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On this intervallet us consider the function xF (x) =f (t) dt,(6.40)aoften called an integral with variable upper limit.Since f ∈ R[a, b], it follows that f |[a,x] ∈ R[a, x] if [a, x] ⊂ [a, b]; thereforethe function x → F (x) is unambiguously defined for x ∈ [a, b].p = q = 2 was first obtained in 1821 by Cauchy and bears hisname. Hölder’s inequality for integrals with p = q = 2 was first discovered in 1859 by the Russianmathematician B.Ya.

Bunyakovskii (1804–1889). This important integral inequality (in the casep = q = 2) is called Bunyakovskii’s inequality or the Cauchy–Bunyakovskii inequality. One alsosome-times sees the less accurate name “Schwarz inequality” after the German mathematicianH.K.A.

Schwarz (1843–1921), in whose work it appeared in 1884.6 The algebraic Hölder inequality for6.3 The Integral and the Derivative361If |f (t)| ≤ C < +∞ on [a, b] (and f , being an integrable function, is boundedon [a, b]), it follows from the additivity of the integral and the elementary estimateof it thatF (x + h) − F (x) ≤ C|h|,(6.41)if x, x + h ∈ [a, b].Actually, we already discussed this while proving Lemma 3 in the precedingsection.It follows in particular from (6.41) that the function F is continuous on [a, b], sothat F ∈ C[a, b].We now investigate the function F more thoroughly.The following lemma is fundamental for what follows.Lemma 1 If f ∈ R[a, b] and the function f is continuous at a point x ∈ [a, b],then the function F defined on [a, b] by (6.40) is differentiable at the point x, andthe following equality holds:F (x) = f (x).Proof Let x, x + h ∈ [a, b].

Let us estimate the difference F (x + h) − F (x). Itfollows from the continuity of f at x that f (t) = f (x) + Δ(t), where Δ(t) → 0as t → x, t ∈ [a, b]. If the point x is held fixed, the function Δ(t) = f (t) − f (x)is integrable on [a, b], being the difference of the integrable function t → f (t) andthe constant f (x). We denote by M(h) the quantity supt∈I (h) |Δ(t)|, where I (h) isthe closed interval with endpoints x, x + h ∈ [a, b].

By hypothesis M(h) → 0 ash → 0.We now write x+h x x+hF (x + h) − F (x) =f (t) dt −f (t) dt =f (t) dt =aaxx+h f (x) + Δ(t) dt ==xx+h=f (x) dt +xwhere we have setx+hΔ(t) dt = f (x)h + α(h)h,xx+hΔ(t) dt = α(h)h.xSincex+hx Δ(t) dt ≤ x Δ(t) dt ≤ x+h xx+hM(h) dt = M(h)|h|,it follows that |α(h)| ≤ M(h), and so α(h) → 0 as h → 0 (in such a way that x + h ∈[a, b]).3626IntegrationThus we have shown that if the function f is continuous at a point x ∈ [a, b], thenfor displacements h from x such that x + h ∈ [a, b] the following equality holds:F (x + h) − F (x) = f (x)h + α(h)h,(6.42)where α(h) → 0 as h → 0.But this means that the function F (x) is differentiable on [a, b] at the point x ∈[a, b] and that F (x) = f (x).A very important immediate corollary of Lemma 1 is the following.Theorem 1 Every continuous function f : [a, b] → R on the closed interval [a, b]has a primitive, and every primitive of f on [a, b] has the form xF(x) =f (t) dt + c,(6.43)awhere c is a constant.Proof We have the implication (f ∈ C[a, b]) ⇒ (f ∈ R[a, b]), so that by Lemma 1the function (6.40) is a primitive for f on [a, b].

But two primitives F(x) and F (x)of the same function on a closed interval can differ on that interval only by a constant; hence F(x) = F (x) + c.For later applications it is convenient to broaden the concept of primitive slightlyand adopt the following definition.Definition 1 A continuous function x → F(x) on an interval of the real line iscalled a primitive (or generalized primitive) of the function x → f (x) defined onthe same interval if the relation F (x) = f (x) holds at all points of the interval, withonly a finite number of exceptions.Taking this definition into account, we can assert that the following theoremholds.Theorem 1 A function f : [a, b] → R that is defined and bounded on a closedinterval [a, b] and has only a finite number of points of discontinuity has a (generalized) primitive on that interval, and any primitive of f on [a, b] has the form(6.43).Proof Since f has only a finite set of points of discontinuity, f ∈ R[a, b], and byLemma 1 the function (6.40) is a generalized primitive for f on [a, b].

Here we havetaken into account, as already pointed out, the fact that by (6.41) the function (6.40)is continuous on [a, b]. If F(x) is another primitive of f on [a, b], then F(x)−F (x)is a continuous function and constant on each of the finite number of intervals intowhich the discontinuities of f divide the closed interval [a, b]. But it then followsfrom the continuity of F(x) − F (x) on all of [a, b] that F(x) − F (x) ≡ const on[a, b].6.3 The Integral and the Derivative3636.3.2 The Newton–Leibniz FormulaTheorem 2 If f : [a, b] → R is a bounded function with a finite number of pointsof discontinuity, then f ∈ R[a, b] andbf (x) dx = F(b) − F(a),(6.44)awhere F : [a, b] → R is any primitive of f on [a, b].Proof We already know that a bounded function on a closed interval having only afinite number of discontinuities is integrable (see Corollary 2 after Proposition 2 inSect.

6.1). The existence of a generalized primitive F(x) of the function f on [a, b]is guaranteed by Theorem 1 , by virtue of which F(x) has the form (6.43). Settingx = a in (6.43), we find that F(a) = c, and so xf (t) dt + F(a).F(x) =aIn particularbf (t) dt = F(b) − F(a),awhich, up to the notation for the variable of integration, is exactly formula (6.44),which was to be proved.Relation (6.44), which is fundamental for all of analysis, is called the Newton–Leibniz formula (or the fundamental theorem of calculus).The difference F(b) − F(a) of values of any function is often written F(x)|ba .

Inthis notation, the Newton–Leibniz formula assumes the formabbf (x) dx = F(x)a .Since both sides of the formula reverse sign when a and b are interchanged, theformula is valid for any relation between the magnitudes of a and b, that is, both fora ≤ b and for a ≥ b.In exercises of analysis the Newton–Leibniz formula is mostly used to computethe integral on the left-hand side, and that may lead to a some-what distorted ideaof its use.

The actual situation is that particular integrals are rarely found using aprimitive; more often one resorts to direct computation on a computer using highlydeveloped numerical methods. The Newton–Leibniz formula occupies a key position in the theory of mathematical analysis itself, since it links integration and3646Integrationdifferentiation.

In analysis it has a very far-reaching extension in the form of theso-called generalized Stokes’ formula.7An example of the use of the Newton–Leibniz formula in analysis itself is provided by the material in the next subsection.6.3.3 Integration by Parts in the Definite Integral and Taylor’sFormulaProposition 1 If the functions u(x) and v(x) are continuously differentiable on aclosed interval with endpoints a and b, thenabu · v (x) dx = (u · v)a −bbv · u (x) dx.(6.45)aIt is customary to write this formula in abbreviated form asbbu dv = u · v −aabv duaand call it the formula for integration by parts in the definite integral.Proof By the rule for differentiating a product of functions, we have(u · v) (x) = u · v (x) + u · v (x).By hypothesis, all the functions in this last equality are continuous, and hence integrable on the interval with endpoints a and b.

Using the linearity of the integral andthe Newton–Leibniz formula, we obtainb(u · v)(x)α =bu · v (x) dx +abu · v (x) dx.aAs a corollary we now obtain the Taylor formula with integral form of the remainder.Suppose on the closed interval with endpoints a and x the function t → f (t) hasn continuous derivatives. Using the Newton–Leibniz formula and formula (6.45),we carry out the following chain of transformations, in which all differentiationsand substitutions are carried out on the variable t: x xf (x) − f (a) =f (t) dt = −f (t)(x − t) dt =a7 G.G.aStokes (1819–1903) – British physicist and mathematician.6.3 The Integral and the Derivative365x= −f (t)(x − t)a +1= f (a)(x − a) −2xaxf (t)(x − t) dt =f (t) (x − t)2 dt =ax 11= f (a)(x − a) − f (t)(x − t)2 a +22xf (t)(x − t)2 dt =a x1 12= f (a)(x − a) + f (a)(x − a) −f (t) (x − t)3 dt =22·3 a= ··· =1= f (a)(x − a) + f (a)(x − a)2 + · · · +21f (n−1) (a)(x − a)n−1 + rn−1 (a; x),+2 · 3 · · · (n − 1)wherern−1 (a; x) =1(n − 1)!xf (n) (t)(x − t)n−1 dt.(6.46)aThus we have proved the following proposition.Proposition 2 If the function t → f (t) has continuous derivatives up to order ninclusive on the closed interval with endpoints a and x, then Taylor’s formula holds:f (x) = f (a) +1 1f (a)(x − a) + · · · +f (n−1) (a)(x − a)n−1 + rn−1 (a; x)1!(n − 1)!with remainder term rn−1 (a; x) represented in the integral form (6.46).We note that the function (x − t)n−1 does not change sign on the closed intervalwith endpoints a and x, and since t → f (n) (t) is continuous on that interval, thefirst mean-value theorem implies that there exists a point ξ such that x1rn−1 (a; x) =f (n) (t)(x − t)n−1 dt =(n − 1)! a x1(n)f (ξ )=(x − t)n−1 dt =(n − 1)!ax111f (n) (ξ ) − (x − t)n = f (n) (ξ )(x − a)n .=(n − 1)!nn!aWe have again obtained the familiar Lagrange form of the remainder in Taylor’stheorem.

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