Proof. If \(C\geq 0\), then \(M_k(Cf) = CM_k(f)\) and \(m_k(Cf) = Cm_k(f)\) on every subinterval, so \(U(P,Cf) = CU(P,f)\) and \(L(P,Cf) = CL(P,f)\) for every partition \(P\). Taking the infimum and supremum over \(P\), \[\overline {\int ^b_a}Cf = C\,\overline {\int ^b_a}f = C\int ^b_af = C\,\underline {\int ^b_a}f = \underline {\int ^b_a}Cf .\]

If \(C<0\), multiplying by \(C\) reverses inequalities, so \(M_k(Cf) = Cm_k(f)\) and \(m_k(Cf) = CM_k(f)\). Hence \(U(P,Cf) = CL(P,f)\) and \(L(P,Cf) = CU(P,f)\), and \begin {align*} \overline {\int ^b_a}Cf &= \inf _P C L(P,f) = C\sup _P L(P,f) = C\int ^b_a f,\\ \underline {\int ^b_a}Cf &= \sup _P CU(P,f) = C\inf _P U(P,f) = C\int ^b_a f . \end {align*}

In both cases \(Cf\in R[a,b]\) and \(\int _a^b Cf = C\int _a^b f\). □

Proof. If \(c = a\) or \(c = b\) there is nothing to prove, so let \(a<c<b\). Partitions \(P_1\) of \([a,c]\) and \(P_2\) of \([c,b]\) combine into a partition \(P_1\cup P_2\) of \([a,b]\), with \begin {align*} U(P_1\cup P_2, f) &= U(P_1,f) + U(P_2,f),\\ L(P_1\cup P_2, f) &= L(P_1,f) + L(P_2,f). \end {align*}

Suppose \(f\in R[a,b]\) and \(\varepsilon >0\). By Riemann’s Criterion there is a partition \(P\) of \([a,b]\) with \(U(P,f) - L(P,f) < \varepsilon \), and by Theorem 5.1.5 the refinement \(P^* = P\cup \{c\}\) satisfies the same inequality. Splitting \(P^*\) at \(c\) into \(P_1\) and \(P_2\), \begin {align*} &\bigl (U(P_1,f) - L(P_1,f)\bigr ) + \bigl (U(P_2,f) - L(P_2,f)\bigr )\\ &\qquad = U(P^*,f) - L(P^*,f) < \varepsilon . \end {align*}

Both brackets are non-negative, so each is less than \(\varepsilon \), and \(f\in R[a,c]\), \(f\in R[c,b]\).

Conversely, if \(f\in R[a,c]\) and \(f\in R[c,b]\), choose \(P_1\) and \(P_2\) with \(U(P_i,f) - L(P_i,f) < \frac {\varepsilon }{2}\); then \(P = P_1\cup P_2\) gives \(U(P,f) - L(P,f) <\varepsilon \), so \(f\in R[a,b]\).

Finally, for any such \(P_1\) and \(P_2\), \[L(P_1,f) + L(P_2,f) \leq \int _a^b f\leq U(P_1,f) + U(P_2,f).\] Taking the supremum of the left side and the infimum of the right side over \(P_1\) and \(P_2\) gives \(\int _a^c f + \int _c^b f\leq \int _a^b f\leq \int _a^c f + \int _c^b f\). □

Proof. Apply the theorem twice: \(f\in R[a,b]\) gives \(f\in R[c,b]\), and then \(f\in R[c,d]\). □

Proof. On any subinterval, \(f(x) + g(x)\leq M_k(f) + M_k(g)\), so \(M_k(f+g)\leq M_k(f) + M_k(g)\); similarly \(m_k(f+g)\geq m_k(f) + m_k(g)\). Hence for every partition \(P\), \[L(P,f) + L(P,g)\leq L(P,f+g)\leq U(P,f+g)\leq U(P,f) + U(P,g).\] Given \(\varepsilon >0\), choose (by Riemann’s Criterion and a common refinement) a partition \(P\) with \(U(P,f) - L(P,f)<\frac {\varepsilon }{2}\) and \(U(P,g) - L(P,g)<\frac {\varepsilon }{2}\). Then \(U(P,f+g) - L(P,f+g)<\varepsilon \), so \(f+g\in R[a,b]\). Moreover both \(\int _a^b(f+g)\) and \(\int _a^bf + \int _a^bg\) lie between \(L(P,f) + L(P,g)\) and \(U(P,f) + U(P,g)\), which differ by less than \(\varepsilon \). Since \(\varepsilon \) is arbitrary, the two numbers are equal. □

Proof. For every partition \(P\), \(m_k(f)\leq m_k(g)\) on each subinterval, so \(L(P,f)\leq L(P,g)\leq \int _a^bg\). Taking the supremum over \(P\) gives \(\int _a^b f\leq \int _a^bg\). □

Proof. Since \(f\) and \(g\) are bounded, there is a \(K> 0\) with \(|f|\leq K\) and \(|g|\leq K\) on \([a,b]\); then \(|fg|\leq K^2\), so \(fg\) is bounded. Let \(\varepsilon >0\), and choose a partition \(P\) with \(U(P,f) - L(P,f) <\varepsilon \) and \(U(P,g) - L(P,g)<\varepsilon \). For \(x', x''\) in a subinterval \([x_{r - 1}, x_r]\), \begin {align*} &|f(x'')g(x'') - f(x')g(x')|\\ &\qquad = \bigl |g(x'')\bigl (f(x'') - f(x')\bigr ) + f(x')\bigl (g(x'') - g(x')\bigr )\bigr |\\ &\qquad \leq K \bigl (M_r(f) - m_r(f)\bigr ) + K \bigl (M_r(g) - m_r(g)\bigr ) . \end {align*}

Taking the supremum over \(x'\) and \(x''\), \(M_r(fg) - m_r(fg)\leq K(M_r(f) - m_r(f)) + K(M_r(g) - m_r(g))\). Multiplying by \(\Delta x_r\) and summing, \begin {align*} U(P,fg) - L(P,fg) &\leq K\bigl (U(P,f) - L(P,f)\bigr ) + K\bigl (U(P,g) - L(P,g)\bigr )\\ &< 2K\varepsilon . \end {align*}

Since \(\varepsilon \) is arbitrary, \(fg\in R[a,b]\) by Riemann’s Criterion. □

Proof. For \(x',x''\) in a subinterval \([x_{r-1},x_r]\), \[\left |\frac {1}{g(x'')} - \frac {1}{g(x')}\right | = \frac {|g(x') - g(x'')|}{|g(x')|\,|g(x'')|}\leq \frac {M_r(g) - m_r(g)}{t^2},\] so \(U\left (P,\frac {1}{g}\right ) - L\left (P,\frac {1}{g}\right )\leq \frac {1}{t^2}\bigl (U(P,g) - L(P,g)\bigr )\). By Riemann’s Criterion \(\frac {1}{g}\in R[a,b]\), and then \(\frac {f}{g} = f\cdot \frac {1}{g}\in R[a,b]\) by Theorem 5.2.6. □

Proof. For \(x',x''\) in a subinterval, \(\bigl ||f(x'')| - |f(x')|\bigr |\leq |f(x'') - f(x')|\), so \(M_r(|f|) - m_r(|f|) \leq M_r(f) - m_r(f)\). Hence \[U(P,|f|) - L(P,|f|)\leq U(P,f) - L(P,f)\] for every \(P\), and \(|f|\in R[a,b]\) by Riemann’s Criterion. Since \(-|f|\leq f\leq |f|\), the previous theorems give \(-\int _a^b|f|\leq \int _a^bf\leq \int _a^b|f|\), that is, \(\left |\int ^b_a f\right | \leq \int ^b_a|f|\). □

Proof. Let \(m\) and \(M\) be the minimum and maximum of \(f\) on \([a,b]\). By the note above, \[m\leq \frac {1}{b - a} \int ^b_af(x)\, dx \leq M .\] By the Intermediate Value Theorem, the continuous function \(f\) takes every value between its minimum and its maximum, so \(f(c) = \frac {1}{b - a}\int ^b_a f(x) \,dx\) for some \(c\in [a,b]\). □

Proof. Let \(m\) and \(M\) be the minimum and maximum of \(f\) on \([a,b]\). Since \(g\geq 0\), \(m g(x) \leq f(x) g(x) \leq M g(x)\) on \([a,b]\), and integrating, \[m \int ^b_ag \leq \int ^b_afg \leq M \int ^b_ag .\] If \(\int _a^bg = 0\), then \(\int _a^bfg = 0\) and any \(c\) will do. Otherwise \(\int _a^bg>0\), and \(\mu = \frac {\int _a^bfg}{\int _a^bg}\) lies in \([m,M]\); by the Intermediate Value Theorem \(f(c) = \mu \) for some \(c\in [a,b]\). □

Solution. Apply the Generalized Mean Value Theorem with \(f(x) = x\) and \(g(x) = \frac {1}{1 + x^2}\geq 0\): a point \(c\in [0,1]\) with the stated property exists. To find it, compute both integrals (using the Fundamental Theorem of Calculus, proved in the next section): \[\int ^1_0\frac {1}{1 + x^2} dx = \tan ^{-1}(1) - \tan ^{-1}(0) = \frac {\pi }{4},\qquad \int ^1_0\frac {x}{1 + x^2} dx = \frac {1}{2}\ln (1 + x^2)\Big |^1_0 = \frac {1}{2}\ln 2 .\] Thus \(\frac {1}{2}\ln 2 = \frac {\pi }{4} c\), and \(c = \frac {2\ln 2}{\pi }\approx 0.441\).

Solution. On \([0,1]\), \(1\leq 1+x^4\leq 2\), so \(\frac {1}{\sqrt {2}}\leq \frac {1}{\sqrt {1+x^4}}\leq 1\). Since the length of the interval is \(1\), the note on bounds gives the first pair of inequalities. (The integral is about \(0.927\).) Similarly \(1\leq e^{x^2}\leq e\) on \([0,1]\). A sharper upper bound comes from \(x^2\leq x\) on \([0,1]\): \(\int _0^1e^{x^2}dx\leq \int _0^1e^x\,dx = e - 1\approx 1.718\).

Solution. \(\int _0^3x^2\,dx = \frac {x^3}{3}\Big |_0^3 = 9\), so the average value is \(\frac {1}{3-0}\cdot 9 = 3\). Solving \(c^2 = 3\) in \([0,3]\) gives \(c = \sqrt {3}\approx 1.732\).

Solution. Let \(f(x) = 1\) for rational \(x\) and \(f(x) = -1\) for irrational \(x\), on \([0,1]\). Then \(|f| = 1\) is integrable. But every subinterval contains both kinds of points, so \(U(P,f) = 1\) and \(L(P,f) = -1\) for every partition, and \(f\) is not integrable.

Solution. \(\int _0^{2\pi }\sin x\,dx = -\cos x\Big |_0^{2\pi } = 0\). On \([0,\pi ]\), \(|\sin x| = \sin x\), and on \([\pi ,2\pi ]\), \(|\sin x| = -\sin x\), so by additivity \[\int _0^{2\pi }|\sin x|\,dx = \int _0^\pi \sin x\,dx - \int _\pi ^{2\pi }\sin x\,dx = 2 - (-2) = 4 .\] The inequality \(|\int f|\leq \int |f|\) is strict here because \(f\) changes sign: positive and negative parts cancel in \(\int f\) but not in \(\int |f|\).