-
Title: Riemann Integral - Examples
-
Series: Real Analysis
-
Chapter: Riemann Integral
-
YouTube-Title: Real Analysis 52 | Riemann Integral - Examples
-
Bright video: https://youtu.be/J9qXHzxeDN4
-
Dark video: https://youtu.be/UcgoXlPuLUs
-
Ad-free video: Watch Vimeo video
-
Quiz: Test your knowledge
-
Dark-PDF: Download PDF version of the dark video
-
Print-PDF: Download printable PDF version
-
Exercise Download PDF sheets
-
Thumbnail (bright): Download PNG
-
Thumbnail (dark): Download PNG
-
Subtitle on GitHub: ra52_sub_eng.srt
-
Timestamps (n/a)
-
Subtitle in English
1 00:00:00,560 –> 00:00:02,259 Hello and welcome back
2 00:00:02,269 –> 00:00:03,140 to real
3 00:00:03,150 –> 00:00:04,300 analysis.
4 00:00:05,019 –> 00:00:06,320 And first, I want to thank
5 00:00:06,329 –> 00:00:07,570 all the nice supporters on
6 00:00:07,579 –> 00:00:08,920 Steady and paypal.
7 00:00:09,789 –> 00:00:11,739 Now, in today’s part 52 we
8 00:00:11,750 –> 00:00:13,189 will talk about examples
9 00:00:13,199 –> 00:00:14,630 for the Riemann integral
10 00:00:15,159 –> 00:00:15,670 for this.
11 00:00:15,680 –> 00:00:16,930 Let’s quickly recall the
12 00:00:16,940 –> 00:00:18,329 definition for Riemann
13 00:00:18,409 –> 00:00:19,809 integrable functions.
14 00:00:20,659 –> 00:00:22,229 What we need is a bounded
15 00:00:22,239 –> 00:00:23,780 function defined on the
16 00:00:23,790 –> 00:00:25,469 compact interval A B.
17 00:00:26,170 –> 00:00:27,600 And we call this function
18 00:00:27,709 –> 00:00:28,989 we man integral.
19 00:00:29,000 –> 00:00:30,659 If the upper and the lower
20 00:00:30,670 –> 00:00:32,060 integral are the same
21 00:00:32,830 –> 00:00:34,509 there, the lower integral
22 00:00:34,520 –> 00:00:36,110 is given when we approximate
23 00:00:36,119 –> 00:00:37,639 the integral with step
24 00:00:37,650 –> 00:00:39,000 functions from below.
25 00:00:39,549 –> 00:00:40,959 And the upper integral is
26 00:00:40,970 –> 00:00:42,560 given when we approximate
27 00:00:42,569 –> 00:00:44,529 the integral by step functions
28 00:00:44,540 –> 00:00:45,279 from above.
29 00:00:46,139 –> 00:00:47,599 Hence, we only get one
30 00:00:47,610 –> 00:00:49,330 well-defined value here
31 00:00:49,340 –> 00:00:51,130 which we call the integral
32 00:00:51,139 –> 00:00:52,319 of the function F.
33 00:00:53,349 –> 00:00:55,049 Now I can tell you it’s possible
34 00:00:55,060 –> 00:00:56,610 to rewrite this definition
35 00:00:56,619 –> 00:00:58,049 here without using
36 00:00:58,060 –> 00:00:59,229 supremo and infimum.
37 00:01:00,299 –> 00:01:01,979 In this alternative form,
38 00:01:01,990 –> 00:01:03,279 you might recognize the
39 00:01:03,290 –> 00:01:05,138 approximation in a better
40 00:01:05,150 –> 00:01:06,760 way to get the
41 00:01:06,769 –> 00:01:07,220 idea.
42 00:01:07,230 –> 00:01:08,500 Let’s simply to a
43 00:01:08,559 –> 00:01:09,980 small graph here.
44 00:01:11,199 –> 00:01:12,959 So you see our mission is
45 00:01:12,970 –> 00:01:14,199 that we approximate this
46 00:01:14,209 –> 00:01:16,190 area here from above
47 00:01:16,199 –> 00:01:17,080 and from below.
48 00:01:17,919 –> 00:01:19,550 Now let’s say this here is
49 00:01:19,559 –> 00:01:21,459 a step function phi and
50 00:01:21,470 –> 00:01:23,069 you see it’s an approximation
51 00:01:23,080 –> 00:01:23,779 form below.
52 00:01:24,510 –> 00:01:26,330 Also, then you can see
53 00:01:26,349 –> 00:01:27,610 without a problem.
54 00:01:27,620 –> 00:01:29,290 And even with the same partition
55 00:01:29,300 –> 00:01:31,279 of the x axis, we can choose
56 00:01:31,290 –> 00:01:33,099 a step function phi that
57 00:01:33,110 –> 00:01:34,790 approximates the integral
58 00:01:34,800 –> 00:01:35,639 from above.
59 00:01:36,559 –> 00:01:38,410 However, the important thing
60 00:01:38,419 –> 00:01:39,900 I want to show you here is
61 00:01:39,910 –> 00:01:41,639 that the area between both
62 00:01:41,650 –> 00:01:43,300 step functions is very
63 00:01:43,309 –> 00:01:44,860 small even
64 00:01:44,870 –> 00:01:45,389 better.
65 00:01:45,400 –> 00:01:47,269 We can make it as small as
66 00:01:47,279 –> 00:01:47,959 we want.
67 00:01:48,910 –> 00:01:50,650 And exactly this fact is
68 00:01:50,660 –> 00:01:52,480 what we can use for an equivalent
69 00:01:52,489 –> 00:01:53,489 formulation here.
70 00:01:54,910 –> 00:01:56,750 Namely for all
71 00:01:56,760 –> 00:01:58,150 epsilon greater than
72 00:01:58,160 –> 00:01:58,800 zero,
73 00:02:00,010 –> 00:02:01,709 we find step functions
74 00:02:01,720 –> 00:02:02,459 Phi MS
75 00:02:03,900 –> 00:02:05,440 with the property that the
76 00:02:05,449 –> 00:02:07,360 one lies below F
77 00:02:07,370 –> 00:02:09,339 and the other one above F.
78 00:02:10,070 –> 00:02:11,929 And moreover, we have
79 00:02:11,940 –> 00:02:13,449 that the difference between
80 00:02:13,460 –> 00:02:15,289 both integrals here is
81 00:02:15,300 –> 00:02:17,130 less than the given epsilon
82 00:02:18,149 –> 00:02:18,619 here.
83 00:02:18,630 –> 00:02:20,509 Please note we know that
84 00:02:20,520 –> 00:02:22,360 the integral of Phi is
85 00:02:22,369 –> 00:02:23,979 always bigger than the integral
86 00:02:23,990 –> 00:02:24,740 of Phi.
87 00:02:25,830 –> 00:02:27,440 So we don’t need an absolute
88 00:02:27,449 –> 00:02:28,190 value here.
89 00:02:28,199 –> 00:02:29,720 We can just calculate the
90 00:02:29,729 –> 00:02:30,479 difference.
91 00:02:31,279 –> 00:02:31,820 OK.
92 00:02:31,830 –> 00:02:33,460 Now this description here
93 00:02:33,589 –> 00:02:35,179 makes it a little bit easier
94 00:02:35,190 –> 00:02:37,089 for us to look at examples.
95 00:02:38,089 –> 00:02:39,869 Therefore, I would say let’s
96 00:02:39,880 –> 00:02:41,389 start with the first one
97 00:02:41,399 –> 00:02:43,309 now and I would like
98 00:02:43,320 –> 00:02:44,539 to start with a
99 00:02:44,550 –> 00:02:45,770 counterexample.
100 00:02:46,589 –> 00:02:48,130 It’s the so-called dela
101 00:02:48,449 –> 00:02:49,009 function.
102 00:02:49,940 –> 00:02:51,360 Indeed, it sounds more
103 00:02:51,369 –> 00:02:52,589 complicated than it really
104 00:02:52,600 –> 00:02:53,119 is.
105 00:02:53,169 –> 00:02:55,000 And I would say let’s define
106 00:02:55,009 –> 00:02:56,399 the function on the interval
107 00:02:56,410 –> 00:02:57,369 01.
108 00:02:58,139 –> 00:02:59,639 And now the common definition
109 00:02:59,649 –> 00:03:01,419 of the function just considers
110 00:03:01,429 –> 00:03:02,309 two cases.
111 00:03:03,320 –> 00:03:04,779 We either get the value
112 00:03:04,789 –> 00:03:06,380 one or zero
113 00:03:06,389 –> 00:03:08,220 depending if the X we
114 00:03:08,229 –> 00:03:09,860 put in is rational or
115 00:03:09,869 –> 00:03:10,339 not.
116 00:03:11,330 –> 00:03:12,940 By using the set names, we
117 00:03:12,949 –> 00:03:14,919 can say it’s one when X
118 00:03:14,929 –> 00:03:16,889 comes from Q and it’s zero
119 00:03:16,899 –> 00:03:18,669 when X comes not from Q.
120 00:03:19,619 –> 00:03:21,339 Now, at first glance, this
121 00:03:21,350 –> 00:03:22,710 looks like a very simple
122 00:03:22,720 –> 00:03:23,229 function.
123 00:03:23,250 –> 00:03:24,800 So let’s tour the graph for
124 00:03:24,809 –> 00:03:25,000 it.
125 00:03:25,770 –> 00:03:27,220 And there you should immediately
126 00:03:27,229 –> 00:03:29,050 see we have infinitely
127 00:03:29,059 –> 00:03:30,559 many rational points for
128 00:03:30,570 –> 00:03:32,449 the value one, but also
129 00:03:32,460 –> 00:03:33,910 infinitely many irrational
130 00:03:33,919 –> 00:03:35,509 points for the value zero.
131 00:03:36,320 –> 00:03:38,100 And moreover, we know the
132 00:03:38,110 –> 00:03:39,570 rational points like
133 00:03:39,580 –> 00:03:41,100 dance in the real number
134 00:03:41,110 –> 00:03:41,539 line.
135 00:03:42,250 –> 00:03:43,759 Please remember this is
136 00:03:43,770 –> 00:03:45,660 exactly how we constructed
137 00:03:45,669 –> 00:03:46,779 the real number line.
138 00:03:47,669 –> 00:03:49,539 For this reason, it’s very
139 00:03:49,550 –> 00:03:51,160 hard to draw this graph of
140 00:03:51,169 –> 00:03:52,649 the function correctly
141 00:03:52,770 –> 00:03:54,149 because you have infinitely
142 00:03:54,160 –> 00:03:55,970 many jumps no matter how
143 00:03:55,979 –> 00:03:57,190 much you zoom in.
144 00:03:58,220 –> 00:03:59,860 And there you might already
145 00:03:59,869 –> 00:04:01,839 see that this function is
146 00:04:01,850 –> 00:04:03,440 not Riemann integrable.
147 00:04:04,350 –> 00:04:05,740 You see this when you want
148 00:04:05,750 –> 00:04:07,119 to choose a step function
149 00:04:07,130 –> 00:04:09,009 si that lies above
150 00:04:09,020 –> 00:04:10,139 the graph of f
151 00:04:11,229 –> 00:04:13,039 such a step function then
152 00:04:13,229 –> 00:04:14,910 also lies essentially
153 00:04:14,919 –> 00:04:15,940 above one.
154 00:04:17,140 –> 00:04:18,850 This is simply because for
155 00:04:18,858 –> 00:04:20,350 any segment you choose on
156 00:04:20,358 –> 00:04:21,959 the real number line, you
157 00:04:21,970 –> 00:04:23,839 always find a rational number.
158 00:04:24,690 –> 00:04:26,670 Hence the value one is
159 00:04:26,679 –> 00:04:28,339 always included in such an
160 00:04:28,350 –> 00:04:29,019 interval.
161 00:04:29,929 –> 00:04:31,619 Indeed, the same holds for
162 00:04:31,630 –> 00:04:33,059 the irrational numbers.
163 00:04:33,070 –> 00:04:33,989 When we want to choose a
164 00:04:34,000 –> 00:04:35,790 step function phi from
165 00:04:35,799 –> 00:04:37,790 below there, the
166 00:04:37,799 –> 00:04:39,429 step function also has to
167 00:04:39,440 –> 00:04:41,070 lie essentially below
168 00:04:41,079 –> 00:04:41,709 zero.
169 00:04:42,730 –> 00:04:44,640 In summary, you see we have
170 00:04:44,649 –> 00:04:46,179 two properties here that
171 00:04:46,190 –> 00:04:48,010 hold for all step functions,
172 00:04:48,079 –> 00:04:49,140 Phi and Phi
173 00:04:49,940 –> 00:04:51,649 and therefore we have immediately
174 00:04:51,660 –> 00:04:53,200 an estimate for the two
175 00:04:53,209 –> 00:04:54,320 integrals here.
176 00:04:55,260 –> 00:04:56,559 And the conclusion will be,
177 00:04:56,570 –> 00:04:58,200 we can’t push the difference
178 00:04:58,209 –> 00:04:59,160 below one.
179 00:05:00,010 –> 00:05:01,529 Of course, the first integral
180 00:05:01,540 –> 00:05:03,209 will always be greater or
181 00:05:03,220 –> 00:05:05,059 equal than one and the other
182 00:05:05,070 –> 00:05:07,029 one always less or equal
183 00:05:07,040 –> 00:05:07,869 than zero.
184 00:05:08,790 –> 00:05:10,359 In other words, we cannot
185 00:05:10,369 –> 00:05:12,179 fulfill this property for
186 00:05:12,190 –> 00:05:13,190 all epsilon.
187 00:05:14,149 –> 00:05:16,019 In fact, this is all we need
188 00:05:16,220 –> 00:05:17,489 in order to show that the
189 00:05:17,500 –> 00:05:19,269 deli function is not
190 00:05:19,279 –> 00:05:20,459 riemann integrable
191 00:05:21,290 –> 00:05:21,519 here.
192 00:05:21,529 –> 00:05:23,000 Again, in the difference,
193 00:05:23,010 –> 00:05:24,429 this first part here is
194 00:05:24,440 –> 00:05:26,209 always greater or equal than
195 00:05:26,220 –> 00:05:26,660 one.
196 00:05:27,609 –> 00:05:29,220 And the second part without
197 00:05:29,230 –> 00:05:30,950 a minus sign is always
198 00:05:30,959 –> 00:05:32,709 less or equal than zero.
199 00:05:34,140 –> 00:05:35,609 Hence the difference of both
200 00:05:35,619 –> 00:05:37,170 numbers is always
201 00:05:37,179 –> 00:05:38,910 greater or equal than one.
202 00:05:40,140 –> 00:05:40,640 OK.
203 00:05:40,679 –> 00:05:42,420 There you see this was our
204 00:05:42,429 –> 00:05:44,029 first counterexample.
205 00:05:44,920 –> 00:05:46,500 Then next, I would say we
206 00:05:46,510 –> 00:05:47,980 look at a function that is
207 00:05:47,989 –> 00:05:49,859 actually Riemann integral.
208 00:05:50,630 –> 00:05:52,209 Of course, for the start,
209 00:05:52,220 –> 00:05:53,929 let’s look at a very simple
210 00:05:53,940 –> 00:05:54,559 example.
211 00:05:55,429 –> 00:05:57,130 And I guess the identity
212 00:05:57,140 –> 00:05:58,890 F of X is equal to X
213 00:05:58,899 –> 00:06:00,690 is a very suitable example.
214 00:06:01,649 –> 00:06:02,970 This is simply because when
215 00:06:02,980 –> 00:06:04,890 we toward the graph, we immediately
216 00:06:04,899 –> 00:06:06,750 see what the integral should
217 00:06:06,760 –> 00:06:07,070 be.
218 00:06:07,809 –> 00:06:09,519 You see the area is given
219 00:06:09,529 –> 00:06:11,239 by this triangle which
220 00:06:11,250 –> 00:06:12,920 means the area should be
221 00:06:12,929 –> 00:06:13,790 one half,
222 00:06:14,730 –> 00:06:16,630 it’s simply half of the square
223 00:06:16,640 –> 00:06:18,029 where we have the sides as
224 00:06:18,040 –> 00:06:18,950 one and one.
225 00:06:19,769 –> 00:06:21,480 However, if we work with
226 00:06:21,489 –> 00:06:22,709 the definition of the Riemann
227 00:06:22,720 –> 00:06:24,519 integral, what we need to,
228 00:06:24,529 –> 00:06:26,239 we can’t use the triangle,
229 00:06:26,250 –> 00:06:28,239 we need to use rectangles.
230 00:06:29,450 –> 00:06:31,130 Hence, here we can actually
231 00:06:31,140 –> 00:06:33,019 see if our approximation
232 00:06:33,029 –> 00:06:33,670 works.
233 00:06:34,570 –> 00:06:35,029 OK.
234 00:06:35,040 –> 00:06:36,500 Now, the question here is
235 00:06:36,510 –> 00:06:38,109 what is a good step function
236 00:06:38,119 –> 00:06:39,220 we can choose here?
237 00:06:40,040 –> 00:06:41,279 Now, the one we see in the
238 00:06:41,290 –> 00:06:42,820 picture has four
239 00:06:42,829 –> 00:06:44,380 steps, the
240 00:06:44,390 –> 00:06:45,890 first height here is
241 00:06:45,899 –> 00:06:47,880 zero, then we go up one
242 00:06:47,890 –> 00:06:49,690 quarter, then the next quarter,
243 00:06:49,700 –> 00:06:51,239 the next quarter and then
244 00:06:51,250 –> 00:06:52,010 it’s the end.
245 00:06:52,809 –> 00:06:53,970 Hence, we have our four
246 00:06:53,980 –> 00:06:55,320 values 0,
247 00:06:55,329 –> 00:06:57,160 1/4 2/4 and
248 00:06:57,170 –> 00:06:58,160 3/4.
249 00:06:59,010 –> 00:07:00,769 Also, it’s not hard to see
250 00:07:00,779 –> 00:07:02,440 that we split the X axis
251 00:07:02,450 –> 00:07:04,299 also in four equal
252 00:07:04,309 –> 00:07:04,869 parts.
253 00:07:05,609 –> 00:07:07,010 So we have to interval 0
254 00:07:07,019 –> 00:07:08,640 to 1 quarter, one quarter
255 00:07:08,649 –> 00:07:10,529 to two quarters and so on.
256 00:07:11,269 –> 00:07:11,619 OK.
257 00:07:11,630 –> 00:07:13,079 So you see this is a well
258 00:07:13,100 –> 00:07:14,320 defined step function.
259 00:07:14,329 –> 00:07:16,119 You can choose for the approximation
260 00:07:16,130 –> 00:07:16,769 from below.
261 00:07:17,529 –> 00:07:19,239 It has exactly four
262 00:07:19,250 –> 00:07:20,570 equidistant steps.
263 00:07:20,579 –> 00:07:22,230 Therefore, let’s put a four
264 00:07:22,239 –> 00:07:23,429 into the index here.
265 00:07:24,250 –> 00:07:25,799 Of course, this tells you
266 00:07:25,809 –> 00:07:27,440 now that the approximation
267 00:07:27,450 –> 00:07:29,329 will get better when we choose
268 00:07:29,339 –> 00:07:31,149 similarly a step function
269 00:07:31,160 –> 00:07:32,309 with more steps.
270 00:07:33,179 –> 00:07:34,609 In fact, this is exactly
271 00:07:34,619 –> 00:07:35,640 what we will do.
272 00:07:35,649 –> 00:07:37,350 But now for an arbitrary
273 00:07:37,359 –> 00:07:38,329 integer N,
274 00:07:39,220 –> 00:07:41,029 hence we have exactly N
275 00:07:41,040 –> 00:07:42,670 steps now, which means the
276 00:07:42,679 –> 00:07:43,720 denominator here.
277 00:07:43,730 –> 00:07:45,510 And here is now N
278 00:07:46,320 –> 00:07:47,850 however, now, instead of
279 00:07:47,859 –> 00:07:49,820 writing in different cases,
280 00:07:49,829 –> 00:07:51,630 I want to put all of them
281 00:07:51,640 –> 00:07:53,350 into one closed formula.
282 00:07:54,160 –> 00:07:55,660 And this is what we can do
283 00:07:55,670 –> 00:07:57,399 with another index K.
284 00:07:58,579 –> 00:08:00,239 So you should see when K
285 00:08:00,250 –> 00:08:02,190 is equal to one, we are in
286 00:08:02,200 –> 00:08:03,230 the first case
287 00:08:03,950 –> 00:08:05,269 K is equal to two, gives
288 00:08:05,279 –> 00:08:07,269 us the next case and so on
289 00:08:07,279 –> 00:08:08,829 until K is equal to
290 00:08:08,839 –> 00:08:10,690 N gives us the last case
291 00:08:10,700 –> 00:08:11,079 here.
292 00:08:12,029 –> 00:08:13,369 Hence, the only thing missing
293 00:08:13,380 –> 00:08:14,600 here is now the value at
294 00:08:14,609 –> 00:08:16,380 the position which is K
295 00:08:16,390 –> 00:08:17,489 minus one,
296 00:08:18,480 –> 00:08:20,079 which definitely fits because
297 00:08:20,089 –> 00:08:21,859 it’s zero in the first case
298 00:08:21,869 –> 00:08:23,299 one, in the second case and
299 00:08:23,309 –> 00:08:23,730 so on.
300 00:08:24,679 –> 00:08:25,209 OK.
301 00:08:25,220 –> 00:08:26,640 So this is a step function
302 00:08:26,649 –> 00:08:27,859 that looks like this.
303 00:08:27,980 –> 00:08:29,619 But now with N steps,
304 00:08:30,579 –> 00:08:31,910 OK, maybe it’s not so
305 00:08:31,920 –> 00:08:33,840 precise because here
306 00:08:33,849 –> 00:08:35,369 we should have chosen an
307 00:08:35,380 –> 00:08:36,520 open interval.
308 00:08:37,299 –> 00:08:39,198 However, then you see it
309 00:08:39,207 –> 00:08:40,558 will clash with the last
310 00:08:40,568 –> 00:08:41,299 case here.
311 00:08:41,957 –> 00:08:43,679 However, we can ignore all
312 00:08:43,688 –> 00:08:45,078 of that because you already
313 00:08:45,088 –> 00:08:46,648 know for the integral, the
314 00:08:46,658 –> 00:08:48,018 boundary points here don’t
315 00:08:48,028 –> 00:08:49,158 make any difference.
316 00:08:49,880 –> 00:08:51,320 Speaking of the integral,
317 00:08:51,330 –> 00:08:53,260 maybe let’s immediately calculate
318 00:08:53,270 –> 00:08:54,979 the integral of phi N.
319 00:08:55,809 –> 00:08:57,739 Now, as we have learned before,
320 00:08:57,750 –> 00:08:59,359 the integral of a step function
321 00:08:59,369 –> 00:09:01,349 is always the sum of
322 00:09:01,359 –> 00:09:02,570 the areas of the
323 00:09:02,580 –> 00:09:03,510 rectangles,
324 00:09:04,599 –> 00:09:06,340 please recall we have N
325 00:09:06,349 –> 00:09:07,159 steps.
326 00:09:07,169 –> 00:09:08,580 Therefore, we have N
327 00:09:08,590 –> 00:09:09,630 rectangles.
328 00:09:10,150 –> 00:09:11,580 Honestly, the first one has
329 00:09:11,590 –> 00:09:12,669 area zero.
330 00:09:12,679 –> 00:09:14,210 So we could ignore it, but
331 00:09:14,219 –> 00:09:15,979 we can include it nevertheless.
332 00:09:16,700 –> 00:09:18,460 So now the area of one
333 00:09:18,469 –> 00:09:20,150 rectangle is simply the
334 00:09:20,159 –> 00:09:22,099 height times the width.
335 00:09:22,929 –> 00:09:24,280 And by construction, this
336 00:09:24,289 –> 00:09:26,030 is for all our rectangles
337 00:09:26,039 –> 00:09:26,909 one over and
338 00:09:28,109 –> 00:09:29,729 you see we have one over
339 00:09:29,739 –> 00:09:31,580 N squared which we can pull
340 00:09:31,590 –> 00:09:32,659 out of the sum.
341 00:09:33,380 –> 00:09:34,900 And then you see the only
342 00:09:34,909 –> 00:09:36,039 thing we have to calculate
343 00:09:36,049 –> 00:09:37,979 now is the sum of
344 00:09:37,989 –> 00:09:39,619 the first N minus one
345 00:09:39,630 –> 00:09:40,440 integers.
346 00:09:41,330 –> 00:09:43,070 And then we can use something
347 00:09:43,080 –> 00:09:44,669 some people call the little
348 00:09:44,679 –> 00:09:45,750 Gauss formula.
349 00:09:46,510 –> 00:09:48,369 In this case here, it’s N
350 00:09:48,380 –> 00:09:49,869 times N minus
351 00:09:49,880 –> 00:09:51,690 one divided by
352 00:09:51,700 –> 00:09:52,090 two.
353 00:09:53,020 –> 00:09:54,119 In the next step, you see,
354 00:09:54,130 –> 00:09:55,729 we can simplify this
355 00:09:55,739 –> 00:09:57,530 into one half
356 00:09:57,539 –> 00:09:59,530 minus one divided by
357 00:09:59,539 –> 00:10:00,359 two N.
358 00:10:01,280 –> 00:10:01,820 OK.
359 00:10:01,830 –> 00:10:03,619 So this is our result here,
360 00:10:03,630 –> 00:10:05,539 the result of this integral.
361 00:10:06,190 –> 00:10:07,349 And what you should immediately
362 00:10:07,359 –> 00:10:09,289 see is that if our approximation
363 00:10:09,299 –> 00:10:10,510 gets better and better.
364 00:10:10,559 –> 00:10:11,770 So if we send N to
365 00:10:11,780 –> 00:10:13,630 infinity, the result is
366 00:10:13,640 –> 00:10:14,489 one half.
367 00:10:15,330 –> 00:10:17,080 However, that’s not enough
368 00:10:17,090 –> 00:10:18,289 for showing that F is we
369 00:10:18,309 –> 00:10:20,049 are integral because we
370 00:10:20,059 –> 00:10:21,750 also have to approximate
371 00:10:21,760 –> 00:10:23,119 the integral from above.
372 00:10:23,890 –> 00:10:25,169 And of course, this is now
373 00:10:25,179 –> 00:10:26,809 what we do with a similar
374 00:10:26,820 –> 00:10:27,659 step functions.
375 00:10:27,750 –> 00:10:29,590 I here I would say
376 00:10:29,599 –> 00:10:31,159 let’s use the same picture
377 00:10:31,169 –> 00:10:33,080 as before to sketch the
378 00:10:33,090 –> 00:10:34,200 new step function.
379 00:10:35,210 –> 00:10:36,250 Of course, it should be the
380 00:10:36,260 –> 00:10:37,880 same staircase as before
381 00:10:37,890 –> 00:10:39,510 but now shifted above the
382 00:10:39,520 –> 00:10:40,000 function.
383 00:10:40,859 –> 00:10:42,070 Therefore, the definition
384 00:10:42,080 –> 00:10:43,750 of PN should look
385 00:10:43,760 –> 00:10:45,489 more or less the same as
386 00:10:45,500 –> 00:10:46,630 the definition of phi.
387 00:10:47,770 –> 00:10:49,030 Of course, the partition
388 00:10:49,039 –> 00:10:50,479 of the X axis should be the
389 00:10:50,489 –> 00:10:50,969 same.
390 00:10:50,979 –> 00:10:52,429 We only have to shift the
391 00:10:52,440 –> 00:10:54,419 values indeed,
392 00:10:54,429 –> 00:10:56,280 instead of K minus one, we
393 00:10:56,289 –> 00:10:57,669 now can choose K.
394 00:10:58,469 –> 00:10:59,919 So you see it’s not hard
395 00:10:59,929 –> 00:11:01,599 at all to define such a step
396 00:11:01,609 –> 00:11:02,049 function.
397 00:11:02,760 –> 00:11:04,169 And indeed in the same way
398 00:11:04,179 –> 00:11:06,049 as before, we can calculate
399 00:11:06,059 –> 00:11:06,900 the integral
400 00:11:07,630 –> 00:11:09,080 again, it’s just a step
401 00:11:09,090 –> 00:11:09,760 function.
402 00:11:09,820 –> 00:11:11,450 So we add up all the
403 00:11:11,460 –> 00:11:12,500 areas of the
404 00:11:12,510 –> 00:11:13,489 rectangles.
405 00:11:14,520 –> 00:11:15,820 And now the only difference
406 00:11:15,830 –> 00:11:17,700 from before is that the height
407 00:11:17,710 –> 00:11:19,669 of the rectangles is slightly
408 00:11:19,679 –> 00:11:20,260 larger.
409 00:11:21,169 –> 00:11:22,809 Still here, we can pull the
410 00:11:22,820 –> 00:11:24,359 fact of one over N squared
411 00:11:24,369 –> 00:11:25,559 out of the sum.
412 00:11:25,619 –> 00:11:27,190 And the only thing that remains
413 00:11:27,200 –> 00:11:28,820 is the sum of the first and
414 00:11:28,830 –> 00:11:30,679 numbers, this
415 00:11:30,690 –> 00:11:31,979 means that we can apply the
416 00:11:31,989 –> 00:11:33,859 same formula as before.
417 00:11:34,020 –> 00:11:35,669 But now we have one additional
418 00:11:35,679 –> 00:11:36,890 number at the end.
419 00:11:37,359 –> 00:11:38,859 For this reason, this sum
420 00:11:38,869 –> 00:11:40,700 is then given as N times
421 00:11:40,710 –> 00:11:42,690 N plus one divided by two.
422 00:11:43,510 –> 00:11:43,940 OK.
423 00:11:43,950 –> 00:11:45,010 And then in the last step,
424 00:11:45,020 –> 00:11:46,690 we can simplify this again
425 00:11:46,809 –> 00:11:48,340 and we get one half
426 00:11:48,349 –> 00:11:49,320 plus
427 00:11:49,330 –> 00:11:50,729 1/2 N.
428 00:11:52,020 –> 00:11:53,390 So with this, you should
429 00:11:53,400 –> 00:11:55,250 see we’ve reached our goal
430 00:11:55,929 –> 00:11:57,590 because the difference between
431 00:11:57,599 –> 00:11:59,380 these two integrals is
432 00:11:59,390 –> 00:12:01,119 exactly one
433 00:12:01,130 –> 00:12:02,309 divided by N.
434 00:12:03,419 –> 00:12:04,940 In other words, we can make
435 00:12:04,950 –> 00:12:06,599 the difference as small as
436 00:12:06,609 –> 00:12:07,260 we want.
437 00:12:08,270 –> 00:12:09,979 So if you recall the epsilon
438 00:12:09,989 –> 00:12:11,710 criterion from above, then
439 00:12:11,719 –> 00:12:13,530 you see that this function
440 00:12:13,539 –> 00:12:15,000 F is Riemann integrable.
441 00:12:16,080 –> 00:12:17,690 And of course, this is our
442 00:12:17,700 –> 00:12:18,570 result here.
443 00:12:19,270 –> 00:12:21,099 Moreover, we also get the
444 00:12:21,109 –> 00:12:22,479 value of the integral of
445 00:12:22,489 –> 00:12:24,440 F which is one half
446 00:12:25,169 –> 00:12:26,549 of course, not a surprise
447 00:12:26,559 –> 00:12:27,150 for you.
448 00:12:27,159 –> 00:12:28,830 But now we have proven it.
449 00:12:29,739 –> 00:12:30,200 OK.
450 00:12:30,210 –> 00:12:31,169 I think that’s good enough
451 00:12:31,179 –> 00:12:32,859 for a first example here
452 00:12:32,869 –> 00:12:34,719 we will consider more complicated
453 00:12:34,729 –> 00:12:35,869 examples later.
454 00:12:36,650 –> 00:12:37,830 Therefore, I hope that I
455 00:12:37,840 –> 00:12:39,070 see you in the next video
456 00:12:39,080 –> 00:12:40,590 when we continue with the
457 00:12:40,599 –> 00:12:41,630 Rayman integral.
458 00:12:42,359 –> 00:12:44,169 Have a nice day and bye.
-
Quiz Content
Q1: What is the an equivalent formulation for a bounded function $f$ being Riemann-integrable?
A1: For all $\varepsilon > 0$, there are step functions $\phi, \psi$ with the property $$ \int_a^b \phi(x) dx = \int_a^b \psi(x) dx$$
A2: For all $\varepsilon > 0$, there are step functions $\phi, \psi$ with the property $$ \int_a^b \phi(x) dx - \int_a^b \psi(x) dx < \varepsilon$$
A3: For all $\varepsilon > 0$, there are step functions $\phi, \psi$ with the property $\phi \leq f \leq \psi$ and $$ \int_a^b \phi(x) dx - \int_a^b \psi(x) dx < \varepsilon$$
A4: For all $\varepsilon > 0$, there are step functions $\phi, \psi$ with the property $\phi \leq f \leq \psi$ and $$ \int_a^b \psi(x) dx - \int_a^b \phi(x) dx < \varepsilon$$
A5: For all $\varepsilon > 0$, there are step functions $\phi, \psi$ with the property $\phi \leq f \leq \psi$ and $$ \int_a^b \psi(x) dx - \int_a^b \phi(x) dx = 0$$
Q2: Is the function $f: [0,5] \rightarrow \mathbb{R}$ given by $$ f(x) = \begin{cases} 1, & x \in { 0, 1, 2, 3, 4, 5 } \ 0, & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$$ Riemann-integrable?
A1: Yes!
A2: No!
Q3: Let $f: [0,5] \rightarrow \mathbb{R}$ be given by $$ f(x) = \begin{cases} 1, & x \in \mathbb{Q} \ 0, & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$$ Is the function Riemann-integrable?
A1: Yes!
A2: No!
-
Last update: 2025-01
