• Title: Riemann Integral for Step Functions

  • Series: Real Analysis

  • Chapter: Riemann Integral

  • YouTube-Title: Real Analysis 49 | Riemann Integral for Step Functions

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    1 00:00:00,620 –> 00:00:02,440 Hello and welcome back to

    2 00:00:02,450 –> 00:00:04,000 real analysis.

    3 00:00:04,489 –> 00:00:05,679 And first, I want to thank

    4 00:00:05,690 –> 00:00:06,739 all the nice people that

    5 00:00:06,750 –> 00:00:08,199 support this channel on Steady

    6 00:00:08,210 –> 00:00:08,960 or paypal.

    7 00:00:09,710 –> 00:00:11,489 Now, in today’s part 49 we

    8 00:00:11,500 –> 00:00:13,029 will continue talking about

    9 00:00:13,039 –> 00:00:14,369 the Riemann integral.

    10 00:00:15,229 –> 00:00:16,920 In particular, we will show

    11 00:00:16,930 –> 00:00:18,350 that the integral for step

    12 00:00:18,360 –> 00:00:20,079 functions is well defined

    13 00:00:20,889 –> 00:00:21,389 for this.

    14 00:00:21,399 –> 00:00:23,360 Please recall as step function

    15 00:00:23,370 –> 00:00:25,149 phi is defined on a compact

    16 00:00:25,159 –> 00:00:26,850 interval and it is piece-

    17 00:00:26,860 –> 00:00:27,909 wisely constant.

    18 00:00:28,840 –> 00:00:30,250 So a typical graph looks

    19 00:00:30,260 –> 00:00:30,979 like this.

    20 00:00:30,989 –> 00:00:32,598 So we have constant parts

    21 00:00:32,610 –> 00:00:34,060 and some jumps involved.

    22 00:00:35,139 –> 00:00:36,299 Therefore, the important

    23 00:00:36,310 –> 00:00:37,610 thing for us is that we can

    24 00:00:37,619 –> 00:00:39,400 choose a partition for the

    25 00:00:39,409 –> 00:00:40,639 interval A B,

    26 00:00:41,689 –> 00:00:42,840 the explicit definition of

    27 00:00:42,849 –> 00:00:44,139 a partition I gave you in

    28 00:00:44,150 –> 00:00:45,549 the last video here.

    29 00:00:45,560 –> 00:00:47,439 Please recall it’s a finite

    30 00:00:47,450 –> 00:00:48,750 set where the elements are

    31 00:00:48,759 –> 00:00:50,500 called X with an index.

    32 00:00:51,389 –> 00:00:52,819 Now corresponding to such

    33 00:00:52,830 –> 00:00:54,650 a partition, we can also choose

    34 00:00:54,659 –> 00:00:56,400 numbers, we call C one C

    35 00:00:56,409 –> 00:00:57,439 two and so on.

    36 00:00:58,299 –> 00:00:59,490 And these are simply the

    37 00:00:59,500 –> 00:01:01,290 important values the function

    38 00:01:01,299 –> 00:01:02,400 phi can have.

    39 00:01:03,119 –> 00:01:03,610 OK.

    40 00:01:03,619 –> 00:01:05,138 Then with this, we can

    41 00:01:05,150 –> 00:01:06,739 define the integral for the

    42 00:01:06,750 –> 00:01:08,139 step function phi

    43 00:01:08,900 –> 00:01:10,139 we just calculate the

    44 00:01:10,150 –> 00:01:11,599 orientated area of a

    45 00:01:11,610 –> 00:01:13,459 rectangle here and sum them

    46 00:01:13,470 –> 00:01:15,389 up and then

    47 00:01:15,400 –> 00:01:17,169 this finite sum is just

    48 00:01:17,180 –> 00:01:19,169 called the integral of Phi

    49 00:01:19,709 –> 00:01:21,209 denote it with this

    50 00:01:21,220 –> 00:01:21,730 symbol.

    51 00:01:22,650 –> 00:01:24,389 Now, the topic for this video

    52 00:01:24,410 –> 00:01:26,069 is this proposition here

    53 00:01:26,879 –> 00:01:28,410 and it tells us that this

    54 00:01:28,419 –> 00:01:29,849 symbol here is well

    55 00:01:29,860 –> 00:01:31,400 defined more

    56 00:01:31,410 –> 00:01:32,919 precisely, this means no

    57 00:01:32,930 –> 00:01:34,209 matter which partition we

    58 00:01:34,220 –> 00:01:36,050 choose with this property,

    59 00:01:36,150 –> 00:01:37,809 we still get out the same

    60 00:01:37,819 –> 00:01:38,330 number.

    61 00:01:39,029 –> 00:01:40,129 You see this because on the

    62 00:01:40,139 –> 00:01:41,500 left hand side, we don’t

    63 00:01:41,510 –> 00:01:42,910 mention the partition at all.

    64 00:01:43,930 –> 00:01:45,180 Therefore, in the proof,

    65 00:01:45,190 –> 00:01:46,540 we just have to consider

    66 00:01:46,550 –> 00:01:48,059 two arbitrarily chosen

    67 00:01:48,069 –> 00:01:48,849 partitions.

    68 00:01:49,699 –> 00:01:51,029 So let’s call the first one,

    69 00:01:51,040 –> 00:01:52,839 P one and the second

    70 00:01:52,849 –> 00:01:53,959 one, P two.

    71 00:01:54,709 –> 00:01:56,260 So for P one, let’s call

    72 00:01:56,269 –> 00:01:57,930 the partition X zero X one

    73 00:01:57,940 –> 00:01:59,529 and so on until we reach

    74 00:01:59,540 –> 00:02:01,010 the last point XN.

    75 00:02:01,819 –> 00:02:03,440 Of course, for P two, we

    76 00:02:03,449 –> 00:02:04,860 need different symbols.

    77 00:02:04,910 –> 00:02:06,199 Therefore, I would suggest

    78 00:02:06,209 –> 00:02:08,119 we take X to the zero.

    79 00:02:09,389 –> 00:02:10,399 Otherwise it should look

    80 00:02:10,407 –> 00:02:10,958 the same.

    81 00:02:10,967 –> 00:02:12,539 So we have X to the one X

    82 00:02:12,548 –> 00:02:13,958 to the two and so on, but

    83 00:02:13,968 –> 00:02:15,248 the size could be different.

    84 00:02:15,259 –> 00:02:16,718 So let’s call it m.

    85 00:02:17,889 –> 00:02:19,039 In other words, the number

    86 00:02:19,050 –> 00:02:20,669 of elements in the two partitions

    87 00:02:20,679 –> 00:02:21,690 could be different

    88 00:02:22,699 –> 00:02:22,990 here.

    89 00:02:23,000 –> 00:02:24,839 Please recall the whole thing

    90 00:02:24,850 –> 00:02:26,029 is a decomposition of the

    91 00:02:26,039 –> 00:02:26,639 X axis.

    92 00:02:26,649 –> 00:02:28,639 So for example, P one, you

    93 00:02:28,649 –> 00:02:30,000 could visualize like this.

    94 00:02:31,119 –> 00:02:32,339 On the other hand, P two,

    95 00:02:32,350 –> 00:02:33,740 you could visualize in the

    96 00:02:33,750 –> 00:02:35,710 same way it could look completely

    97 00:02:35,720 –> 00:02:36,630 differently.

    98 00:02:36,639 –> 00:02:38,119 But of course, it could also

    99 00:02:38,130 –> 00:02:39,179 look similarly

    100 00:02:40,289 –> 00:02:41,589 and maybe the first case

    101 00:02:41,600 –> 00:02:43,070 we should consider would

    102 00:02:43,080 –> 00:02:44,529 be that they are very

    103 00:02:44,539 –> 00:02:45,139 similar.

    104 00:02:46,039 –> 00:02:47,539 However, here, please don’t

    105 00:02:47,550 –> 00:02:49,380 forget both partitions should

    106 00:02:49,389 –> 00:02:50,770 work for our chosen step

    107 00:02:50,779 –> 00:02:51,660 function Phi

    108 00:02:52,559 –> 00:02:53,690 for the first one, we call

    109 00:02:53,699 –> 00:02:55,020 the value CJ.

    110 00:02:55,130 –> 00:02:56,979 And for the second one, maybe

    111 00:02:56,990 –> 00:02:57,699 DJ.

    112 00:02:58,589 –> 00:03:00,000 Of course, in some sense,

    113 00:03:00,009 –> 00:03:01,259 the values should be the

    114 00:03:01,270 –> 00:03:01,910 same.

    115 00:03:01,979 –> 00:03:03,850 But in general, the indices

    116 00:03:03,860 –> 00:03:05,089 don’t coincide.

    117 00:03:05,889 –> 00:03:07,070 The important thing you should

    118 00:03:07,080 –> 00:03:08,309 remember here is that the

    119 00:03:08,320 –> 00:03:10,009 step function phi is

    120 00:03:10,020 –> 00:03:10,949 always the same.

    121 00:03:11,669 –> 00:03:13,240 For example, the craft could

    122 00:03:13,250 –> 00:03:14,089 look like this.

    123 00:03:15,119 –> 00:03:17,100 And then you see one possibility

    124 00:03:17,110 –> 00:03:18,339 for a partition would be

    125 00:03:18,350 –> 00:03:20,130 X one and X two here.

    126 00:03:21,130 –> 00:03:22,460 However, of course, you can

    127 00:03:22,470 –> 00:03:23,740 also think of other

    128 00:03:23,750 –> 00:03:25,399 possibilities for a partition.

    129 00:03:26,470 –> 00:03:27,559 So maybe here on the right

    130 00:03:27,570 –> 00:03:29,529 hand side, you just see more

    131 00:03:29,539 –> 00:03:31,190 points on the X axis.

    132 00:03:32,020 –> 00:03:33,279 We still have points at the

    133 00:03:33,289 –> 00:03:34,419 jump points here.

    134 00:03:34,429 –> 00:03:35,750 But otherwise we have more

    135 00:03:35,759 –> 00:03:36,199 points.

    136 00:03:36,210 –> 00:03:37,830 So it’s a finer partition

    137 00:03:37,839 –> 00:03:39,160 than on the left hand side.

    138 00:03:40,029 –> 00:03:41,639 But otherwise this property

    139 00:03:41,649 –> 00:03:43,089 here that is important is

    140 00:03:43,100 –> 00:03:43,970 still fulfilled.

    141 00:03:44,860 –> 00:03:45,279 OK.

    142 00:03:45,289 –> 00:03:47,080 As promised, this is exactly

    143 00:03:47,089 –> 00:03:48,399 the first case we want to

    144 00:03:48,410 –> 00:03:48,860 consider.

    145 00:03:48,869 –> 00:03:50,320 Now more

    146 00:03:50,330 –> 00:03:52,000 precisely it means that partition

    147 00:03:52,009 –> 00:03:53,610 two just adds more

    148 00:03:53,619 –> 00:03:55,080 points to partition one.

    149 00:03:55,880 –> 00:03:57,750 This means as a set P

    150 00:03:57,759 –> 00:03:59,539 two is a super set of P

    151 00:03:59,550 –> 00:03:59,979 one.

    152 00:04:01,020 –> 00:04:02,889 In words, we would say partition

    153 00:04:02,899 –> 00:04:04,500 two is finer than

    154 00:04:04,509 –> 00:04:05,419 partition one.

    155 00:04:06,429 –> 00:04:07,820 What this means in the picture

    156 00:04:07,830 –> 00:04:09,630 you can see above because

    157 00:04:09,639 –> 00:04:11,570 X one is exactly

    158 00:04:11,580 –> 00:04:13,410 the same point as X tilde

    159 00:04:13,419 –> 00:04:14,050 three.

    160 00:04:14,809 –> 00:04:16,760 Moreover X two is

    161 00:04:16,769 –> 00:04:18,230 the same point as X two,

    162 00:04:18,238 –> 00:04:19,029 the five.

    163 00:04:19,760 –> 00:04:21,119 Now on the left hand side,

    164 00:04:21,130 –> 00:04:22,649 there is no point between

    165 00:04:22,660 –> 00:04:23,329 the two.

    166 00:04:23,339 –> 00:04:24,760 But on the right hand side,

    167 00:04:24,769 –> 00:04:26,220 we find an additional one.

    168 00:04:27,279 –> 00:04:27,649 OK?

    169 00:04:27,660 –> 00:04:28,890 Maybe it’s a good idea to

    170 00:04:28,899 –> 00:04:30,679 use exactly this example

    171 00:04:30,690 –> 00:04:32,179 to explain the general concept

    172 00:04:32,190 –> 00:04:32,480 here.

    173 00:04:33,329 –> 00:04:34,489 Hence, we can write this

    174 00:04:34,500 –> 00:04:35,480 equalities and

    175 00:04:35,489 –> 00:04:37,209 inequalities where we have

    176 00:04:37,220 –> 00:04:39,079 X one is equal to X to the

    177 00:04:39,089 –> 00:04:40,989 three and X to the five is

    178 00:04:41,000 –> 00:04:42,109 equal to X two.

    179 00:04:43,079 –> 00:04:44,540 And in the middle, we still

    180 00:04:44,549 –> 00:04:46,170 find X to the four.

    181 00:04:46,869 –> 00:04:48,410 Now, going back to the picture

    182 00:04:48,420 –> 00:04:49,730 on the left hand side, we

    183 00:04:49,739 –> 00:04:51,190 have our value, we would

    184 00:04:51,200 –> 00:04:52,359 call C two

    185 00:04:53,299 –> 00:04:54,720 because this is the value

    186 00:04:54,730 –> 00:04:56,359 of the second interval here.

    187 00:04:57,239 –> 00:04:58,510 However, on the right hand

    188 00:04:58,519 –> 00:04:59,869 side, this would be D

    189 00:04:59,880 –> 00:05:01,640 four and D

    190 00:05:01,649 –> 00:05:02,549 five.

    191 00:05:03,859 –> 00:05:05,170 However, the important thing

    192 00:05:05,179 –> 00:05:06,950 is this is still equal to

    193 00:05:06,959 –> 00:05:07,519 C two.

    194 00:05:08,250 –> 00:05:09,989 And of course, this is important

    195 00:05:10,000 –> 00:05:11,220 when we want to calculate

    196 00:05:11,230 –> 00:05:12,899 the area of the corresponding

    197 00:05:12,910 –> 00:05:13,750 rectangles

    198 00:05:14,769 –> 00:05:15,059 there.

    199 00:05:15,070 –> 00:05:16,470 Please keep in mind on the

    200 00:05:16,480 –> 00:05:18,200 right hand side, we have

    201 00:05:18,209 –> 00:05:19,459 two rectangles.

    202 00:05:20,220 –> 00:05:21,589 So we have to one area on

    203 00:05:21,600 –> 00:05:23,059 the left and the other one

    204 00:05:23,070 –> 00:05:23,799 on the right.

    205 00:05:24,779 –> 00:05:26,190 And maybe not so surprising.

    206 00:05:26,200 –> 00:05:27,179 In the end, this should be

    207 00:05:27,190 –> 00:05:28,929 the same area as we find

    208 00:05:28,940 –> 00:05:30,160 it on the left hand side.

    209 00:05:30,970 –> 00:05:32,220 However, the important thing

    210 00:05:32,230 –> 00:05:34,109 for us here is this comes

    211 00:05:34,119 –> 00:05:35,510 out of the calculation,

    212 00:05:36,399 –> 00:05:37,660 namely on the right hand

    213 00:05:37,670 –> 00:05:39,299 side, the first rectangle

    214 00:05:39,309 –> 00:05:41,160 is the four times the

    215 00:05:41,170 –> 00:05:42,380 length of the interval

    216 00:05:43,320 –> 00:05:45,100 which is X to the four

    217 00:05:45,109 –> 00:05:46,549 minus X to the

    218 00:05:46,559 –> 00:05:47,220 three

    219 00:05:48,100 –> 00:05:49,839 plus the second rectangle,

    220 00:05:49,850 –> 00:05:51,320 which is the five

    221 00:05:51,329 –> 00:05:53,040 times the length of the interval.

    222 00:05:53,950 –> 00:05:55,619 And this one here is X to

    223 00:05:55,630 –> 00:05:57,410 the five minus X to the

    224 00:05:57,420 –> 00:05:57,850 four.

    225 00:05:58,790 –> 00:06:00,200 And now I see because the

    226 00:06:00,209 –> 00:06:01,660 four and the five are the

    227 00:06:01,670 –> 00:06:03,450 same, we can factor it

    228 00:06:03,459 –> 00:06:05,440 out, then we

    229 00:06:05,450 –> 00:06:06,940 can just call it C two

    230 00:06:06,950 –> 00:06:08,500 because it’s the same value.

    231 00:06:09,489 –> 00:06:10,880 And now we put all the X

    232 00:06:10,910 –> 00:06:12,839 tilde into the parentheses

    233 00:06:12,869 –> 00:06:14,440 and then you see X tilde

    234 00:06:14,500 –> 00:06:16,059 four will cancel out.

    235 00:06:16,899 –> 00:06:18,760 Moreover, we also know

    236 00:06:18,769 –> 00:06:20,250 X 25 is

    237 00:06:20,260 –> 00:06:21,519 exactly X two.

    238 00:06:22,220 –> 00:06:23,730 So we can substitute this

    239 00:06:23,739 –> 00:06:24,519 immediately.

    240 00:06:25,200 –> 00:06:26,769 And the same we can do for

    241 00:06:26,779 –> 00:06:28,619 X to the three which is X

    242 00:06:28,630 –> 00:06:29,019 one.

    243 00:06:30,029 –> 00:06:31,510 Hence, we find this whole

    244 00:06:31,519 –> 00:06:33,149 thing here is X two

    245 00:06:33,160 –> 00:06:34,929 times X two minus

    246 00:06:34,940 –> 00:06:36,799 X one which

    247 00:06:36,809 –> 00:06:38,609 is as you know the area

    248 00:06:38,619 –> 00:06:40,109 here on the left hand side.

    249 00:06:41,160 –> 00:06:41,619 OK.

    250 00:06:41,630 –> 00:06:42,980 Now with this calculation,

    251 00:06:42,989 –> 00:06:44,779 you should see this works

    252 00:06:44,790 –> 00:06:46,579 exactly the same for all

    253 00:06:46,589 –> 00:06:47,540 other points here.

    254 00:06:48,600 –> 00:06:49,859 In other words, splitting

    255 00:06:49,869 –> 00:06:51,540 up the rectangle into more

    256 00:06:51,549 –> 00:06:53,130 rectangles here does not

    257 00:06:53,140 –> 00:06:54,220 change the area.

    258 00:06:55,350 –> 00:06:57,209 Therefore, also summing up

    259 00:06:57,220 –> 00:06:59,109 all rectangles gives us the

    260 00:06:59,119 –> 00:07:00,100 same result.

    261 00:07:00,820 –> 00:07:02,000 So the left hand side here

    262 00:07:02,010 –> 00:07:03,480 is the area we calculate

    263 00:07:03,489 –> 00:07:04,609 with the first partition.

    264 00:07:04,709 –> 00:07:05,839 And on the right hand side,

    265 00:07:05,850 –> 00:07:07,640 you see the area we calculate

    266 00:07:07,649 –> 00:07:08,720 with the second partition.

    267 00:07:09,600 –> 00:07:10,630 And by the argument from

    268 00:07:10,640 –> 00:07:12,260 above the sum is the

    269 00:07:12,269 –> 00:07:12,690 same.

    270 00:07:13,540 –> 00:07:14,609 And this is what we want

    271 00:07:14,619 –> 00:07:16,459 to show that the sum does

    272 00:07:16,470 –> 00:07:17,839 not depend on the choice

    273 00:07:17,850 –> 00:07:18,600 of the partition.

    274 00:07:19,480 –> 00:07:21,089 However, please here keep

    275 00:07:21,100 –> 00:07:22,739 in mind this only worked

    276 00:07:22,750 –> 00:07:24,089 because one partition was

    277 00:07:24,100 –> 00:07:25,279 finer than the other.

    278 00:07:26,190 –> 00:07:27,579 Therefore, the second case

    279 00:07:27,589 –> 00:07:29,230 should be that we don’t have

    280 00:07:29,239 –> 00:07:29,649 this.

    281 00:07:30,579 –> 00:07:32,279 So we don’t have this subset

    282 00:07:32,290 –> 00:07:33,200 relation there.

    283 00:07:34,399 –> 00:07:36,380 So P one is not a subset

    284 00:07:36,390 –> 00:07:37,179 of P two

    285 00:07:38,220 –> 00:07:40,040 and of course, also not

    286 00:07:40,049 –> 00:07:41,459 the other way around because

    287 00:07:41,470 –> 00:07:42,869 then we would be in the first

    288 00:07:42,880 –> 00:07:44,760 case again, it would just

    289 00:07:44,769 –> 00:07:46,700 be renaming the partition.

    290 00:07:47,649 –> 00:07:49,510 However, in this case, we

    291 00:07:49,519 –> 00:07:51,510 still can use the first case

    292 00:07:51,519 –> 00:07:52,859 because we could define a

    293 00:07:52,869 –> 00:07:54,329 new partition P

    294 00:07:54,339 –> 00:07:54,869 three.

    295 00:07:55,769 –> 00:07:57,109 It just needs to include

    296 00:07:57,119 –> 00:07:58,589 all the points of both

    297 00:07:58,600 –> 00:07:59,390 partitions,

    298 00:08:00,380 –> 00:08:01,980 which means we have to union

    299 00:08:01,989 –> 00:08:03,799 P one with P two.

    300 00:08:04,769 –> 00:08:06,690 Now with this P three is

    301 00:08:06,700 –> 00:08:08,589 clearly a superset of P one

    302 00:08:08,600 –> 00:08:09,380 and P two.

    303 00:08:10,359 –> 00:08:12,059 Hence, here in both

    304 00:08:12,070 –> 00:08:13,980 cases, we can use the first

    305 00:08:13,989 –> 00:08:14,450 case.

    306 00:08:15,470 –> 00:08:16,779 Therefore, we immediately

    307 00:08:16,790 –> 00:08:18,160 get two results out.

    308 00:08:19,040 –> 00:08:20,899 First we get when we calculate

    309 00:08:20,910 –> 00:08:22,399 this sum for P one.

    310 00:08:22,480 –> 00:08:24,260 So let’s write Sigma with

    311 00:08:24,269 –> 00:08:24,940 P one.

    312 00:08:25,920 –> 00:08:27,399 Then you know, I just mean

    313 00:08:27,410 –> 00:08:29,019 this sum with the correct

    314 00:08:29,029 –> 00:08:29,619 partition.

    315 00:08:30,649 –> 00:08:32,000 Then the first case tells

    316 00:08:32,010 –> 00:08:33,630 us now this is exactly the

    317 00:08:33,640 –> 00:08:35,510 same as calculating the sum

    318 00:08:35,520 –> 00:08:37,099 with the partition three.

    319 00:08:38,030 –> 00:08:38,530 OK.

    320 00:08:38,539 –> 00:08:40,330 And a similar result we get

    321 00:08:40,340 –> 00:08:42,210 when we apply this inclusion

    322 00:08:42,219 –> 00:08:43,330 for the first case

    323 00:08:44,169 –> 00:08:45,698 there we get that the sum

    324 00:08:45,739 –> 00:08:47,578 using P two is the

    325 00:08:47,588 –> 00:08:49,338 same as the sum using

    326 00:08:49,348 –> 00:08:50,028 P three.

    327 00:08:51,150 –> 00:08:52,739 However, of course, the sum

    328 00:08:52,750 –> 00:08:54,299 with P three is always the

    329 00:08:54,309 –> 00:08:54,869 same.

    330 00:08:54,989 –> 00:08:56,729 So here we see the sum with

    331 00:08:56,739 –> 00:08:58,530 P one is exactly the

    332 00:08:58,539 –> 00:09:00,250 same as the sum with P two.

    333 00:09:01,070 –> 00:09:02,650 Hence also here, the

    334 00:09:02,659 –> 00:09:04,570 result is the sum

    335 00:09:04,580 –> 00:09:06,289 does not depend on the choice

    336 00:09:06,299 –> 00:09:07,090 of the partition.

    337 00:09:07,989 –> 00:09:09,409 And this is exactly what

    338 00:09:09,419 –> 00:09:10,570 we wanted to prove.

    339 00:09:10,580 –> 00:09:12,450 The integral symbol is well

    340 00:09:12,460 –> 00:09:13,010 defined.

    341 00:09:13,960 –> 00:09:14,409 OK.

    342 00:09:14,419 –> 00:09:16,159 So you see this was a technical

    343 00:09:16,169 –> 00:09:17,900 proof but we had to do it.

    344 00:09:18,619 –> 00:09:20,039 However, now in the next

    345 00:09:20,049 –> 00:09:21,440 video, we can talk about

    346 00:09:21,450 –> 00:09:23,020 the important properties

    347 00:09:23,030 –> 00:09:24,630 the integral has for step

    348 00:09:24,640 –> 00:09:25,239 functions.

    349 00:09:26,169 –> 00:09:27,349 Therefore, I hope I see you

    350 00:09:27,359 –> 00:09:28,789 there and have a nice day.

    351 00:09:28,880 –> 00:09:29,549 Bye.

  • Quiz Content

    Q1: The function $\phi : [0,1] \rightarrow \mathbb{R}$ with $$ \phi(x) = \begin{cases} 1 , , \text{ for } x = 0 \ 2 , , \text{ for } x \neq 0 \end{cases} $$ is a step function. What is possible partition you can choose for calculating the integral?

    A1: ${0,1}$

    A2: ${0,2}$

    A3: ${1,2}$

    Q2: The function $\phi : [0,2] \rightarrow \mathbb{R}$ with $$ \phi(x) = \begin{cases} 1 , , \text{ for } x \in [0,1) \ 2 , , \text{ for } x \in [1,2]\end{cases} $$ is a step function. What is possible partition you can choose for calculating the integral?

    A1: ${0,1}$

    A2: ${0,2}$

    A3: ${1,2,3}$

    A4: ${0,1,2}$

    Q3: The function $\phi : [0,2] \rightarrow \mathbb{R}$ with $$ \phi(x) = \begin{cases} 1 , , \text{ for } x \in [0,1) \ 2 , , \text{ for } x \in [1,2]\end{cases} $$ is a step function. What is the integral value $\int_0^2 \phi(x) , dx$?

    A1: 0

    A2: 1

    A3: 2

    A4: 3

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