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It follows thatlim F (δ)(x) = f (x),δ→+0which is completely natural.6.3.6 Problems and Exercises1. Using the integral, findna) limn→∞ [ (n+1)2 + ··· +b) limn→∞1α +2α +···+nαnα+1n];(2n)2, if α ≥ 0.2. a) Show that any continuous function on an open interval has a primitive on thatinterval.b) Show that if f ∈ C (1) [a, b], then f can be represented as the difference oftwo nondecreasing functions on [a, b] (see Problem 4 of Sect. 6.1).3.
Show that if the function g is smooth, then the second mean-value theorem (Theorem 6 of Sect. 6.2) can be reduced to the first mean-value theorem through integration by parts.4. Show that if f ∈ C(R), then for any fixed closed interval [a, b], given ε > 0 onecan choose δ > 0 so that the inequality |Fδ (x) − f (x)| < ε holds on [a, b], whereFδ is the average of the function studied in Example 6.5. Show that x2 t1 2edt ∼ 2 ex as x → +∞.tx1 x+16. a) Verify that the function f (x) = x sin(t 2 ) dt has the following representation as x → ∞: 1cos(x 2 ) cos(x + 1)2−+O 2 .f (x) =2x2(x + 1)xb) Find limx→∞ xf (x) and limx→∞ xf (x).7. Show that if f : R → R is a periodic function that is integrable on every closedinterval [a, b] ⊂ R, then the function xF (x) =f (t) dtacan be represented as the sum of a linear function and a periodic function.6.3 The Integral and the Derivative3738.
a) Verify that for x > 1 and n ∈ N the function#n1 πPn (x) =x + x 2 − 1 cos ϕ dϕπ 0is a polynomial of degree n (the nth Legendre polynomial).b) Show thatdψ1 πPn (x) =.√π 0 (x − x 2 − 1 cos ψ)n9. Let f be a real-valued function defined on a closed interval [a, b] ⊂ R andξ1 , . . . , ξm distinct points of this interval. The values of the Lagrange interpolating polynomial of degree m − 1Lm−1 (x) :=mf (ξj )3 x − ξiξj − ξii=jj =1are equal to the values of the function at the points ξ1 , . .
. , ξm (the nodes of theinterpolation), and if f ∈ C (m) [a, b], then1 (m) fζ (x) ωm (x),m!f (x) − Lm−1 (x) =(where ωm (x) = mi=1 (x − ξi ) and ζ (x) ∈ ]a, b[ (see Exercise 11 in Sect. 5.3).b−aLet ξi = b+a+22 θi ; then θi ∈ [−1, 1], i = 1, . . . , m.a) Show thatbmb−a ci f (ξi ),2Lm−1 (x) dx =ai=1whereci =1 3−1 i=jt − θiθj − θidt.In particularbα1 ) a L0 (x) dx = (b − a)f ( a+b2 ), if m = 1, θ1 = 0;bb−aα2 ) a L1 (x) dx = 2 [f (a) + f (b)], if m = 2, θ1 = −1, θ2 = 1;ba+bα3 ) a L2 (x) dx = b−a6 [f (a) + 4f ( 2 ) + f (b)], if m = 3, θ1 = −1, θ2 = 0,θ3 = 1.b) Assuming that f ∈ C (m) [a, b] and setting Mm = maxx∈[a,b] |f (m) (x)|, estimate the magnitude Rm of the absolute error in the formulabaf (x) dx =abLm−1 (x) dx + Rm(*)3746Integrationband show that |Rm | ≤ Mmm! a |ωm (x)| dx.c) In cases α1 ), α2 ), and α3 ) formula (*) is called respectively the rectangular,trapezoidal, and parabolic rule.
In the last case it is also called Simpson’s rule.8Show that the following formulas hold in cases α1 ), α2 ), and α3 ):R1 =f (ξ1 )(b − a)2 ,4R3 = −R2 = −f (ξ2 )(b − a)3 ,12f (4) (ξ3 )(b − a)5 ,2880where ξ1 , ξ2 , ξ3 ∈ [a, b] and the function f belongs to a suitable class C (k) [a, b].d) Let f be a polynomial P . What is the highest degree of polynomials P forwhich the rectangular, trapezoidal, and parabolic rules respectively are exact?Let h = b−an , xk = a + hk (k = 0, 1, .
. . , n), and yk = f (xk ).e) Show that in the rectangular rulebf (x) dx = h(y0 + y1 + · · · + yn−1 ) + R1athe remainder has the form R1 = f 2(ξ ) (b − a)h, where ξ ∈ [a, b].f) Show that in the trapezoidal rulebf (x) dx =a*h)(y0 + yn ) + 2(y1 + y2 + · · · + yn−1 ) + R22the remainder has the form R2 = − f 12(ξ ) (b − a)h2 , where ξ ∈ [a, b].g) Show that in Simpson’s rule (the parabolic rule)abf (x) dx =h)(y0 + yn ) + 4(y1 + y3 + · · · + yn−1 ) +3*+ 2(y2 + y4 + · · · + yn−2 ) + R3 ,which can be written for even values of n, the remainder R3 has the formR3 = −f (4) (ξ )(b − a)h4 ,180where ξ ∈ [a, b].h) Starting from the relationπ =4108 T.Simpson (1710–1761) – British mathematician.dx,1 + x26.4 Some Applications of Integration375compute π within 10−3 , using the rectangular, trapezoidal, and parabolic rules. Notecarefully the efficiency of Simpson’s rule, which is, for that reason, the most widelyused quadrature formula.
(That is the name given to formulas for numerical integration in the one-dimensional case, in which the integral is identified with the areaof the corresponding curvilinear trapezoid.)10. By transforming formula (6.46), obtain the following forms for the remainderterm in Taylor’s formula, where we have set h = x − a: 1 (n)hna) (n−1)!(a + τ h)(1 − τ )n−1 dτ ;0 f√n1b) hn! 0 f (n) (x − h n t) dt.11. Show that the important formula (6.48) for change of variable in an integralremains valid without the assumption that the function in the substitution is monotonic.6.4 Some Applications of IntegrationThere is a single pattern of ideas that often guides the use of integration in applications; for that reason it is useful to expound this pattern once in its pure form. Thefirst subsection of this section is devoted to that purpose.6.4.1 Additive Interval Functions and the IntegralIn discussing the additivity of the integral over intervals in Sect.
6.2 we introducedthe concept of an additive (oriented) interval function. We recall that this is a function (α, β) → I (α, β) that assigns a number I (α, β) to each ordered pair of points(α, β) of a fixed closed interval [a, b], in such a way that the following equalityholds for any triple of points α, β, γ ∈ [a, b]:I (α, γ ) = I (α, β) + I (β, γ ).(6.49)It follows from (6.49) when α = β = γ that I (α, α) = 0, while for α = γ wefind that I (α, β) + I (β, α) = 0, that is, I (α, β) = −I (β, α). This relation shows theeffect of the order of the points α, β.SettingF(x) = I (a, x),by the additivity of the function I we haveI (α, β) = I (a, β) − I (a, α) = F(β) − F(α).Thus, every additive oriented interval function has the formI (α, β) = F(β) − F(α),(6.50)3766Integrationwhere x → F(x) is a function of points on the interval [a, b].It is easy to verify that the converse is also true, that is, any function x → F(x)defined on [a, b] generates an additive (oriented) interval function by formula(6.50).We now give two typical examples.xExample 1 If f ∈ R[a, b], the function F(x) = a f (t) dt generates via formula(6.50) the additive function βf (t) dt.I (α, β) =αWe remark that in this case the function F(x) is continuous on the closed interval[a, b].Example 2 Suppose the interval [0, 1] is a weightless string with a bead of unit massattached to the string at the point x = 1/2.Let F(x) be the amount of mass located in the closed interval [0, x] of the string.Then by hypothesis!0 for x < 1/2,F(x) =1 for 1/2 ≤ x ≤ 1.The physical meaning of the additive functionI (α, β) = F(β) − F(α)for β > α is the amount of mass located in the half-open interval ]α, β].Since the function F is discontinuous, the additive function I (α, β) in this casecannot be represented as the Riemann integral of a function – a mass density.
(Thisdensity, that is, the limit of the ratio of the mass in an interval to the length of theinterval, would have to be zero at any point of the interval [a, b] except the pointx = 1/2, where it would have to be infinite.)We shall now prove a sufficient condition for an additive interval function to begenerated by an integral, one that will be useful in what follows.Proposition 1 Suppose the additive function I (α, β) defined for points α, β of aclosed interval [a, b] is such that there exists a function f ∈ R[a, b] connected withI as follows: the relationinf f (x)(β − α) ≤ I (α, β) ≤ sup f (x)(β − α)x∈[α,β]x∈[α,β]holds for any closed interval [α, β] such that a ≤ α ≤ β ≤ b. Then bI (a, b) =f (x) dx.a6.4 Some Applications of Integration377Proof Let P be an arbitrary partition a = x0 < · · · < xn = b of the closed interval[a, b], let mi = infx∈[xi−1 ,xi ] f (x), and let Mi = supx∈[xi−1 ,xi ] f (x).For each interval [xi−1 , xi ] of the partition P we have by hypothesismi Δxi ≤ I (xi−1 , xi ) ≤ Mi Δxi .Summing these inequalities and using the additivity of the function I (α, β), weobtainnmi Δxi ≤ I (a, b) ≤i=1nMi Δxi .i=1The extreme terms in this last relation are familiar to us, being the upper andlower Darboux sums of the function f corresponding to the partition P of the closedinterval [a, b].
As λ(P ) → 0 they both have the integral of f over the closed interval[a, b] as their limit. Thus, passing to the limit as λ(P ) → 0, we find thatbI (a, b) =f (x) dx.aLet us now illustrate Proposition 1 in action.6.4.2 Arc LengthSuppose a particle is moving in space R3 and suppose its law of motion is known tobe r(t) = (x(t), y(t), z(t)), where x(t), y(t), and z(t) are the rectangular Cartesiancoordinates of the point at time t.We wish to define the length l[a, b] of the path traversed during the time intervala ≤ t ≤ b.Let us make some concepts more precise.Definition 1 A path in R3 is a mapping t → (x(t), y(t), z(t)) of an interval of thereal line into R3 defined by functions x(t), y(t), z(t) that are continuous on theinterval.Definition 2 If t → (x(t), y(t), z(t)) is a path for which the domain of the parameter t is the closed interval [a, b] then the pointsA = x(a), y(a), z(a) and B = x(b), y(b), z(b)in R3 are called the initial point and terminal point of the path.Definition 3 A path is closed if it has both an initial and terminal point, and thesepoints coincide.3786IntegrationDefinition 4 If Γ : I → R3 is a path, the image Γ (I ) of the interval I in R3 iscalled the support of the path.The support of an abstract path may turn out to be not at all what we would liketo call a curve.
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