[PDF] [PDF] Math 554 – Riemann Integration

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Theorem If f is continuous on [a, b], then f is Riemann-integrable on [a, b] Proof We use the condition (*) to prove that f is Riemann-integrable If ϵ > 0, we set 

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Math 554 { Riemann IntegrationHandout #9b (Dec. 4) Corollary.Iffis Riemann integrable on [a;b], then so is€fand Z b a(€f(x))dx=€Z b af(x)dx:

Proof.

We use the condition (*) and note that

U (€f;P) =€L(f;P); L(€f;P) =€U(f;P) and so b a(€f(x))dx=€Rb af(x)dx:2 Note. This result shows that for a function to be Riemann integrable it is enough to find, for each of one another. In this case the Riemann integral is within Examples:You should go through the following two examples on your own to make sure you understand the mechanics. 1. R b ak dx=k(b€a). 2. R b axdx=1 2 (b2€a2). [Hint: Use the result proved earlier:Pni=1i=n(n+1)=2.] Theorem.Iffis continuous on [a;b], thenfis Riemann-integrable on [a;b].

Proof.

is a

Ž >0 such thatjf(y)

j

U(f;P)€L(f;P) =n

X i The following indented material (definition and resulting corollary) are included for completeness and are not required for the further development. You will not be responsible for these two on the

Final.

Defn.ARiemann sumforffor a partitionPof an interval [a;b] is defined by R (f;P;˜) :=nX j=1f(˜j)Δxj where the ˜j, satisfyingxj€1"˜j"xj(1"j"n), are arbitrary.

Corollary.

Suppose thatfis Riemann integrable on [a;b], then there is a unique number( =Rb such that if

PžP1;P2, then

i: iii: where R(f;P1;˜) is any Riemann sum offfor the partitionP1. In this sense, we can interpretZb af(x)dx= limkPk!0R(f;P;˜): although we would actually need to show a little more to be precise.

Proof.Since

L (f;P2)""U(f;P1) for all partitions, we see that parts i.) and ii.) follow from the definition of the Riemann integral. To see part iii.), we observe thatmj"f(˜j)"Mj and hence that L (f;P1)"R(f;P1;˜)"U(f;P1):

But we also know that both

L (f;P1)""U(f;P1) and condition (*) hold, from which part iii.) follows.2 Note.The following theorem is also included for completeness, but will not be needed for our development. Theorem.Iffis monotone on [a;b], thenfis Riemann-integrable on [a;b]. Proof.Iffis constant, then we are done. We prove the case forfmonotone increasing. The case for monotone decreasing is similar. We again use the condition (*) to prove that partitionPwithkPk< Ž. Sincefis monotone increasing on [a;b], thenMi=f(xi) andmi=f(xi€1). Hence

U(f;P)€L(f;P) =nX

i =1(Mi€mi)Δxi n X i =1(f(xi)€f(xi€1))Δxi " k PknX i =1(f(xi)€f(xi€1)) Theorem.(Monotone Property of the Riemann Integral) Suppose thatfandgare Riemann integrable and kis a real number, then i.) g"fimpliesRb ag dx"Rb af dx. ii.)jRb af dxj "Rb ajfjdx Proof.Property i.) follows directly from the definition of the upper and lower integrals using the inequalities sup Ig(x)"supIf(x) and infIg(x)"infIf(x) for each subintervalI. Property ii.) is proved by applying property i.) to the inequality €j fj "f" jfj; to obtain€Rb ajfjdx"Rb af dx"Rb ajfjdx. But this inequality is equivalent to property ii.).2 Defn. We extend the definition of the integral to include general limits of integration. These are consistent with our earlier definition. 1. R a af(x)dx= 0. 2. R a bf(x)dx=€Rb af(x)dx. Theorem.Iffis Riemann integrable on [a;b], then it is Riemann integrable on each subinterval [c;d]"[a;b]. Moreover, ifc2[a;b], then (3) Z b af(x)dx=Z c af(x)dx+Z b cf(x)dx:

Proof.

applied tofover the interval [a;b], we have that there exists a partitionPof [a;b] such that condition (*) holds. Let˜Pbe the refinement obtained fromPwhich contains the pointscandd. LetPƒbe the partition obtained by restricting the partition˜Pto the interval [c;d], then and sofis Riemann integrable over [c;d]. To prove the identity (3), we use the fact that condition (*) holds whenfis Riemann integrable. Let respectively, to obtain partitions

PIwhich satisfy

(4) 0 "UI(f;PI)€Z I

We let

Pbe the partition of [a;b] formed by the union of the two partitionsP[a;c];P[c;b], and˜Pbe the common refinement of

PandP[a;b]. Observing that

(5)U[a;b](f;˜P) =U[a;c](f;˜P1) +U[c;b](f;˜P2); we can combine with inequality (4) to obtain

ŒŒŒRc

af dx+Rb cf dx€Rb af dxŒŒŒ"ŒŒŒU[a;c](f;˜P)€Rc af dxŒŒŒ+ŒŒŒU[c;b](f;˜P)€Rb cf dxŒŒŒ

ŒŒŒU[a;b](f;˜P)€Rb

af dxŒŒŒ Corollary.Each bounded, piece-wise continuous function with left and right hand limits at each point of an interval [a;b] is Riemann integrable. Moreover, its integral is the sum of the integrals of the "pieces". Theorem.(Intermediate Value Theorem for Integrals) Iffis continuous on [a;b], then there exists

˜betweenaandbsuch thatZb

af(x)dx=f(˜)(b€a):

Proof.Sincefis continuous on [a;b] and for‘:=R

b af dx b€athere holds min [a;b]f(x)"‘"max[a;b]f(x); then by the Intermediate Value Theorem for continuous functions, there exists a

˜2[a;b] such that

f(˜) =‘.2 Theorem.(Fundamental Theorem of Calculus, I. Derivative of an Integral) Suppose thatfis continuous on [ a;b] and setF(x) :=Rx af(y)dy, thenFis differentiable andF0(x) =f(x) for a < x < b

Proof.

Notice that

F (x0+h)€F(x0) h =R x0+h x 0f dx h =f(˜) for some˜betweenx0andx0+h. Hence, ash!0, then˜=˜hconverges tox0and so the displayed difference quotient has a limit of f(x0) ash!0.2

Theorem.

(Fundamental Theorem of Calculus, Part II. Integral of a Derivative) Suppose thatF is function with a continuous derivative on [ a;b], then Z b aF0(y)dy=F(x)jx=bx=a:=F(b)€F(a)

Proof.DefineG(x) :=Rx

aF0(y)dy, and setH:=F€G. Since the derivative ofHis identically zero (Part I of the Fundamental Theorem of Calculus), then the Mean Value Theorem implies that H (b)€H(a) = 0(b€a) = 0. Expressing this in terms ofFandGgives F (b)€F(a) =G(b)€G(a) =Z b aF0(y)dy; which establishes the theorem. 2 Defn.For a functionf, we call any functionF, whose derivative isf, anantiderivative off. Theorem.(Linearity Property of the Riemann Integral) Suppose thatfandgare Riemann inte- grable and kis a real number, then i.) Rb ak f(x)dx=kRb af(x)dx ii.)Rb af+g dx=Rb af dx+Rb ag dx

Proof.

It is a good exercise to prove these directly from the definition and to use the condition (*).

The serious student should go through this in detail just for additional practice . We present simpler

proofs using the Fundamental Theorems of Calculus. For part ii.) we letFbe an antiderivative of f andGbe an antiderivative ofg, thenH:=F+Gis an antiderivative off+g. Therefore R b af+g dx=H(b)€H(a) =F(b)€F(a) +G(b)€G(a) =Rb af dx+Rb ag dx: Part i.) is proved similarly using the corresponding property of differentiation. 2quotesdbs_dbs14.pdfusesText_20