1
00:00:00,900 --> 00:00:05,750
So, last section for today.
integration by substitution.

2
00:00:09,850 --> 00:00:14,450
So we did integration by parts, and that
was the product rule running backwards.

3
00:00:16,710 --> 00:00:18,930
Integration by substitution we can think
of

4
00:00:18,930 --> 00:00:21,420
as just working the chain rule backwards.

5
00:00:26,750 --> 00:00:31,260
So I'm going to start out by letting
capital F of x just be the

6
00:00:31,260 --> 00:00:36,130
antiderivative of little f of x.
And

7
00:00:36,130 --> 00:00:40,409
then I'm going to think of x as being a
function of another variable, u.

8
00:00:42,280 --> 00:00:48,200
Then by the chain rule, if I took the
derivative of capital F of g of u,

9
00:00:50,880 --> 00:00:55,400
I would end up with f prime evaluated at g
of u.

10
00:00:55,400 --> 00:00:59,080
And then the chain rule then says I have
to take this function

11
00:00:59,080 --> 00:01:02,279
on the inside and multiply by its
derivative to, so I end up with.

12
00:01:03,540 --> 00:01:06,460
A g prime of u showing up on the outside.

13
00:01:06,460 --> 00:01:10,390
And then because of the way I've defined
little f of x to be the, the

14
00:01:10,390 --> 00:01:12,700
derivative of capital F or rather a
capital

15
00:01:12,700 --> 00:01:15,640
F to be an anti derivative of little f.

16
00:01:15,640 --> 00:01:15,930
I can

17
00:01:15,930 --> 00:01:20,570
write that as F of g of u times g prime of
u.

18
00:01:27,780 --> 00:01:27,780
[SOUND].

19
00:01:27,780 --> 00:01:31,060
And then if I integrate both sides with
respect to u.

20
00:01:32,830 --> 00:01:35,570
Well, on this side, I, I just had d of u.

21
00:01:36,950 --> 00:01:37,880
d by du.

22
00:01:37,880 --> 00:01:40,790
So the derivative with respect to u of f
of g of u.

23
00:01:41,840 --> 00:01:47,160
So that's going to be the complementary
operation to my anti derivative.

24
00:01:47,160 --> 00:01:50,020
So that's just going to be f capital f of
g of u.

25
00:01:51,180 --> 00:01:52,960
Is going to be equal to the integral

26
00:01:54,710 --> 00:01:58,980
of little f of g of u times g prime of u
du.

27
00:02:07,220 --> 00:02:12,470
And so my rule for integration by
substitution then, is if x, f of

28
00:02:12,470 --> 00:02:15,340
x is an integrable function and g

29
00:02:15,340 --> 00:02:19,220
of x is invertible and continuously
differentiable.

30
00:02:19,220 --> 00:02:22,460
Then I can make the substitution x equals
g of u.

31
00:02:23,660 --> 00:02:26,120
And it's going to change my integral in
the following way.

32
00:02:26,120 --> 00:02:32,450
So if I replace the variable x in my
function f of x with g of u,

33
00:02:33,660 --> 00:02:38,869
then I have to multiply by g prime of u in
order to get the same answer.

34
00:02:44,020 --> 00:02:46,040
And so I think the easiest way to see how

35
00:02:46,040 --> 00:02:49,330
this is going to, work is just to try an
example.

36
00:02:50,510 --> 00:02:53,450
So if I look at this, I see

37
00:02:57,590 --> 00:03:00,469
the top of this fraction as the
exponential function.

38
00:03:01,490 --> 00:03:06,140
And then, the argument to that exponential
function is the square root of x.

39
00:03:10,560 --> 00:03:14,030
So, I don't know how to integrate this
function just looking at that.

40
00:03:14,030 --> 00:03:18,970
I'm not going to make any progress if I
keep it in terms of x.

41
00:03:18,970 --> 00:03:21,510
But, it would be easier to integrate.

42
00:03:21,510 --> 00:03:28,060
If, instead of having a square root in the
exponent here, I just had a single letter.

43
00:03:28,060 --> 00:03:31,080
So if this was, say, e to the u, for
instance.

44
00:03:32,220 --> 00:03:36,350
So, the way I can do that, is make a
substitution.

45
00:03:36,350 --> 00:03:39,540
I'm going to define u to be the square
root of x.

46
00:03:42,500 --> 00:03:47,600
And then if I do that, I could also think
of it as x being equal to u squared.

47
00:03:49,660 --> 00:03:51,220
And then if I were to take the derivative
of

48
00:03:51,220 --> 00:03:56,240
that, I would find that dx is equal to 2u
du.

49
00:04:00,060 --> 00:04:02,660
And so let's see what happens when I
start.

50
00:04:02,660 --> 00:04:06,560
Plugging in these these substitutions
into, into my integral.

51
00:04:10,920 --> 00:04:16,060
So on the bottom, I can write square root

52
00:04:16,060 --> 00:04:21,000
of x, so this square root of x in the
denominator just becomes a u.

53
00:04:23,470 --> 00:04:26,790
In the numerator my e to the square root
of x.

54
00:04:26,790 --> 00:04:32,470
That's going to be e to the u, and then my
dx term over here

55
00:04:32,470 --> 00:04:36,510
this is going to be the part that sort of
picks up my bit from the chain roll.

56
00:04:38,230 --> 00:04:44,075
dx is equal to 2udu So I'm going to put
that right here in place of the dx.

57
00:04:46,020 --> 00:04:48,850
And then what's going to happen that's
very convenient,

58
00:04:48,850 --> 00:04:50,570
is that this u here and the u

59
00:04:50,570 --> 00:04:52,770
in the denominator, they're going to
cancel each other out.

60
00:04:56,190 --> 00:04:58,910
And then because 2 is a constant, I can
use my,

61
00:04:58,910 --> 00:05:03,340
my scaling property just to move that 2
outside of the integral.

62
00:05:03,340 --> 00:05:08,600
And so I can rewrite this as 2 times the
indefinite integral of e to the u d u.

63
00:05:08,600 --> 00:05:09,100
And

64
00:05:11,890 --> 00:05:17,270
now since I'm integrating, this d u tells
me the variable that I'm integrating with.

65
00:05:17,270 --> 00:05:19,500
This e to the u is going to be a very easy

66
00:05:19,500 --> 00:05:23,689
function to integrate with respect to u,
because it's its own antiderivative.

67
00:05:24,690 --> 00:05:30,700
So I end up with 2 e to the u plus c but
that's not quite finished yet,

68
00:05:32,820 --> 00:05:36,470
because the answer to the question that I
asked, should be

69
00:05:36,470 --> 00:05:39,500
given in terms of x because this is a
function of x.

70
00:05:41,130 --> 00:05:43,060
But I know that u is equal to the square
root

71
00:05:43,060 --> 00:05:46,220
of x so all I have to do in my final step.

72
00:05:46,220 --> 00:05:48,940
Is just replaced this u with the square
root of x.

73
00:05:50,190 --> 00:05:54,050
So this would be in, in my opinion

74
00:05:54,050 --> 00:05:56,400
probably the simplest way to write down
the answers.

75
00:05:56,400 --> 00:05:57,830
So to have 2 times e to the

76
00:05:57,830 --> 00:06:01,010
square root of x plus an arbitrary
constant.

77
00:06:01,010 --> 00:06:05,400
If you take the derivative of that, you're
going to get this e

78
00:06:05,400 --> 00:06:07,970
to the square root of x divided by the
square root of x.

79
00:06:09,212 --> 00:06:16,140
and here I've assumed that x is greater
than 0 just

80
00:06:16,140 --> 00:06:18,010
because I want to be able to take the
square root of it.

81
00:06:23,170 --> 00:06:27,530
So now let's consider trying a little bit
more complicated function.

82
00:06:33,390 --> 00:06:38,480
So this, this doesn't really look very,
very much like it's a, a function

83
00:06:38,480 --> 00:06:41,920
that I'm going to be able to write down as
a function of another function,

84
00:06:45,730 --> 00:06:50,950
but it is something that I can still use
this rule to, to simplify quite a bit.

85
00:06:50,950 --> 00:06:57,670
So, suppose I define u to be e to the x
minus e to the minus x.

86
00:06:57,670 --> 00:07:02,570
So it's going to be the denominator as my
integrand.

87
00:07:04,540 --> 00:07:09,189
So integrand is just a word that means the
function that I'm trying to integrate.

88
00:07:12,330 --> 00:07:17,320
Well, if I take the derivative of this
function with respect to, with respect to

89
00:07:17,320 --> 00:07:25,311
x, I'm going to get du is equal to e to
the x, and then this

90
00:07:25,311 --> 00:07:32,190
one, by the chain rule, I'm going to have
a minus sign pop down, so this, oh, sorry.

91
00:07:32,190 --> 00:07:38,540
This side, when I take its derivative, I'm
going to get e to the minus x times

92
00:07:38,540 --> 00:07:40,180
the derivative of the exponent.

93
00:07:40,180 --> 00:07:44,200
The derivative of minus x is negative 1,
so that's going to have that's going

94
00:07:44,200 --> 00:07:48,010
to have the effect of just changing this
minus sign into a plus sign.

95
00:07:49,560 --> 00:07:52,510
So that's just a little application of the
chain rule here.

96
00:07:52,510 --> 00:07:53,010
And

97
00:07:55,570 --> 00:07:58,490
now if I go ahead and make these
substitutions,

98
00:08:03,220 --> 00:08:09,590
I end up with e to the x plus e to the
minus x times dx.

99
00:08:09,590 --> 00:08:12,310
But that's just what I've said, that's my
du.

100
00:08:13,660 --> 00:08:18,540
And I have 1 over e to the x minus e to
the minus x.

101
00:08:18,540 --> 00:08:23,050
And that's what I've called my, that's the
variable I've chosen fort my substitution.

102
00:08:24,150 --> 00:08:28,310
So this whole thing simplifies into just
finding the anti-derivative of 1

103
00:08:28,310 --> 00:08:29,020
over u du.

104
00:08:29,020 --> 00:08:34,630
And that again was in my little dictionary
of anti-derivatives.

105
00:08:34,630 --> 00:08:36,060
So that's the log of u.

106
00:08:40,060 --> 00:08:43,720
And then to be finished I have to
substitute back

107
00:08:43,720 --> 00:08:47,240
so u is equal to e to the x minus x.

108
00:08:49,360 --> 00:08:52,910
So the value of this anti-derivative here
is going to be log of e

109
00:08:52,910 --> 00:08:57,360
to the x minus e to the minus x plus an
arbitrary constant.

110
00:09:03,710 --> 00:09:05,760
And then there's one more trick we can use

111
00:09:08,960 --> 00:09:12,859
with integration by substitution when
we're working with definite integrals.

112
00:09:16,260 --> 00:09:18,510
So by the Fundamental Theorem of Calculus,

113
00:09:22,520 --> 00:09:26,914
let's see, if I get to where I used that.
When

114
00:09:26,914 --> 00:09:31,190
I integrate, oops.
When I

115
00:09:31,190 --> 00:09:36,960
integrate f of x, I should just have

116
00:09:36,960 --> 00:09:42,650
capital f of x evaluated at its upper and
lower endpoints.

117
00:09:42,650 --> 00:09:47,570
But if I'm going to change from x into u,
so I'm going to make some

118
00:09:47,570 --> 00:09:50,390
sort of substitution, the endpoints have
to

119
00:09:50,390 --> 00:09:53,560
change, too, because what I'm really
saying.

120
00:09:53,560 --> 00:09:56,840
When I write these limits, oops, the
limits of integration,

121
00:09:56,840 --> 00:10:02,090
so from a to b, I'm talking about values
of x.

122
00:10:02,090 --> 00:10:06,720
So I'm talking about x equals a to x
equals b.

123
00:10:06,720 --> 00:10:09,070
When there's only one variable generally
you don't write

124
00:10:09,070 --> 00:10:13,410
this notation just because there's no, no
possibility of ambiguity.

125
00:10:13,410 --> 00:10:15,580
Because there's only one variable that
you're talking about.

126
00:10:18,020 --> 00:10:22,520
But when I make my substitution, so when I
want to use the

127
00:10:22,520 --> 00:10:27,102
integration by substitution formula.
Here my

128
00:10:27,102 --> 00:10:31,510
intergram is now a function of u.
So the limits on

129
00:10:31,510 --> 00:10:36,930
my integral also need to be in terms of u,
rather than in terms of x.

130
00:10:36,930 --> 00:10:43,370
So I need to take g inverse.
So that's the, if I have u

131
00:10:43,370 --> 00:10:45,160
is equal to g of x.

132
00:10:47,830 --> 00:10:52,730
then I need to use that function backwards
to get what my limits should

133
00:10:52,730 --> 00:10:57,479
be once I've changed from sort of
integrating in x space into u space.

134
00:11:07,020 --> 00:11:10,040
And so if we just follow this calculation
all the way through, what

135
00:11:10,040 --> 00:11:12,230
we're eventually going to end up with, so

136
00:11:12,230 --> 00:11:15,480
I used my integration by substitution
formula.

137
00:11:17,480 --> 00:11:19,450
Then once I have this guy here, I can

138
00:11:19,450 --> 00:11:24,840
run the chain rule backwards which gives
me this.

139
00:11:24,840 --> 00:11:27,230
The integral and the derivative.

140
00:11:27,230 --> 00:11:29,360
So I, if I take the antiderivative of a

141
00:11:29,360 --> 00:11:31,500
derivative, I just end up with the same
function.

142
00:11:31,500 --> 00:11:32,210
So I have

143
00:11:32,210 --> 00:11:36,220
f of g of u.
I've defined my limits.

144
00:11:36,220 --> 00:11:41,200
To be g inverse b and g inverse of a so
when I evaluate

145
00:11:41,200 --> 00:11:45,830
this g of u at g inverse of b and g
inverse of a.

146
00:11:45,830 --> 00:11:50,040
So that it's getting kind of sloppy here,
but g and g

147
00:11:50,040 --> 00:11:54,760
inverse their functions that sort of undo
whatever the first function did.

148
00:11:54,760 --> 00:11:58,560
So g of g inverse of x is just equal to x.

149
00:11:58,560 --> 00:12:02,950
So all this is really going to end up
being is f of b minus f of a.

150
00:12:02,950 --> 00:12:05,220
Which is what we were after in the first
place.

151
00:12:09,930 --> 00:12:11,490
And so, this ends up being the

152
00:12:12,940 --> 00:12:17,000
integration by substitution rule, for a
definite integral.

153
00:12:17,000 --> 00:12:20,130
And then I'll go through a quick example
of, of using this guy.

154
00:12:20,130 --> 00:12:20,130
[INAUDIBLE]

155
00:12:20,130 --> 00:12:20,142
.
>>

156
00:12:20,142 --> 00:12:20,642
[INAUDIBLE].

157
00:12:25,920 --> 00:12:26,310
>> Yeah.

158
00:12:26,310 --> 00:12:31,000
So it's, it's essentially a shortcut that
avoids you having to go back.

159
00:12:31,000 --> 00:12:34,120
So you could just compute the anti
derivative.

160
00:12:34,120 --> 00:12:34,770
>> Right.

161
00:12:34,770 --> 00:12:36,228
>> And once you have the
anti-derivative.

162
00:12:36,228 --> 00:12:38,010
Derivative.
That's all you need.

163
00:12:38,010 --> 00:12:40,550
It's just a trick to get you the
anti-derivative.

164
00:12:40,550 --> 00:12:43,350
But if you change the limits, you don't
have to substitute

165
00:12:43,350 --> 00:12:48,280
backwards cause you already have limits
in, in the right numbers.

166
00:12:48,280 --> 00:12:51,288
Or in, in the right scale I suppose for
the right variable.

167
00:12:51,288 --> 00:12:55,956
>> Right.
>> So suppose I

168
00:12:55,956 --> 00:13:01,540
want to integrate

169
00:13:01,540 --> 00:13:07,050
and suppose I want to do this by
substitution.

170
00:13:07,050 --> 00:13:10,900
Because if I looked at this, I would
probably just do the multiplication.

171
00:13:10,900 --> 00:13:16,170
And then use the anti power rule.
But,

172
00:13:16,170 --> 00:13:18,530
what I'm going to do is recognize that the
derivative

173
00:13:18,530 --> 00:13:21,080
of this, is going to have an x squared in
it.

174
00:13:21,080 --> 00:13:27,720
So I could make some sort of substitution
for this x cubed minus a constant piece.

175
00:13:29,140 --> 00:13:33,240
So I'm going to let u equals, equal x
cubed minus 1.

176
00:13:33,240 --> 00:13:37,130
Then du is going to be 3x squared dx.

177
00:13:37,130 --> 00:13:41,570
So I already have an x squared d x here,
so to get a three I just need

178
00:13:41,570 --> 00:13:46,640
to multiply by one, use the very clever
choice of one of being

179
00:13:46,640 --> 00:13:51,925
one third times three, and that's going to
give me one

180
00:13:51,925 --> 00:13:57,020
third, oops.

181
00:13:57,020 --> 00:13:59,210
There are supposed to be a fourth power up

182
00:13:59,210 --> 00:14:01,160
here which made it a little bit more
complicated.

183
00:14:08,190 --> 00:14:12,560
so I'm going to put my 3x squared dx over
on the right hand side.

184
00:14:12,560 --> 00:14:16,859
And that's going to be just equal to the
derivative of my substitution.

185
00:14:18,060 --> 00:14:21,630
In order to put this 3 here I had to
introduce a 1 3rd,

186
00:14:21,630 --> 00:14:25,680
but because that's a constant value I can
move that out of the integral sign.

187
00:14:27,950 --> 00:14:30,090
And then I'll go ahead and make my
substitution,

188
00:14:32,350 --> 00:14:34,670
but I'm also going to change the, the
limits when I

189
00:14:34,670 --> 00:14:38,250
do this too, so, when I look at the lower
limit.

190
00:14:38,250 --> 00:14:43,490
When I put negative 1 into here, I have
negative 1 cubed.

191
00:14:43,490 --> 00:14:48,470
Is minus 1 times minus 1 is 1 times minus
1 again is negative 1.

192
00:14:48,470 --> 00:14:49,810
Minus 1.

193
00:14:49,810 --> 00:14:52,780
So that becomes negative 2.

194
00:14:52,780 --> 00:14:56,920
And then 0 cubed is obviously going to be
0.

195
00:14:56,920 --> 00:14:57,455
Minus

196
00:14:57,455 --> 00:15:00,392
1.
So that gets me at negative one up here.

197
00:15:00,392 --> 00:15:07,502
Because the 3 X squared

198
00:15:07,502 --> 00:15:13,550
is what I can replace

199
00:15:13,550 --> 00:15:17,270
by the, by the chain rule backwards.
Becomes my D U term.

200
00:15:20,220 --> 00:15:23,500
And so I can use my anti-power rule to
figure

201
00:15:23,500 --> 00:15:26,010
out what an anti-derivative for u to the
fourth is.

202
00:15:27,380 --> 00:15:33,310
And then I just end up with 1 15th times
negative 1 to the fifth power.

203
00:15:35,870 --> 00:15:36,390
Oops.

204
00:15:40,170 --> 00:15:41,160
What have I done here?

205
00:15:46,570 --> 00:15:50,370
Yeah, so this should be a 1 15th too, boy.
Lots of typos on these slides.

206
00:15:50,370 --> 00:15:51,880
I don't know what happened this time
around.

207
00:15:53,740 --> 00:15:55,240
at the bottom limit squared.

208
00:15:57,820 --> 00:16:01,370
And if you do the math, you end up with 31
over 15.

209
00:16:01,370 --> 00:16:05,811
Sorry, minus 31 over 15.