B.4 Integral Theorems and Proofs

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 \(\Vert P_{2}\Vert < \delta_{2}.\)

NOTE

The partition \(P_{1}\) is \(x_{0}=a,\ldots ,x_{k}=b.\) The partition \(P_{2}\) is \(x_{k}=b,\ldots ,x_{n}=c.\)

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*}

B-9

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}

B-10

Need To Review

The Mean Value Theorem for Integrals is discussed in Section 5.4, p. 372.

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*}

The bounds on an integral theorem is proved by contradiction.

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*}

B-11

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.