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By Problem 2d) of Sect. 6.2, we may assume that ξ lies in the open intervalwith endpoints a and x.3666IntegrationThis reasoning can be repeated, taking the expression f (n) (ξ )(x − ξ )n−k , wherek ∈ [1, n], outside the integral in (6.46). The Cauchy and Lagrange forms of theremainder term that result correspond to the values k = 1 and k = n.6.3.4 Change of Variable in an IntegralOne of the basic formulas of integral calculus is the formula for change of variablein a definite integral.
This formula is just as important in integration theory as theformula for differentiating a composite function is in differential calculus. Undercertain conditions, the two formulas can be linked by the Newton–Leibniz formula.Proposition 3 If ϕ : [α, β] → [a, b] is a continuously differentiable mapping of theclosed interval α ≤ t ≤ β into the closed interval a ≤ x ≤ b such that ϕ(α) = a andϕ(β) = b, then for any continuous function f (x) on [a, b] the function f (ϕ(t))ϕ (t)is continuous on the closed interval [α, β], andbaf (x) dx =βf ϕ(t) ϕ (t) dt.(6.47)αProof Let F(x) be a primitive of f (x) on [a, b].
Then, by the theorem on differentiation of a composite function, the function F(ϕ(t)) is a primitive of the function f (ϕ(t))ϕ (t), which is continuous, being the composition and product of continuous functions on the closed interval [α, β]. By the Newton–Leibniz formulabβa f (x) dx = F(b) − F(a) and α f (ϕ(t))ϕ (t)dt = F(ϕ(β)) − F(ϕ(α)). But byhypothesis ϕ(α) = a and ϕ(β) = b, so that Eq. (6.47) does indeed hold.It is clear from formula (6.47) how convenient it is that we have not just the symbol for the function, but the entire differential f (x) dx in the symbol for integration,which makes it possible to obtain the correct integrand automatically when the newvariable x = ϕ(t) is substituted in the integral.So as not to complicate matters with a cumbersome proof, in Proposition 3 wedeliberately shrank the true range of applicability of (6.47) and obtained it by theNewton–Leibniz formula.
We now turn to the basic theorem on change of variable,whose hypotheses differ somewhat from those of Proposition 3. The proof of thistheorem will rely directly on the definition of the integral as the limit of Riemannsums.Theorem 3 Let ϕ : [α, β] → [a, b] be a continuously differentiable strictly monotonic mapping of the closed interval α ≤ t ≤ β into the closed interval a ≤ x ≤ bwith the correspondence ϕ(α) = a, ϕ(β) = b or ϕ(α) = b, ϕ(β) = a at theendpoints. Then for any function f (x) that is integrable on [a, b] the function6.3 The Integral and the Derivative367f (ϕ(t))ϕ (t) is integrable on [α, β] andϕ(β)βf (x) dx =ϕ(α)f ϕ(t) ϕ (t) dt.(6.48)αProof Since ϕ is a strictly monotonic mapping of [α, β] onto [a, b] with endpointscorresponding to endpoints, every partition Pt (α = t0 < · · · < tn = β) of the closedinterval [α, β] generates a corresponding partition Px of [a, b] by means of the images xi = ϕ(ti ) (i = 0, .
. . , n); the partition Px may be denoted ϕ(Pt ). Here x0 = aif ϕ(α) = a and x0 = b if ϕ(α) = b. It follows from the uniform continuity of ϕ on[α, β] that if λ(Pt ) → 0, then λ(Px ) = λ(ϕ(Pt )) also tends to zero.Using Lagrange’s theorem, we transform the Riemann sum σ (f ; Px , ξ ) as follows:nf (ξi )Δxi =i=1nf (ξi )(xi − xi−1 ) =i=1nn =f ϕ(τi ) ϕ (τ̃i )(ti − ti−1 ) =f ϕ(τi ) ϕ (τ̃i )Δti .i=1i=1Here xi = ϕ(ti ), ξi = ϕ(τi ), ξi lies in the closed interval with endpoints xi−1and xi , and the points τi and τ̃i lie in the interval with endpoints ti−1 and ti (i =1, .
. . , n).Nextnnf ϕ(τi ) ϕ (τ̃i )Δti =f ϕ(τi ) ϕ (τi )Δti +i=1i=1+nf ϕ(τi ) ϕ (τ̃i ) − ϕ (τi ) Δti .i=1Let us estimate this last sum. Since f ∈ R[a, b], the function f is bounded on[a, b]. Let |f (x)| ≤ C on [a, b]. Then nn f ϕ(τ ) ϕ (τ̃i ) − ϕ (τi ) Δti ≤ C ·ω ϕ ; Δi Δti ,i=1i=1where Δi is the closed interval with endpoints ti−1 and ti .This last sum tends to zero as λ(Pt ) → 0, since ϕ is continuous on [α, β].Thus we have shown thatni=1f (ξi )Δxi =nf ϕ(τi ) ϕ (τi )Δti + α,i=13686Integrationwhere α → 0 as λ(Pt ) → 0.
As already pointed out, if λ(Pt ) → 0, then λ(Px ) → 0also. But f ∈ R[a, b], so that as λ(Px ) → 0 the sum on the left-hand side of this ϕ(β)last equality tends to the integral ϕ(α) f (x) dx. Hence as λ(Pt ) → 0 the right-handside of the equality also has the same limit.But the sum ni=1 f (ϕ(τi ))ϕ (τi )Δti can be regarded as a completely arbitraryRiemann sum for the function f (ϕ(t))ϕ (t) corresponding to the partition Pt withdistinguished points τ = (τ1 , . . .
, τn ), since in view of the strict monotonicity of ϕ,any set of points τ can be obtained from some corresponding set ξ = (ξ1 , . . . , ξn ) ofdistinguished points in the partition Px = ϕ(Pt ).Thus, the limit of this sum is, by definition, the integral of the functionf (ϕ(t))ϕ (t) over the closed interval [α, β], and we have simultaneously provedboth the integrability of f (ϕ(t))ϕ (t) on [α, β] and formula (6.48).6.3.5 Some ExamplesLet us now consider some examples of the use of these formulas and the theoremson properties of the integral proved in the last two sections.Example 11−1#1 − x 2 dx==π/2#−π/212π/2−π/21 − sin t cos t dt =2(1 + cos 2t) dt =π/2−π/2cos2 t dt =π/21π1= .t + sin 2t 222−π/2In computing this integral we made the change of variable x = sin t and then,after finding a primitive for the integrand that resulted from this substitution, weapplied the Newton–Leibniz formula.Of course, we could have√ proceeded differently.
We could have√found the rathercumbersome primitive 12 x 1 − x 2 + 12 arcsin x for the function 1 − x 2 and thenused the Newton–Leibniz formula. This example shows that in computing a definite integral one can fortunately sometimes avoid having to find a primitive for theintegrand.Example 2 Let us show that πa)sin mx cos nx dx = 0,−π πc)for m, n ∈ N.−πcos2 nx dx = πb)π−πsin2 mx dx = π,6.3 The Integral and the Derivativea)πsin mx cos nx dx =−π=12369π−πsin(n + m)x − sin(n − m)x dx =π111−cos(n + m)x +cos(n − m)x =2n+mn−m−π= 0,if n − m = 0.
The case when n − m = 0 can be considered separately, and in thiscase we obviously arrive at the same result.b)π π11 π12x−sin 2mx = π.sin mx dx =(1 − cos 2mx) dx =2 −π22m−π−πc)π−πcos2 nx dx =12π−π(1 + cos 2nx) dx =π11x+sin 2nx = π.22n−πExample 3 Let f ∈ R[−a, a]. We shall show that! a a2 0 f (x) dx, if f is an even function,f (x) dx =0,if f is an odd function.−aIf f (−x) = f (x), then 0af (x) dx =f (x) dx +−a=−a a=2f (−t) dt +0f (x) dx =00af (x) dx =0f (x) dx =0aaaf (−t)(−1) dt +af (−x) + f (x) dx =0af (x) dx.0If f (−x) = −f (x), we obtain from the same computations that a a af (−x) + f (x) dx =f (x) dx =0 dx = 0.−a00Example 4 Let f be a function defined on the entire real line R and having period T ,that is f (x + T ) = f (x) for all x ∈ R.If f is integrable on each finite closed interval, then for any a ∈ R we have theequality a+T Tf (x) dx =f (x) dx,a03706Integrationthat is, the integral of a periodic function over an interval whose length equals theperiod T of the function is independent of the location of the interval of integrationon the real line:a+T0f (x) dx =aTf (x) dx +0f (x) dx +f (x) dx +0aa+Tf (x) dx =T0f (x) dx +af (t + T ) · 1 dt =0aT=T0a=f (x) dx +0f (x) dx +af (t) dt =0Tf (x) dx.0Here we have made the change of variable x = t + T and used the periodicity ofthe function f (x).1Example 5 Suppose we need to compute the integral 0 sin(x 2 ) dx, for examplewithin 10−2 .We know that the primitive sin(x 2 ) dx (the Fresnel integral) cannot be expressed in terms of elementary functions, so that it is impossible to use the Newton–Leibniz formula here in the traditional sense.We take a different approach.
When studying Taylor’s formula in differentialcalculus, we found as an example (see Example 11 of Sect. 5.3) that on the interval[–1, 1] the equalitysin x ≈ x −1 3 1 5x + x =: P (x)3!5!holds within 10−3 .But if | sin x − P (x)| < 10−3 on the interval [–1, 1], then | sin(x 2 ) − P (x 2 )| <−310 also, for 0 ≤ x ≤ 1.Consequently,1 sin x 2 dx −001 1 2 2 2 sin x − P x dx <P x dx ≤0110−3 dx < 10−3 .01Thus, to compute the integral 0 sin(x 2 ) dx with the required precision, it suffices1to compute the integral 0 P (x 2 ) dx. But01 111x 2 − x 3 + x 10 dx =3!5!01 31 71 11 1 111=x −x +x = 3 − 3!7 + 5!11 =3!3!75!110 P x 2 dx == 0.310 ± 10−3 ,6.3 The Integral and the Derivative371and therefore1 sin x 2 dx = 0.310 ± 2 × 10−3 = 0.31 ± 10−2 .0b1Example 6 The quantity μ = b−aa f (x) dx is called the integral average value ofthe function on the closed interval [a, b].Let f be a function that is defined on R and integrable on any closed interval.We use f to construct the new function1Fδ (x) =2δx+δf (t) dt,x−δwhose value at the point x is the integral average value of f in the δ-neighborhoodof x.We shall show that Fδ (x) (called the average of f ) is, compared to f , more regular.
More precisely, if f is integrable on any interval [a, b], then Fδ (x) is continuouson R, and if f ∈ C(R), then Fδ (x) ∈ C (1) (R).We verify first that Fδ (x) is continuous: x+δ+h x−δ≤Fδ (x + h) − Fδ (x) = 1 f(t)dt+f(t)dt2δ x+δx−δ+h C1C|h| + C|h| = |h|,2δδ≤if |f (t)| ≤ C, for example, in the 2δ-neighborhood of x and |h| < δ. It is obviousthat this estimate implies the continuity of Fδ (x).Now if f ∈ C(R), then by the rule for differentiating a composite functionddxaϕ(x)df (t) dt =dϕϕf (t) dt ·adϕ= f ϕ(x) ϕ (x),dxso that from the expressionFδ (x) =12δx+δaf (t) dt −12δx−δf (t) dtawe find thatf (x + δ) − f (x − δ).2δAfter the change of variable t = x + u in the integral, the function Fδ (x) can bewritten as1 δFδ (x) =f (x + u) du.2δ −δFδ (x) =3726IntegrationIf f ∈ C(R), then, applying the first mean-value theorem, we find thatFδ (x) =1f (x + τ ) · 2δ = f (x + τ ),2δwhere |τ | ≤ δ.
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