1. Let0
Assume first that c is not a partition point of P. In this case, c is contained in exactly one partition interval [xj−1,xj]. Then
1 ifi=j Mi :=sup{f(x):x∈[xi−1,xi]}= 0 ifi̸=j
n WethushavethatU(f,P)=Mi(xi−xi−1)=xj −xj−1 <δ.
Ontheotherhand,ifcis apartitionpointofP,sayx=xj for1≤j≤n−1,weobservethat
c is now contained in two partition intervals: [xj−1,xj] and [xj,xj+1]. Thus 1 ifi=jori=j+1
Mi :=sup{f(x):x∈[xi−1,xi]}= 0 ifi̸=j
Consequently,U(f,P)=Mi(xi−xi−1)=(xj −xj−1)+(xj+1−xj)<2δ.
In all possible cases we thus have the estimate 0 ≤ U(f,P) < 2δ if ||P|| < δ.
Finally, we need to determine L(f) and U(f): L(f) = sup{L(f,P) : P ∈ P[0,1]} = sup{0} = 0. Furthermore,0≤U(f)≤U(f,P)<2δwhichimpliesthat∀δ>0:0≤U(f)<2δ. Consequently (as proven in MATH 242), U(f) = 0.
We now have that L(f) = U(f) = 0, which proves that f is Riemann integrable on [0,1] and 1
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2. Letf:[0,1]→Rbedefinedby
−1 ifx=0
f(x) := 0 if 0 < x < 1 1 ifx=1
Prove directly from the definition of Riemann integrability that f is Riemann integrable on [a, b] b
Let δ > 0 be arbitrary and let P = {x0 = 0,…,xn = 1} be a partition of [0,1] of mesh less than δ. Note that
Consequently,
0 if1≤i
0 if1≤i≤j−1 Mi :=sup{f(x):x∈[xi−1,xi]}= 1 ifj≤i≤n
U(f,P)=Mi(xi −xi−1)=(xi −xi−1)=xn −xj−1 =b−xj−1
i=1 i=j nn
L(f,P)=mi(xi −xi−1)= (xi −xi−1)=xn −xj =b−xj i=1 i=j +1
mi :=inf{f(x):x∈[xi−1,xi]}=
0 if1≤i≤j
1 ifj+1≤i≤n
which implies that
U(f,P)−L(f,P)=(b−xj−1)−(b−xj)=xj −xj−1 <δ Consequently,
0 ≤ U (f ) − L(f ) ≤ U (f, P ) − L(f, P ) < δ
i.e. ∀δ > 0 : 0 ≤ U(f)−L(f) < δ which implies that U(f)−L(f) = 0 and thus that U(f) = L(f). This proves that f is Riemann integrable. We still need to determine the value of the integral. Since xj−1 < c ≤ xj, we have that
U(f,P)=b−xj−1 >b−c⇒U(f)=inf{U(f,P):P ∈P[a,b]}≥b−c L(f,P)=b−xj ≤b−c ⇒L(f)=sup{L(f,P):P ∈P[a,b]}≤b−c
Thus b − c ≤ U(f) = f = L(f) ≤ b − c which implies that f = b − c.
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4. Let f : [a,b] → R be Riemann integrable and non-negative (i.e. f(x) ≥ 0 for all x ∈ [a,b]). b
Hint: Prove first that for any partition P of [a,b] it holds that U(f,P) ≥ 0.
1. Solution: Using upper sums
LetP ={x0 =a,…,xn =b}beanypartitionof[a,b]. Then∀1≤i≤n: Mi := sup{f(x) : x ∈ [xi−1,xi]} ≥ f(xi) ≥ 0
and thus U(f,P) = Mi(xi − xi−1) ≥ 0. This means that 0 is a lower bound for the set
{U(f,P) : P ∈ P[a,b]}. And since the infimum of this set is its greatest lower bound, we conclude
that U(f) = inf {U(f,P) : P ∈ P[a,b]} ≥ 0.
Finally, since f is Riemann integrable we conclude that
2. Solution: Using lower sums
Prove that
LetP ={x0 =a,…,xn =b}beanypartitionof[a,b]. Then∀1≤i≤nweknowthattheset {f(x) : x ∈ [xi−1,xi]} is bounded from below by 0. Since the infimum of this set is its greatest lower bound, it follows that
∀1≤i≤n: mi :=inf{f(x):x∈[xi−1,xi]}≥0
and thus L(f, P ) = mi(xi − xi−1) ≥ 0 for all partitions P of [a, b]. Let P1 be any partition
of [a,b]. Then L(f) = sup{L(f,P) : P ∈ P[a,b]} ≥ L(f,P1) ≥ 0.
Finally, since f is Riemann integrable we conclude that
f = L(f) ≥ 0.
f = U(f) ≥ 0.