Proof. Since \(f\in R[a,b]\) then \(\displaystyle {\overline {\int ^b_a}f\, = \,\underline {\int ^b_a}f}\,\) and \(f\) is bounded on \([a,b]\). Now \(\begin {vmatrix} Cf\\ \end {vmatrix} = \begin {vmatrix} C\\ \end {vmatrix}\,\begin {vmatrix} f\\ \end {vmatrix}\,\) which implies that \(Cf\) is bounded on \([a,b]\). Now, for \(C = 0\), the result is obvious.
If \(C>0\), let \(P\) be a partition of \([a,b]\). Then \begin {align*} U\big (P,Cf\big ) & = \sum ^n_{k = 1}C\, M_k(f)\, \Delta x_k\\\\ & = C\, \sum ^n_{k = 1} M_k(f)\, \Delta x_k\\\\ & = C\, U\big (P,f\big ) \end {align*}

Thus, \(\, \inf \, U\big (P,Cf\big ) = C\, \inf \, U\big (P, f\big )\,\) and also we have \(\, \sup \, L\big (P, Cf\big ) = C\, \sup \, L\big (P,f\big )\). Therefore \[\overline {\int ^b_a}Cf = \inf \Big (U\big (P,Cf\big )\Big ) = C\, \inf U\big (P,f\big ) = C\, \overline {\int ^b_a}f\] Since \(f\) in Riemann integrable, we have that \begin {align*} C\,\overline {\int ^b_a}f = C\,\underline {\int ^b_a}f & = C\, \sup L\big (P,f\big ) = \sup L \big (P, Cf\big ) = \underline {\int ^b_a}Cf \end {align*}

Thus \(\, \displaystyle {\overline {\int ^b_a}Cf = \underline {\int ^b_a}Cf}\,\) and \(Cf\in R[a,b]\).

Since \(\, \displaystyle {\overline {\int ^b_a}Cf = C\,\underline {\int ^b_a}f}\,\) we have \(\, \displaystyle {\int ^b_aCf \, = \, C\, \int ^b_af}\).

If \(C< 0\), then \(\, C\, M_k(f) = \inf \big \{C\,f(x):\, x\in [x_{k-1},x_k]\big \}\,\) and

\(\, C\, m_k(f) = \sup \big \{C\,f(x):\, x\in [x_{k-1},x_k]\big \}\,\) and so

\(\, \sup \big \{U\big (P,Cf\big ):\, P\in P[a,b]\big \} = C\, \inf \big \{U\big (P,f\big ):\, P\in P[a,b]\big \}\,\) and

\(\inf \big \{L\big (P,Cf\big ):\, P\in P[a,b]\big \} = C\, \sup \big \{L\big (P,f\big ):\, P\in P[a,b]\big \}.\,\) Therefore \begin {align*} \overline {\int ^b_a} Cf & = \inf \big \{L\big (P,Cf\big ):\, P\in P[a,b]\big \} = C\, \sup \big \{L\big (P,f\big ):\, P\in P[a,b]\big \}\\\\ & = C\, \underline {\int ^b_a}f = C\, \int ^b_af = C\, \overline {\int ^b_a}f\\\\ & = C\, \inf \big \{U\big (P,f):\, P\in P[a,b]\big \} = \sup \big \{U\big (P,Cf\big ):\, P\in P[a,b]\big \}\\\\ & = \underline {\int ^b_a}Cf \end {align*}

Hence again \(\, Cf\in R[a,b]\,\) and \(\, \displaystyle {\int ^b_a Cf\, = \, C\, \int ^b_af}\).


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Proof. For \(c\in [a,b]\) we have that \(f\) is bounded on \([a,c]\) and on \([c,b]\) if and only if it is bounded on \([a,b]\). Since \(f\in R[a,b]\) given \(\varepsilon >0\) there exists a partition \(P\) such that \[U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \] Let \(P^* = P\cup \big \{c\big \}\,\) be a refinement of \(P\) by adding a point, then \(P^*\in [a,b]\) and \[U\big (P^*,f\big ) - L\big (P^*,f\big )\leq U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \] We break the partition \(P^*\) on \([a,b]\) to \(P_1\) on \([a,c]\) and \(P_2\) on \([c,b]\). Then we have \begin {align*} U\big (P^*,f\big ) - L\big (P^*,f\big ) & = U\big (P_1,f\big ) + U\big (P_2,f\big ) - L\big (P_1,f\big ) - L\big (P_2,f\big )\\ & = \Big (U\big (P_1,f\big ) - L\big (P_1,f\big )\Big ) + \Big (U\big (P_2,f\big ) - L\big (P_2,f\big )\Big )\\ & < \varepsilon \end {align*}

It follows that \(\, U\big (P_1,f\big ) - L\big (P_1,f\big ) < \varepsilon \,\) and \(\, U\big (P_2,f\big ) - L\big (P_2,f\big ) < \varepsilon .\,\) Thus \(f\in R[a,c]\) and \(f\in R[c,b]\).
Now it can be shown that \(\, U\big (P_1,f\big ) + U\big (P_2,f\big ) = U\big (P^*,f\big )\,\) implies that \[\inf \, U\big (P_1,f\big ) \, + \, \inf \, U\big (P_2,f\big )\, = \, \inf \, U\big (P^*,f\big )\] This then implies that \(\, \displaystyle {\int ^c_af\, + \, \int ^b_cf \, = \, \int ^b_af}.\,\)

Conversely, in \(c\in (a,b)\) and \(f\in R[a,c]\) and \(f\in R[c,d]\), then there exists partitions \(P_2\in P[a,c]\) and \(P_2\in P[c,b]\) such that \(\, U\big (P_1,f\big ) - L\big (P_1,f\big ) < \varepsilon /2\,\) and \(\, U\big (P_2,f\big ) - L\big (P_1,f\big )<\varepsilon /2\). Then for \(P= P_1\cup P_2\,\) we have \begin {align*} U\big (P,f\big ) - L\big (P,f\big ) & = U\big (P_1,f\big ) + U\big (P_2,f\big ) - L\big (P_1,f\big ) - L\big (P_2,f\big )\\ & = U\big (P_1,f\big ) - L\big (P_1,f\big ) + U\big (P_2,f\big ) - L\big (P_2,f\big )\\ & < \varepsilon \end {align*}

Hence \(f\in R[a,b]\) and the required relation of integrals follow as before.

□

Proof. Let \(\varepsilon >0\). Since \(f\in R[a,b]\), Riemann’s criterion gives a partition \(P\) of \([a,b]\) with \(U(f,P)-L(f,P)<\varepsilon \). Refine \(P\) by adjoining the points \(c\) and \(d\); refinement never increases the difference, so the refined partition \(P'\) also satisfies \(U(f,P')-L(f,P')<\varepsilon \).

Let \(Q\) be the part of \(P'\) lying in \([c,d]\), a partition of \([c,d]\). Every term of \(U(f,Q)-L(f,Q)\) appears among the terms of \(U(f,P')-L(f,P')\), and all terms are non-negative, so \[U(f,Q)-L(f,Q) \leq U(f,P')-L(f,P') < \varepsilon .\] By Riemann’s criterion again, \(f\in R[c,d]\). □

Proof. Since \(f,\, g\in R[a,b]\), then there functions are bounded on \([a,b]\). Therefore there exists on \(M>0\) such that \(\begin {vmatrix} f(x)\\ \end {vmatrix} \leq M\) and \(\begin {vmatrix} g(x)\\ \end {vmatrix} \leq M\,\) for all \(x\in [a,b]\). Thus \[\begin {vmatrix} f(x) + g(x)\\ \end {vmatrix} \leq \begin {vmatrix} f(x)\\ \end {vmatrix} + \begin {vmatrix} g(x)\\ \end {vmatrix}\leq 2M\] for all \(x\in [a,b]\). This shows that \(\, f + g\,\) is bounded on \([a,b]\). Therefore \(\, \displaystyle {\overline {\int ^b_a}\big (f + g\big )}\,\) and \(\, \displaystyle {\underline {\int ^b_a}\big (f + g\big )}\,\) exists.

But if \(\, A\subseteq [a,b]\,\) then we have \(\, \big \{f(x) + g(x):\, x\in A\big \} \subset \big \{f(x_1) + g(x_2):\, x_1,\, x_2\in A\big \}\). Thus for any partition \(P\) of \([a,b]\), we have \(\, U\big (P,f+g\big )\leq U\big (P,f\big ) + U\big (P,g\big )\,\) and
\(\, L\big (P,f+g\big )\geq L\big (P,f\big ) + L\big (P,g\big )\). Consequently, \begin {align*} \sup L\big (P,f\big ) + \sup L \big (P,g\big ) & \leq \sup L\big (P, f+g\big ) \leq \inf U\big (P,f+g\big ) \leq \inf U\big (P,f\big ) \, + \, \inf U\big (P,g\big ) \end {align*}

Thus \[\underline {\int ^b_a}f\, + \, \underline {\int ^b_a}g \, \leq \, \underline {\int ^b_a}\big (f + g\big )\, \leq \overline {\int ^b_a}\big (f+ g)\, \leq \,\overline {\int ^b_a}f\, + \, \overline {\int ^b_a}g\] and since \(f\) and \(g\) are Riemann integrable on \([a,b]\), we have \(\, \displaystyle {\underline {\int ^b_a}f = \int ^b_af = \overline {\int ^b_a}f}\,\) and
\(\, \displaystyle {\underline {\int ^b_a}g = \int ^b_a g = \overline {\int ^b_a}g}\,\hspace {0.3cm}\) so that \[\int ^b_a f\, + \, \int ^b_a\, \leq \, \int ^b_a\big (f + g\big )\, \leq \,\overline {\int ^b_a}\big (f + g\big )\, \leq \, \int ^b_af \, + \, \int ^b_a g\] This shows that all the inequalities are in fact equalities. Hence \(f + g \in R[a,b]\) and \[\int ^b_a \big (f + g\big ) \, = \, \overline {\int ^b_a}\big (f + g\big ) \, = \, \int ^b_a f \, + \, \int ^b_a g\]
□

Proof. For any partition \(P\) of \([a,b]\) we have \(\, L\big (P,f\big )\leq U\big (P,f\big ) \leq U\big (P,g\big ).\hspace {0.2cm}\) This implies that \begin {align*} &\sup _P\big \{L\big (P,f\big ) - U\big (P,g\big )\big \} \leq 0\\ \implies \hspace {0.3cm} & \sup _P\, L\big (P,f\big ) - \inf \, U\big (P,g\big ) \leq 0\\ \implies \hspace {0.3cm} &\int ^b_a f\, - \, \int ^b_a g \leq 0\\\\ \implies \hspace {0.3cm} & \int ^b_af \leq \int ^b_a g\\\\ \end {align*} □

Proof. Since \(f, \, g\in R[a,b]\), they are bounded on \([a,b]\). Therefore, there exists \(K> 0\, \ni \\ \, \begin {vmatrix} fg\\ \end {vmatrix} = \begin {vmatrix} f\\ \end {vmatrix}\, \begin {vmatrix} g\\ \end {vmatrix}\, = \, K^2\,\) on \([a,b]\).
This implies that \(fg\) is also bounded on \([a,b]\). Thus for any partition \(P\) of \([a,b]\) both \(U\big (P,fg\big )\) and \(L\big (P,fg\big )\) exists and are bounded
We also have that for any \(\varepsilon >0\, \exists \, \) a partition \(P\) of \([a,b]\) such that \(\, U\big (P,f\big ) - L\big (P,f\big ) <\varepsilon \,\) and \(\, U\big (P,g\big ) - L\big (P,g\big )<\varepsilon \).

Let \(\, M_r(f)\, ,\, m_r(f)\, \) be bounds of \(f\) on \([x_{r-1},x_r]\)

\(\, M_r(g)\, ,\, m_r(g)\, \) be bounds of \(g\) on \([x_{r-1},x_r]\)

\(\, M_r(fg)\, ,\, m_r(fg)\, \) be bounds of \(fg\) on \([x_{r-1},x_r]\)

Then, for all \(x',\, x''\in [x_{r - 1}, x_r]\,\) we have \begin {align*} \begin {vmatrix} fg(x'') - fg(x')\\ \end {vmatrix} & = \begin {vmatrix} g(x'')\, f(x'') - f(x') + f(x')\,g(x'') - g(x')\\ \end {vmatrix}\\ & \leq \begin {vmatrix} g(x'')\\ \end {vmatrix}\, \begin {vmatrix} f(x'') - f(x')\\ \end {vmatrix} + \begin {vmatrix} f(x')\\ \end {vmatrix}\, \begin {vmatrix} g(x'') - g(x')\\ \end {vmatrix}\\ &\leq K\, \begin {vmatrix} M_r(f) - m_r(f)\\ \end {vmatrix} + K\, \begin {vmatrix} M_r(g) - m_r(g)\\ \end {vmatrix} \end {align*}

Thus \(\, \Big (M_r(fg) - m_r(fg)\Big )\leq K\Big (M_r(f) - m_r(f)\Big ) + K\Big (M_r(g) - m_r(g)\Big ).\,\) Hence \[\sum ^n_{r=1}\Big (M_r(fg) - m_r(fg)\Big )\, \Delta x_r\, \leq \, K\,\sum ^n_{r=1}\Big (M_r(f) - m_r(f)\Big )\, \Delta x_r + K\,\sum ^n_{r=1}\Big (M_r(g) - m_r(g)\Big )\, \Delta x_r\] Thus \(\, U\big (P,fg\big ) - L\big (P,fg\big ) < K\, \varepsilon + K\, \varepsilon = 2K\varepsilon \,.\,\) This implies that \(fg\in R[a,b]\).


□

Proof. The proof is similar to Theorem 5 but in this case \(\,\begin {vmatrix} f/g\\ \end {vmatrix}\leq K/t\,\) on \([a,b]\) and if \(\, M_r(f/g)\, ,\, m_r(f/g)\,\) denotes the bounds of \(f/g\) on \([x_{r-1}, x_r]\) then \[M_r(f/g) - m_r(f/g) \leq \dfrac {K}{t^2}\, \Big (M_r(f) - m_r(f)\Big ) \, + \, \dfrac {K}{t^2}\, \Big (M_r(g) - m_r(g)\Big )\]

□

Proof. Since \(f\in R[a,b],], \exists \, K\, \ni \, \begin {vmatrix} f\\ \end {vmatrix} < K\,\) on \([a,b]\) and so \(\begin {vmatrix} f\\ \end {vmatrix}\) is also bounded on \([a,b]\). Also given \(\varepsilon > 0\) there exist a partition \(P\) of \([a,b]\) such that \(\, U\big (P,f\big ) - L\big (P,f\big ) < \varepsilon \).
Let \(\, M_r(f)\, , \, m_r(f)\, ,\, M_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big )\, \) and \(\, m_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big )\,\) so the bounds of \(f\) and \(\begin {vmatrix} f\\ \end {vmatrix}\) respectively on \([x_{r-1},x_r]\), then for all \(x'\, , \, x''_r\in [x_{r-1},x_r]\,\) we have \(\, \begin {vmatrix} \begin {vmatrix} f(x'')\\ \end {vmatrix} - \begin {vmatrix} f(x')\\ \end {vmatrix} \end {vmatrix} \leq \begin {vmatrix} f(x'') - f(x')\\ \end {vmatrix}\,\) and by taking supremum we have \(\, M_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big ) - m_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big ) \leq M_r(f) - m_r(f)\,.\,\) Thus \begin {align*} U\big (P,\begin {vmatrix} f\\ \end {vmatrix}\big ) - L\big (P,\begin {vmatrix} f\\ \end {vmatrix} & = \sum ^n_{r=1}\Big (M_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big ) - m_r\big (\begin {vmatrix} f\\ \end {vmatrix}\big )\Big )\, \Delta x_r\\\\ & \leq \sum ^n_{r=1}\Big (M_r(f) - m_r(f)\Big )\, \Delta x_r\\\\ & = U\big (P,f\big ) - L\big (L,f\big )\\ & < \varepsilon \end {align*}

Hence \(\, \begin {vmatrix} f\\ \end {vmatrix}\in R[a,b]\,\) since \(\begin {vmatrix} f\\ \end {vmatrix}> f\,\) on \([a,b]\) we have \[\begin {vmatrix} \displaystyle {\sum ^n_{r=1}M_r(f)\, \Delta x_r}\\ \end {vmatrix}\leq \sum ^n_{r=1}M_r \, \begin {vmatrix} f\\ \end {vmatrix}\, \Delta x_r\] That is \(\, \begin {vmatrix} U\big (P,f\big )\\ \end {vmatrix}\leq U\big (P,f\big )\,\) for all the partition \(\, P\in P[a,b]\). Let \(\, \begin {vmatrix} \begin {vmatrix} P\\ \end {vmatrix} \end {vmatrix} \longrightarrow 0\, \) to get \[\begin {vmatrix} \displaystyle {\int ^b_a f}\\ \end {vmatrix}\, \leq \, \int ^b_a \begin {vmatrix} f\\ \end {vmatrix}\]

□

Proof. Since \(m\leq f \leq M\) on \([a,b]\) , then for \(f\in R[a,b]\) we have \[\int ^b_a m \leq \int ^b_a f \leq \int ^b_a M\]

\[\implies \hspace {0.2cm} m\,(b - a) \leq \int ^b_a f\leq M\,(b-a)\] These inequalities also follow from the fact that \[m\, (b - a) \leq L(P , f) \leq \int ^b_a f \leq U(P , f) \leq M\, (b - a)\] for any partition \(P\).


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Proof. From the above Remark; if \(M\) and \(m\) are bounds for \(f\), then \[m\, (b -a )\leq \int ^b_af(x)\, dx \leq M\, (b - a)\]

\[\implies \hspace {0.3cm} m\leq \dfrac {1}{(b - a)}\, \int ^b_af(x)\, dx \leq M\] This shows that \(\hspace {0.2cm}\displaystyle {\dfrac {1}{b - a}\,\int ^b_a f(x)\, dx}\hspace {0.2cm}\) is a value between \(m\) and \(M\) of a continuous function on \([a,b]\). Therefore, by the intermediate value theorem for continuous functions on \([a,b]\, f\) takes this value at some point \(c\in c[a,b]\). Thus \(\hspace {0.2cm}\displaystyle {\dfrac {1}{b - a}\, \int ^b_a f(x) \,dx = f(c)}\hspace {0.1cm}\) for some \(c\) in \([a,b]\). Hence \[\int ^b_af(x)\, dx = (b - a)\, f(c)\]

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Proof. Let \(m\) and \(M\) be bounds of \(f\) in \([a,b]\), then \(m\leq f(x) \leq M\) for all \(x\in [a,b]\). Since \(g(x)\geq 0\) for all \(x\in [a,b]\), we have \(\, m\, g(x) \leq f(x)\, g(x) \leq M\, g(x)\,\) for all \(x\in [a,b]\). This implies that \[m\, \int ^b_ag(x)\, dx \leq \int ^b_af(x)\, g(x)\, dx \leq M\, \int ^b_ag(x)\,dx\] Thus, there is a number \(\mu \) with \(\, m\leq \mu \leq M\,\) such that \[\int ^b_af(x)\, g(x)\, dx = \mu \, \int ^b_ag(x)\, dx\] But since \(f\) is continuous on \([a,b]\), by the Intermediate Value Theorem there exists \(c\in [a,b]\) such that \(f(c) = \mu \). This implies that \[\int ^b_af(x)\, g(x)\, dx = f(c)\, \int ^b_ag(x)\,dx\hspace {0.5cm}\text {as required}\]

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Solution. Let \(f(x) = x\) and \(g(x) = \dfrac {1}{1 + x^2}\). Then \(f\) and \(g\) satisfy the condition of the Generalized Mean Value Theorem. Therefore, there exists a point \(c\in [0,1]\) such that \begin {align*} \int ^1_0x\,\Big (\dfrac {1}{1 + x^2}\Big )\, dx & = f(c)\, \int ^1_0\dfrac {1}{1 + x^2}\, dx\\\\ & = c\, \int ^1_0\dfrac {1}{1 + x^2}\, dx \end {align*}

But \(\hspace {0.2cm} \displaystyle {\int ^1_0\dfrac {1}{1 + x^2}\, dx = \tan ^{-1}(x)\Big |^1_0 = \tan ^{-1}(1) - \tan ^{-1}(0) = \dfrac {\pi }{4} - 0 = \dfrac {\pi }{4}}\hspace {0.2cm}\) and also

\(\displaystyle {\int ^1_0\dfrac {x}{1 + x^2}\, dx = \dfrac {1}{2}\, \ln (1 + x^2)\Big |^1_0 = \dfrac {1}{2}\, \ln (2)}\)

Thus \(\hspace {0.3cm}\dfrac {1}{2}\, \ln (2) = \dfrac {\pi }{4}\, c\implies c = \dfrac {2}{\pi }\, \ln (2)\)