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, n), and then pass to the limit as λ(P ) → 0.Theorem 4 can be interpreted as asserting that the integral is monotonic as afunction of the integrand.Theorem 4 has a number of useful corollaries.Corollary 1 If a ≤ b, f ∈ R[a, b]; and m ≤ f (x) ≤ M at each x ∈ [a, b], then bf (x) dx ≤ M · (b − a),(6.22)m · (b − a) ≤aand, in particular, if 0 ≤ f (x) on [a, b], then b0≤f (x) dx.aProof Relation (6.22) is obtained by integrating each term in the inequality m ≤f (x) ≤ M and using Theorem 4.Corollary 2 If f ∈ R[a, b], m = infx∈[a,b] f (x), and M = supx∈[a,b] f (x), thenthere exists a number μ ∈ [m, M] such that bf (x) dx = μ · (b − a).(6.23)ab1Proof If a = b, the assertion is trivial.
If a = b, we set μ = b−aa f (x) dx. It thenfollows from (6.22) that m ≤ μ ≤ M if a < b. But both sides of (6.23) reverse signif a and b are interchanged, and therefore (6.23) is also valid for b < a.Corollary 3 If f ∈ C[a, b], there is a point ξ ∈ [a, b] such that bf (x) dx = f (ξ )(b − a).a(6.24)3546IntegrationProof By the intermediate-value theorem for a continuous function, there is apoint ξ on [a, b] at which f (ξ ) = μ ifm = min f (x) ≤ μ ≤ max f (x) = M.x∈[a,b]x∈[a,b]Therefore (6.24) follows from (6.23).The equality (6.24) is often called the first mean-value theorem for the integral.We, however, reserve that name for the following somewhat more general proposition.Theorem 5 (First mean-value theorem for the integral) Let f, g ∈ R[a, b], m =infx∈[a,b] f (x), and M = supx∈[a,b] f (x).
If g is nonnegative (or nonpositive) on[a, b], then b b(f · g)(x) dx = μg(x) dx,(6.25)aawhere μ ∈ [m, M].If, in addition, it is known that f ∈ C[a, b], then there exists a point ξ ∈ [a, b]such that b b(f · g)(x) dx = f (ξ )g(x) dx.(6.26)aaProof Since interchanging the limits of integration leads to a simultaneous signreversal on both sides of Eq. (6.25), it suffices to verify this equality for the casea < b. Reversing the sign of g(x) also reverses the signs of both sides of (6.25), sothat we may assume without loss of generality that g(x) ≥ 0 on [a, b].Since m = infx∈[a,b] f (x) and M = supx∈[a,b] f (x), we have, for g(x) ≥ 0,mg(x) ≤ f (x)g(x) ≤ Mg(x).Since m · g ∈ R[a, b], f · g ∈ R[a, b], and M · g ∈ R[a, b], applying Theorem 4and Theorem 1, we obtain b b bmg(x) dx ≤f (x)g(x) dx ≤ Mg(x) dx.(6.27)aaabIf a g(x) dx = 0, it is obvious from these inequalities that (6.25) holds.bIf a g(x) dx = 0, then, settingμ=bg(x) dxa−1 b·(f · g)(x) dx,awe find by (6.27) thatm ≤ μ ≤ M,but this is equivalent to (6.25).6.2 Linearity, Additivity and Monotonicity of the Integral355The equality (6.26) now follows from (6.25) and the intermediate-value theoremfor a function f ∈ C[a, b], if we take account of the fact that when f ∈ C[a, b], wehavem = min f (x)x∈[a,b]and M = max f (x).x∈[a,b]We remark that (6.23) results from (6.25) if g(x) ≡ 1 on [a, b].c.
The Second Mean-Value Theorem for the IntegralThe so-called second mean-value theorem4 is significantly more special and delicatein the context of the Riemann integral.So as not to complicate the proof of this theorem, we shall carry out a usefulpreparatory discussion that is of independent interest.Abel’stransformation. Thisis the name given to the following transformation ofthe sum ni=1 ai bi . Let Ak = ki=1 ai ; we also set A0 = 0. Thenmai bi =i=lnnn(Ai − Ai−1 )bi =Ai bi −Ai−1 bi =i=1=ni=1Ai bi −i=1n−1i=1Ai bi+1 = An bn − A0 b1 +i=0n−1Ai (bi − bi+1 ).i=1Thusnai bi = (An bn − A0 b1 ) +i=1n−1Ai (bi − bi+1 ),(6.28)i=1or, since A0 = 0,nai bi = An bn +i=1n−1Ai (bi − bi+1 ).(6.29)i=1Abel’s transformation provides an easy verification of the following lemma.Lemma 2 If the numbers Ak = ki=1 ai (k = 1, .
. . , n) satisfy the inequalities m ≤Ak ≤ M and the numbers bi (i = 1, . . . , n) are nonnegative and bi ≥ bi+1 for i =1, . . . , n − 1, thenmb1 ≤nai bi ≤ Mb1 .(6.30)i=14 Under an additional hypothesis on the function, one that is often completely acceptable, Theorem 6 in this section could easily be obtained from the first mean-value theorem. On this point, seeProblem 3 at the end of Sect.
6.3.3566IntegrationProof Using the fact that bn ≥ 0 and bi − bi+1 ≥ 0 for i = 1, . . . , n − 1, we obtainfrom (6.29),nai bi ≤ Mbn +i=1n−1M(bi − bi+1 ) = Mbn + M(b1 − bn ) = Mb1 .i=1The left-hand inequality of (6.30) is verified similarly.Lemma 3 If f ∈ R[a, b], then for any x ∈ [a, b] the function xf (t) dtF (x) =(6.31)ais defined and F (x) ∈ C[a, b].Proof The existence of the integral in (6.31) for any x ∈ [a, b] is already knownfrom Proposition 4 of Sect. 6.1; therefore it remains only for us to verify that thefunction F (x) is continuous.
Since f ∈ R[a, b], we have |f | ≤ C < ∞ on [a, b].Let x ∈ [a, b] and x + h ∈ [a, b]. Then, by the additivity of the integral and inequalities (6.19) and (6.20) we obtain x x+hF (x + h) − F (x) = f (t) dt −f (t) dt =a= xax+h f (t) dt ≤ x f (t) dt ≤ C|h|.x+h Here we have used inequality (6.20) taking account of the fact that for h < 0 wehave x+h x x f (t) dt = −f (t) dt =f (t) dt.
xx+hx+hThus we have shown that if x and x + h both belong to [a, b], thenF (x + h) − F (x) ≤ C|h|(6.32)from which it obviously follows that the function F is continuous at each point of[a, b].We now prove a lemma that is a version of the second mean-value theorem.Lemma 4 If f, g ∈ R[a, b] and g is a nonnegative nonincreasing function on [a, b]then there exists a point ξ ∈ [a, b] such thatbaξ(f · g)(x) dx = g(a)f (x) dx.a(6.33)6.2 Linearity, Additivity and Monotonicity of the Integral357Before turning to the proof, we note that, in contrast to relation (6.26) of the firstmean-value theorem, it is the function f (x) that remains under the integral sign in(6.33), not the monotonic function g.Proof To prove (6.33), as in the cases considered above, we attempt to estimate thecorresponding Riemann sum.Let P be a partition of [a, b].
We first write the identityb(f · g) dx =n xi(f · g)(x) dx =i=1 xi−1a=nxif (x) dx +g(xi−1 )xi−1i=1n )*g(x) − g(xi−1 ) f (x) dxxii=1 xi−1and then show that the last sum tends to zero as λ(P ) → 0.Since f ∈ R[a, b], it follows that |f (x)| ≤ C < ∞ on [a, b]. Then, using theproperties of the integral already proved, we obtain n n xi xi ) *g(x) − g(xi )f (x) dx ≤g(x) − g(xi−1 ) f (x) dx ≤xi−1xi−1i=1i=1≤Cn xig(x) − g(xi−1 ) dx ≤i=1 xi−1≤Cnω(g; Δi )Δxi → 0i=1as λ(P ) → 0, because g ∈ R[a, b] (see Proposition 2 of Sect. 6.1). Thereforebn(f · g)(x) dx = limλ(P )→0axif (x) dx.g(xi−1 )xi−1i=1We now estimate the sum on the right-hand side of (6.34).
SettingF (x) =xf (t) dt,aby Lemma 3 we obtain a continuous function on [a, b].Letm = min F (x)x∈[a,b]and M = max F (x).x∈[a,b](6.34)358Since6 xixi−1Integrationf (x) dx = F (xi ) − F (xi−1 ), it follows thatnxig(xi−1 )f (x) dx =nF (xi ) − F (xi−1 ) g(xi−1 ).xi−1i=1(6.35)i=1Taking account of the fact that g is nonnegative and nonincreasing on [a, b], andsettingai = F (xi ) − F (xi−1 ),bi = g(xi−1 ),we find by Lemma 2 thatnmg(a) ≤F (xi ) − F (xi−1 ) g(xi−1 ) ≤ Mg(a),(6.36)i=1sinceAk =kai = F (xk ) − F (x0 ) = F (xk ) − F (a) = F (xk ).i=1Having now shown that the sums (6.35) satisfy the inequalities (6.36), and recalling relation (6.34), we havemg(a) ≤b(f · g)(x) dx ≤ Mg(a).(6.37)aIf g(a) = 0, then, as inequalities (6.37) show, the relation to be proved (6.33) isobviously true.If g(a) > 0, we set b1(f · g)(x) dx.μ=g(a) a x It follows from (6.37) that m ≤ μ ≤ M, and from the continuity of F (x) =a f (t) dt on [a, b] that there exists a point ξ ∈ [a, b] at which F (ξ ) = μ.
But thatis precisely what formula (6.33) says.Theorem 6 (Second mean-value theorem for the integral) If f, g ∈ R[a, b| and gis a monotonic function on [a, b], then there exists a point ξ ∈ [a, b| such thatab(f · g)(x) dx = g(a)aξbf (x) dx + g(b)f (x) dx.(6.38)ξThe equality (6.38) (like (6.33), as it happens) is often called Bonnet’s formula.55 P.O. Bonnet (1819–1892) – French mathematician and astronomer. His most important mathematical works are in differential geometry.6.2 Linearity, Additivity and Monotonicity of the Integral359Proof Let g be a nondecreasing function on [a, b]. Then G(x) = g(b) − g(x) isnonnegative, nondecreasing, and integrable on [a, b]. Applying formula (6.33), wefind b ξ(f · G)(x) dx = G(a)f (x) dx.(6.39)aaButbb(f · G) dx = g(b)αf (x) dx −aξG(a)a(f · g)(x) dx,aξf (x) dx = g(b)bξf (x) dx − g(a)af (x) dx.aTaking account of these relations and the additivity of the integral, we obtain theequality (6.38), which was to be proved, from (6.39).If g is a nonincreasing function, setting G(x) = g(x) − g(b), we find that G(x)is a nonnegative, nonincreasing, integrable function on [a, b].
We then obtain (6.39)again, and then formula (6.38).6.2.4 Problems and Exercises1. Show that if f ∈ R[a, b] and f (x) ≥ 0 on [a, b], then the following statementsare true.a) If the function f (x) assumes a positive value f (x0 ) > 0 at a point of continuity x0 ∈ [a, b], then the strict inequality bf (x) dx > 0aholds.bb) The condition a f (x) dx = 0 implies that f (x) = 0 at almost all points of[a, b].2.
Show that if f ∈ R[a, b], m = inf]a,b[ f (x), and M = sup]a,b[ f (x), thenba) a f (x) dx = μ(b − a), where μ ∈ [m, M] (see Problem 5a of Sect. 6.1);b) if f is continuous on [a, b], there exists a point ξ ∈ ]a, b[ such that bf (x) dx = f (ξ )(b − a).a3.
Show that if f ∈ C[a, b], f (x) ≥ 0 on [a, b], and M = max[a,b] f (x), thenblimn→∞1/nnf (x) dxa= M.3606Integration4. a) Show that if f ∈ R[a, b], then |f |p ∈ R[a, b] for p ≥ 0.b) Starting from Hölder’s inequality for sums, obtain Hölder’s inequality forintegrals:6 b (f · g)(x) dx ≤ab1/p |f | (x) dx·pab1/q|g| (x) dxq,aif f, g ∈ R[a, b], p > 1, q > 1, and p1 + q1 = 1.c) Starting from Minkowski’s inequality for sums, obtain Minkowski’s inequality for integrals:b1/p|f + g| (x) dxpa≤b1/p|f | (x) dxpa+b1/p|g| (x) dxp,aif f, g ∈ R[a, b] and p ≥ 1. Show that this inequality reverses direction if 0 <p < 1.d) Verify that if f is a continuous convex function on R and ϕ an arbitrarycontinuous function on R, then Jensen’s inequality c11 c fϕ(t) dt ≤f ϕ(t) dtc 0c 0holds for c = 0.6.3 The Integral and the Derivative6.3.1 The Integral and the PrimitiveLet f be a Riemann-integrable function on a closed interval [a, b].
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