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Learn about the Riemann integral, its definition, properties, and how to evaluate integrals using upper and lower sums. Includes examples and a criterion for integrability.
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Chapter 7 Integration
Section7.1 The Riemann Integral
Definition 7.1.1 Let [a,b] be an interval in . A partitionP of [a,b] is a finite set of points {x0, x1,…, xn} in [a,b] such that a=x0<x1< … <xn=b. If P and Qare two partitions of [a,b] with PQ, then Q is called a refinement of P. Definition 7.1.2 Suppose thatfis a bounded function defined on [a,b] and that P ={x0,…,xn} is a partition of [a,b]. For each i= 1, …,nwe let mi(f ) = inf {f (x) : x [xi –1, xi]} and Mi(f) = sup {f(x): x [xi –1, xi]}. Lettingxi = xi – xi–1 (i = 1,…, n), we define the lower sum offwith respect to P to be and the upper sum offwith respect to P to be The lower sum The upper sum M1 m1 x1 x1
Since f is bounded on [a,b], there exist numbers m and M such that m f (x) M for all x [a,b]. Thus for any partition P of [a, b] we have m(b – a) L( f, P) U( f, P) M(b– a). This implies that the upper and lower sums forfform a bounded set, and it guarantees the existence of the following upper and lower integrals off. The lower sum The upper sum M M M1 m1 m m x1 x1
Definition 7.1.3 Letfbe a bounded function defined on [a,b]. Then U( f )=inf {U(f,P): P is a partition of [a,b]} is called the upper integral offon [a,b]. Similarly, L( f )=sup {L(f,P): P is a partition of [a,b]} is called the lower integral offon [a,b]. If these upper and lower integralsareequal,thenwesaythatfis (Riemann)integrableon [a,b], andwedenote their common value by or by Thatis, ifL(f ) = U(f ), then is the (Riemann) integral offon [a,b].
y y x x Theorem 7.1.4 Letfbe a bounded function on [a,b]. If P and Q are partitions of [a,b] and Q is a refinement of P, then L( f,P) L( f,Q) U(f,Q) U(f,P). Idea of Proof: In a refinement, new points are added to the partition. These additional points increase the lower sums and decrease the upper sums. the lower sum the upper sum y=f(x) y=f(x) Mk area subtracted area added mk mk(xk –xk–1) Mk(xk –xk–1) xk–1 xk–1 xk xk x* x*
Practice 7.1.5 Letfbe a bounded function on [a,b]. If P and Q are partitions of [a,b], prove that L(f,P)U(f,Q). Theorem 7.1.6 Letfbe a bounded function on [a,b]. Then L( f )U(f ). Proof: If P and Q are partitions of [a,b], then by Practice 7.1.5 we have L(f,P)U(f,Q). Thus U( f,Q) is an upper bound for the setS = {L( f,P): P is a partition of [a,b]}. It follows that U( f,Q) is greater than or equal to sup S = L(f). That is, L(f)U( f,Q) for each partition Q of [a,b]. But then, L( f )inf{U( f,Q): Q is a partition of [a,b]} = U( f ).
Example 7.1.7 Let’s use upper and lower sums to evaluate For each n , let Pn be the partition in which xi = 1/n for each i = 1, 2,…,n. Sincef(x) = x2 is an increasing function on [0,1], on any subinterval [(i – 1)/n,i/n], the supremum occurs at the right endpoint and we have from Exercise 3.1.3 Thus, as n .
Likewise, sincef(x) = x2 is an increasing function on [0,1], on any subinterval [(i – 1)/n,i/n], the infimum occurs at the left endpoint and we have Thus, as n . Since limnU(f,Pn) = 1/3 and limnL(f,Pn) = 1/3, we must have U(f)1/3andL(f)1/3. ButsinceL( f )U( f )byTheorem7.1.6,thismeans that L( f) = U( f) = 1/3, so that
Example 7.1.8 Using upper and lower sums to evaluate integrals is messy, but they can be helpful in showing a given function is notintegrable. Consider the function g: [0,2] defined by LetP ={x0,x1,…,xn} be any partition of [0,2]. Since each subinterval [xi–1, xi] contains both rational and irrational numbers, we have Mi = 1 and mi = 0 for all i = 1,…,n. Thus, and Since the upper and lower integralsofgon[0,2] arenotequal, weconcludethatgisnotintegrable on [0,2]. It follows that U(g) = 2 and L(g) = 0. In our next theorem we establish a criterion for determining when a function is integrable. It will be useful in the next section.
Theorem 7.1.9 Letfbe a bounded function on [a,b]. Thenfis integrableiff for each > 0 there exists a partition Pof [a,b] such that U( f,P) – L( f,P)<. Proof: Suppose thatfis integrable, so that L( f) = U( f). Given> 0, there exists a partition P1 of [a,b] such that L( f, P1) > L( f ) – /2. (ThisfollowsfromthedefinitionofL( f )asasupremum.) Similarly, there exists a partition P2 of [a,b] such that U( f, P2) < U( f ) + /2 Let P =P1P2 . Then P is a refinement of both P1 and P2, so Theorem 7.1.4 implies U(f, P) – L( f, P) U(f, P2) – L( f, P1) < U( f) + /2 – L( f ) – /2 = U( f) – L( f) + = .
Conversely, given > 0, suppose that there exists a partition P of [a,b] such that U( f,P) <L( f,P)+. Then we have U( f ) U( f,P) <L( f,P)+ L ( f ) + . Since > 0 is arbitrary, we must have U( f ) L( f ). But then Theorem 7.1.6 implies that L( f ) = U( f ), so thatfis integrable.
Example (not in book) Let f (x) = 2x on [1, 4] and let P = {1, 2, 3.5, 4}. 8 = (2)(1) + (4)(1.5) + (7)(0.5) = 2 + 6 + 3.5 = 11.5 6 4 = (4)(1) + (7)(1.5) + (8)(0.5) = 4 + 10.5 + 4 = 18.5 2 1 2 3 4