riemann sums for \(f\) on \(P\)*. Since \(\Vert P\hbox{*}\Vert < \delta \) implies that \(\Vert P_{1}\Vert < \delta \) and \( \Vert P_{2}\Vert < \delta ,\) it follows from (3) that \begin{equation*} \left\vert \sum\limits_{i=1}^{n}f( u_{i}) \Delta x_{i}-\left[ \displaystyle\int_{a}^{b}f(x)\,dx+\int_{b}^{c}f( x) \,dx\right] \right\vert < \varepsilon \end{equation*}

for every Riemann sum \(\sum\limits_{i=1}^{n}f( u_{i}) \Delta x_{i}\) for \(f\) on \([ a,c] \) whose partition \(P\)* of \([ a,c ] \) has \(b\) as an endpoint of a subinterval of the partition and has norm \(\Vert P\hbox{*}\Vert < \delta .\) Therefore, \begin{equation*} \displaystyle\int_{a}^{c}f(x)\,dx=\displaystyle\int_{a}^{b}f(x)\,dx+\displaystyle\int_{b}^{c}f(x)\,dx \end{equation*}

Part 2 There are six possible orderings (permutations) of the numbers \(a,\) \(b\,,\) and \(c\): \begin{equation*} \begin{array}{lllllllllll} a < b < c & \quad & a < c < b & \quad & b < a < c & \quad & b < c < a & \quad & c < a < b & \quad & c < b < a \end{array} \end{equation*}

In Part 1, we showed that the theorem is true for the order \(a < b < c.\) Now consider any other order, say, \(b < c < a.\) From Part 1, \begin{equation} \displaystyle\int_{b}^{c}f(x)\,dx+\displaystyle\int_{c}^{a}f(x)\,dx = \displaystyle\int_{b}^{a}f(x)\,dx\tag{4} \end{equation}

But, \begin{equation*} \begin{array}{lll} \displaystyle\int_{c}^{a}f(x)\,dx = -\displaystyle\int_{a}^{c}f(x)\,dx &\quad \hbox{and}\quad & \displaystyle\int_{b}^{a}f(x)\,dx = -\displaystyle\int_{a}^{b}f(x)\,dx \end{array} \end{equation*}

Now we substitute this into (4). \begin{eqnarray*} \displaystyle\int_{b}^{c}f(x)\,dx-\displaystyle\int_{a}^{c}f(x)\,dx &=&-\displaystyle\int_{a}^{b}f(x)\,dx \\[5pt] \displaystyle\int_{a}^{b}f(x)\,dx+\displaystyle\int_{b}^{c}f(x)\,dx &=&\displaystyle\int_{a}^{c}f(x)\,dx \end{eqnarray*}

proving the theorem for \(b < c < a.\)

The proofs for the remaining four permutations of \(a\,,\) \(b,\) and \(c\) are similar.

THEOREM Fundamental Theorem of Calculus, Part 1

Let \(f\) be a function that is continuous on a closed interval \([a,b]\) . The function \(I\) defined by \begin{equation*} I\,(x) = \displaystyle\int_{a}^{x}f(t)\,dt \end{equation*} has the property that it is continuous on \([a,b]\) and differentiable on \((a,b) \). Moreover, \begin{equation*} I^{\prime} (x) = \dfrac{d}{dx}\left[ \int_{a}^{x}f(t)\,dt\right] = f(x) \end{equation*} for all \(x\) in \((a,b).\)

Proof

Let \(x\) and \(x + h\), \(h\neq 0\), be in the interval \(( a,b) \). Then \begin{equation*} I( x) = \int_{a}^{x}f(t) \,dt\qquad I(x+h) =\int_{a}^{x + h}f(t)\, dt \end{equation*}and \begin{eqnarray*} I(x+h)-I(x)&=& \int_{a}^{x+h}f(t)\,dt+\int_{x}^{a}f(t)\,dt \qquad {\color{#0066A7}{\hbox{\(\int_{x}^{a}{f}(t)\, dt = - \int_{a}^{x}{f}(t)\, {dt}\)}}}\\[5pt] &=&\int_{x}^{a}f(t)\,dt+ \int_{a}^{x+h}f(t)\,dt=\displaystyle\int_{x}^{x+h}f(t)\,dt\\[-10pt] \end{eqnarray*} Dividing both sides by \(h\neq 0,\) we get \begin{equation} {\dfrac{{I(x+h)-I(x)}}{{h}}}={\dfrac{{1}}{{h}}}{\int_{x}^{x+h}{f(t)\,dt}}\tag{5} \end{equation}

Now we use the Mean Value Theorem for Integrals in the integral on the right. There are two possibilities: either \(h > 0\) or \(h < 0\).

If \(h > 0\), there is a number \(u\), where \(x \leq u \leq x + h\), for which \begin{eqnarray*} \displaystyle\int_{x}^{x+h}f(t)\,{dt} \,&=&\, f(u) h \\[4pt] \dfrac{{1}}{{h}}\displaystyle\int_{x}^{x + h}f(t)\,{dt} \,&=&\, f(u) \\[6pt] {\dfrac{{I(x+h)-I(x)}}{{h}}} \,&=&\, {f}(u)\quad\quad\quad\quad\quad {\color{#0066A7}{\hbox{From (5)}}} \end{eqnarray*}

Since \(x\leq u\leq x + h\), as \(h\rightarrow 0^{+}\), \(u\) approaches \(x^{+},\) so \begin{eqnarray*} \lim\limits_{h\rightarrow 0^{+}}\dfrac{I (x + h) -I( x) }{h} = \lim\limits_{h\rightarrow 0^{+}}f(u) =\lim\limits_{u\rightarrow x^{+}}f(u) \underset{\underset{\color{#0066A7}{\hbox{\(f\) is continuous}}}{\color{#0066A7}{\uparrow}}}{=} f( x) \end{eqnarray*}

Using a similar argument for \(h < 0\), we obtain \begin{equation*} \lim\limits_{h\rightarrow 0^{-}}\dfrac{I (x+h) -I ( x) }{h}=f(x) \end{equation*}

Since the two one-sided limits are equal, \begin{equation*} \lim\limits_{h\rightarrow 0}\dfrac{I (x+h)-I (x)}{h}=f(x) \end{equation*}

The limit is the derivative of the function \(I\), meaning \(I^{\prime} (x)=f(x)\) for all \(x\) in \((a,b)\).

THEOREM Bounds on an Integral

If a function \(f\) is continuous on a closed interval \([a,b]\) and if \( m\) and \(M\) denote the absolute minimum and absolute maximum values of \(f\) on \([a,b]\), respectively, then \begin{eqnarray*} m(b-a)\leq \displaystyle\int_{a}^{b}f(x)\,dx\leq M (b-a) \end{eqnarray*}

Proof

Part 1 \(\ m(b-a)\leq \displaystyle\int_{a}^{b}f(x)\,dx\) Assume \begin{equation} m(b-a) > \displaystyle\int_{a}^{b}f(x)\,dx\tag{6} \end{equation}

Since \(f\) is continuous on \([a,b] ,\) \begin{equation*} \lim\limits_{\Vert P\Vert \rightarrow 0}\sum\limits_{i = 1}^{n}f( u_{i}) \Delta x_{i} = \int_{a}^{b}f(x) dx \end{equation*}

By (6), \(\ m(b-a)-\displaystyle\int_{a}^{b}f(x)\,dx > 0.\) We choose \(\varepsilon\), so that \begin{equation} \varepsilon = m(b-a)-\displaystyle\int_{a}^{b}f(x)\,dx > 0 \tag{7} \end{equation}

Then there is a number \(\delta > 0\), so that for all partitions \(P\) of \( [a,b] \) with norm \(\Vert P\Vert < \delta,\) we have \begin{equation*} \left\vert \sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i}-\int_{a}^{b}f( x) \,dx\right\vert < \varepsilon \end{equation*}

which is equivalent to \begin{equation*} \int_{a}^{b}f( x) \,dx-\varepsilon < \sum\limits_{i = 1}^{n}f( u_{i}) \Delta x_{i} < \int_{a}^{b}f( x) \,dx + \varepsilon \end{equation*}

By (7), the right inequality can be expressed as \begin{eqnarray*} \sum\limits_{i = 1}^{n}f( u_{i}) \Delta x_{i} < \int_{a}^{b}f( x) \,dx + \varepsilon &=& \int_{a}^{b}f( x) \,dx + \left[ m(b-a)-\displaystyle\int_{a}^{b}f(x)\,dx\right]\\[5pt] &=& m ( b-a) \end{eqnarray*}

Consequently, \begin{equation*} \sum\limits_{i = 1}^{n}f( u_{i}) \Delta x_{i} < m( b-a) = \sum\limits_{i = 1}^{n}m\,\Delta x_{i} \end{equation*}

implying that for every partition \(P\) of \([ a,b] \) with \( \Vert P\Vert < \delta ,\) \begin{equation*} f( u_{i}) < m \end{equation*}

for some \(u_{i}\) in \([a,b] .\) But this is impossible because \(m\) is the absolute minimum of \(f\) on \([a,b] .\) Therefore, the assumption \(m(b-a) > \displaystyle\int_{a}^{b}f(x)\,dx\) is false. That is, \begin{equation*} \ m(b-a)\leq \displaystyle\int_{a}^{b}f(x)\,dx \end{equation*}

Part 2 To prove \(\displaystyle\int_{a}^{b}f(x)\,dx\leq M (b-a),\) use a similar argument.