Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous calculus classes...) You may use the idetity i 3 = 4 (4 + 2 3 + 2 ). Solutio: Let N ad defie the dissectio D = {,, 2,...,, }. The fuctio f(x) = x 3 is icreasig betwee ad. Therefore the supremum of the values o a iterval (x i, x i ) is f(x i ) = x 3 i, ad the ifimum is f(x i ) = x 3 i. Thus we ca calculate the lower ad upper sums of f with respect to D : L(f, D ) = A similar calculatio gives ( i )3. = 4 (i ) 3 = 4 4 (4 + 2 3 + 2 ) 3 4 = 4 2 + 4 2 U(f, D ) = 4 + 2 + 4 2. As we have L(f, D ) 4 ad also U(f, D ) 4. From this it follows that f is itegrable o [, ] ad that as required. x 3 dx = 4,
(2) Suppose that g(x) is a cotiuous fuctio o a iterval [a, b] such that g(x) > for all x. Show that Solutio b a g(x)dx >. Sice g(x) o [a, b] the fuctio g is defied ad cotiuous o [a, b]. Hece there is M > so that g(x) < M for all x. This meas that g(x) > M > for all x i [a, b]. Let D = {a, b}. The L(f, D) b a f. However, so we re doe. L(f, D) > b a >, M (3) Let f : [, 3] R be defied by: if x 2 f(x) = if x (2, 3] Q if x (2, 3] Q Prove that f is ot Riema itegrable. Solutio: f is itegrable o [, 3] if ad oly if it is itegrable o [, 2] ad also o [2, 3]. Thus, it is eough to show that f is ot itegrable o [2, 3]. Well, for ay dissectio D = {2 = x,..., x = 3} of [2, 3] we have sup{f(x) x (x i, x i )} =, ad if{f(x) x (x i, x i )} =. Therefore L(f, D) =, ad 3 f =. Similarly, ad 2 U(f, D) =, 3 2 f =. This shows that f is ot itegrable o [2, 3] ad hece it is ot itegrable o [, 3].
(4) Defie p : [, 2] R as follows: p(x) = { x, if x if x > Prove that p(x) is Riema itegrable o [, 2] ad determie Solutio: p(x)dx. f is cotiuous so itegrable o [, 2]. We have f = Howie works out f = 2. O [, 2], f is idetically, so it is easy to see that all lower ad upper sums (with respect to ay dissectio) are equal to, which meas that Therefore, f = 2. f + f =. (5) Suppose that f : [a, b] R is (Riema) itegrable o [a, b]. Prove that b a b a f(x)dx = lim f f. ( a + i b a ). Solutio: Fix N. Cosider the followig dissectio: D = {a, a+ b a 2 +b a, a+b a 2 +2(b a),..., a+ b a 2 )(b a) +(, b} This is a partitio of [a, b] ito itervals, oe of legth b a 2, oe of legth b a ad 2 of legth b a 2 + b a. It is more atural to defie the partitio without the + b a 2 wat the poit a + i(b a) is. Therefore, it the itervals are I,,..., I,, ad poits, but we to be i the iterior of the i th iterval, which it m i, = if{f(x) x I i, } M i, = sup{f(x) x I i, },
from which it follows that (for i < ) m i, f(a + i b a ) M i,. Therefore, the terms of the lower ad the upper sum of D match up to give a iequality with the required sum b a f ( ), except for the first ad lat terms where the legths of the itervals are wrog. The best thig to do is ote that these terms are small, ad goig to as. Therefore, if we prove that there is a umber L so that L = lim L(f, D ) = lim U(f, D ) () the we ll see that L = b a f ad that L = lim b a f ( ), which is what we are required to prove. Thus, we are left to prove Equatio (). To do this, ote that we kow that f is itegrable. Let ɛ > be arbitrary. There is a dissectio D of [a, b] so that U(f, D) L(f, D) < ɛ 2. Suppose that there are k itervals i D ad let these itervals be J,..., J k. Suppose also that M = sup{ f(x) x [a, b]}. (We ll assume that M >, sice otherwise there s almost othig to prove.) For ay, ay give term i the sum b a f ( ) has size at most M b a. Choose N N so that N > 2kM(b a) ɛ. The, if > N we cosider U(f, D ) L(f, D ). This has terms i it, ivolvig the differece betwee the supremum ad the ifimum of f o a give iterval. All but at most k of the itervals I i, are etirely cotaied withi some iterval J s. Sice the itervals J s are bigger, the differece betwee the supremum ad ifimum of f o the iterval J s ca oly be bigger tha o the correspodig I i,. The choice of guaratees that each of the other terms i the differece (those correspodig to itervals ot etirely cotaied i a J s ) have size at most ɛ 2k, ad there are at most k of these terms. Thus, we split the sum U(f, D ) L(f, D ) ito two sets of terms. The first set of terms has total size o bigger tha U(f, D) L(f, D) < ɛ 2,
while the secod set of terms has size at most k. ɛ 2k = ɛ 2. Thus, puttig these together we see that U(f, D ) L(f, D ) < ɛ which proves Equatio (), ad fiishes the proof.