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Department
Mathematics
Course
MATH 118
Professor
Steve Spencer
Semester
Winter

Description
Chapter 1. Review of the Riemann Integral The Riemann Integral 1.1 De▯nition: A partition of the closed interval [a;b] is a set X = fx ;x ;▯▯▯;x g 0 1 n with a = x < x < x < ▯▯▯ < x = b: 0 1 2 n The intervals [xi▯1;xi] are called the subintervals of [a;b], and we write ▯ i = x i x i▯1 th for the size of the i subinterval. Note that Xn ▯ i = b ▯ a: i=1 The size of the partition X, denoted by jXj is ▯ ▯ jXj = max ▯ x 1 ▯ i ▯ n : i 1.2 De▯nition: Let X be a partition of [a;b], and let f : [a;b] ! R be bounded. A Riemann sum for f on X is a sum of the form Xn S = f(t )▯ x for some t 2 [x ;x ]: i i i i▯1 i i=1 The points tiare called sample points. 1.3 De▯nition: Let f : [a;b] ! R be bounded. We say that f is (Riemann) integrable on [a;b] when there exists a number I with the property that for every ▯ > 0 there exists ▯ > 0 such that for every partition X of [a;b] with jXj < ▯ we have jS ▯ Ij < ▯ for every Riemann sum for f on X, that is ▯n ▯ ▯ ▯ ▯ f(ti)▯ i ▯ I ▯ ▯: ▯i=1 ▯ for every choice of i 2 [i▯1;x i The number I can be shown to be unique. It is called the (Riemann) integral of f on [a;b], and we write Z Z b b I = f , or I = f(x)dx: a a 1.4 Example: Show that the constant function f(x) = c is integrable on any interval Z b [a;b] and we have c dx = c(b ▯ a). a Solution: The solution is left as an exercise. 1 1.5 Example: Show that the identity function f(x) = x is integrable on any interval Z b 1 2 2 [a;b], and we have x dx = 2(b ▯ a ). a Solution: Let ▯ > 0. Choose ▯ = 2▯ . Let X be any partition of [a;b] with jXj < ▯. Let n b▯a n P P 1 2 2 ti2 [xi▯1 ;xi] and set S = f(ti)▯ix = ti▯ i. We must show that jS▯ 2 (b ▯a )j < ▯. i=1 i=1 Notice that n n Pn X X 2 2 (xi+ x i▯1)▯ i = (xi+ x i▯1)(xi▯ x i▯1) = xi ▯ xi▯1 i=1 i=1 i=1 = (x12 ▯ x 0 + (x 22▯ x 1 + ▯▯▯ + (x n▯12 ▯ xn▯2 2) + (xn2 ▯ x n▯12) = ▯x 02 + (x1 2▯ x 1 + ▯▯▯ + (x n▯1 2 ▯ xn▯1 2) + xn2 = x n2▯ x 02 = b ▯ a 2 ▯ 1 ▯ 1 1 and that when t i [x i▯1;x i we have t i 2(xi+ x i▯1 ) ▯ 2(xi▯ x i▯1) = 2 x,iand so ▯ ▯ ▯ 1 2 2 ▯ ▯ P 1Pn ▯ S ▯ 2(b ▯ a ) = ▯ ti▯ix ▯ 2 (xi+ x i▯1 )▯ix▯ i=1 i=1 ▯ P ▯ ▯ ▯ = ▯ ti▯ (x +ix i+1) ▯ i ▯ i=1 2 P ▯ 1 ▯ ▯ ti▯ (2 + i i+1) ▯ix i=1 P P ▯ 1▯ i▯ xi▯ 1▯▯ i i=1 2 i=1 2 = ▯(b ▯ a) = ▯: 2 ▯ 1.6 Example: Let f(x) = 1 if x 2 Q Show that f is not integrable on [0;1]. 0 if x = Q: R 1 Solution: Suppose, for a contradiction, that f is integrable on [0;1], and write I = 0 f. Let ▯ = . Choose ▯ so that for every partition X with jXj < ▯ we have jS▯Ij < 1for every 2 n2 P Riemann sum S for f on X. Choose a partition X with jXj < ▯. Let S = 1 f(ti)▯ i i=1 Pn where each t i [x i▯1 ;xi] is chosen with ti2 Q, and let S =2 f(si)▯ i where each i=1 si2 [x i▯1;x i is chosen with si2= Q. Note that we have jS ▯ 1j < 1 and jS 2 Ij < 1. n n 2 2 P P Since each ti2 Q we have f(t ) i 1 and so S = 1 f(ti)▯ix = ▯ i = 1 ▯ 0 = 1, and in1 i=1 P 1 since each si2= Q we have f(s )i= 0 and so S =2 f(si)▯ i = 0. Since jS 1 Ij < 2 we i=1 have j1 ▯ Ij < 1 and so 1 < I < 3, and since jS2▯ Ij < 1 we have j0 ▯ Ij < 1 and so 1 1 2 2 2 2 2 ▯ 2 < I < ,2giving a contradiction. 2 Integrals of Continuous Functions 1.7 Theorem: (Continuous Functions are Integrable) Let f : [a;b] ! R be continuous. Then f is integrable on [a;b]. Proof: We omit the proof, which is quite di▯cult. 1.8 Note: Let f be integrable on [a;b]. Lnt X be any sequence of partitions of [a;b] with n!1m jXnj = 0. Let n be any Riemann sum for f on Xn. Then fSng converges with Z b lim S = f(x)dx: n!1 n a R b Proof: Write I = a f. Given ▯ > 0, choose ▯ > 0 so that for every partition X of [a;b] with jXj < ▯ we have jS ▯ Ij < ▯ for every Riemann sum S for f on X, and then choose N so that n > N =) jX n < ▯. Then we have n > N =) jS n Ij < ▯. 1.9 Note: Let f be integrable on [a;b]. If we let n be the partition of [a;b] into n equal-sized subintervals, and we net S be the Riemann sum on X using right-endpoints, then by the above note we obtain the formula Z n b X b▯a b▯a f(x)dx = lim f(x n;i▯n;i , where xn;i a + n i and ▯n;i = n : a n!1 i=1 Z 2 1.10 Example: Find 2 dx. 0 Solution: Let f(x) = 2 . Note that f is continuous and hence integrable, so we have Z 2 Xn Xn ▯ ▯▯ ▯ Xn ▯ ▯ 2 dx = lim f(x )▯ x = lim f 2i 2 = lim 22i=n 2 n!1 n;i n;i n!1 n n n!1 n 0 i=1 i=1 i=1 1=n = lim 2 ▯ 4 ▯ 4 ▯ 1 , by the formula for the sum of a geometric sequence n!1 n 41=n▯ 1 1 ▯ 1=n▯▯ 1 ▯ n x = lim 6 ▯ 4 lim ▯ 1=n ▯ = 6 lim 1=n = 6 lim x n!1 n!1 n 4 ▯ 1 n!1 4 ▯ 1 x!0 4 ▯ 1 = 6 lim 1 , by l’opital’s Rule x!0 ln4 ▯ 4 6 3 = ln 4= ln 2 3 1.11 Lemma: (Summation Formulas) We have X X Xn Xn 2 2 n(n + 1) 2 n(n + 1)(2n + 1) 3 n (n + 1) 1 = n ; i = 2 ; i = 6 ; i = 4 i=1 i=1 i=1 i=1 Proof: These formulas could be proven by induction, but we give a more constructive P P P ▯2 2▯ proof. It is obvious tha1 = 1+1+▯▯▯1 = n. To ▯nd i, consider i ▯(i▯1) . i=1 i=1 n=1 On the one hand, we have P ▯ 2 2▯ 2 2 2 2 2 2 2 2 i ▯ (i ▯ 1) = (1 ▯ 0 ) + (2 ▯ 1 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 2) ) + (n ▯ (n ▯ 1) ) i=1 2 2 2 2 2 2 2 2 = ▯0 + (1 ▯ 1 ) + (2 ▯ 2 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 1) ) + n 2 = n and on the other hand, n n n n n P ▯ 2 2▯ P ▯ 2 2 ▯ P P P i ▯ (i ▯ 1) = i ▯ (i ▯ 2i + 1) = (2i ▯ 1) = 2 i ▯ 1 i=1 i=1 i=1 i=1 i=1 Pn Pn Equating these gives n = 2 i ▯ 1 and so i=1 i=1 Pn P 2 i = n + 1 = n + n = n(n + 1); i=1 i=1 P 2 P ▯3 3▯ as required. Next, to ▯ndi , consider i ▯ (i ▯ 1) . On the one hand we have n=1 i=1 n P ▯3 3▯ 3 3 3 3 3 3 3 3 i ▯ (i ▯ 1)= (1 ▯ 0 ) + (2 ▯ 1 ) + (3 ▯ 2 ) + ▯▯▯ + (n ▯ (n ▯ 1) ) i=1 = ▯0 + (1 ▯ 1 ) + (2 ▯ 2 ) + ▯▯▯ + ((n ▯ 1) ▯ (n ▯ 1) ) + n = n 3 and on the other hand, Pn ▯ ▯ P ▯ ▯ i ▯ (i ▯ 1)= i ▯ (i ▯ 3i + 3i ▯ 1) i=1 i=1 P 2 Pn 2 Pn Pn = (3i ▯ 3i + 1) = 3 i ▯ 3 i + 1: i=1 i=1 i=1 i=1 n n n Equating these gives n = 3 i ▯ 3P i +P 1 and so i=1 i=1 i=1 P Pn P 6 i = 2n + 6 i ▯ 2 1 = 2n + 3n(n + 1) ▯ 2n = n(n + 1)(2n + 1) i=1 i=1 i=1 Pn 3 Pn ▯4 4▯ as required. Finally, to ▯nd , consider i ▯ (i ▯ 1) . On the one hand we have i=1 i=1 n P ▯4 4▯ 4 i ▯ (i ▯ 1) = n ; i=1
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