BAM!!!
Mr. Tarrou.
Well we just got done looking at finding derivatives of functions that involve the natural exponential
function.
So now of course we are going to turn that around and talk about integrals.
We are going to do four examples in this video.
The first two are just going to be indefinite integrations.
We will have that plus c at the end of our answers.
We will do a definite integration example.
Then our last example is going to be finding the specific integral through a given point,
or basically solving a differential equation given a particular solution.
We are going to find the indefinite integral and then be given a point that integral function
is supposed to go through which will allow us to solve for C and find that particular
solution.
Integration Rule for the Natural Exponential function, well the derivative of e^x is itself
e^x.
So we have the same relationship with the integration.
The indefinite integral of e to the x dx is e^x plus c.
Ok.
Most of our problems are going to be more sophisticated or more complicated than this
so we have our U substitution.
The indefinite integral of e^u du is equal to e^u+c.
So we just have to find that U substitution that is going to allow us to write our problem
to specifically or exactly fit this rule.
Then we can apply it.
Over here we have the indefinite integral of e to the x the fifth power times x to the
fourth dx.
Now sometimes a lit bit tricky to figure out how to set up your U substitution but I am
pretty sure it is not going to be x^4 because there is really not anything going on here.
Our main operation is this e, this natural base, raised to the x to the fifth power.
So I am going to let U equal x^5 and see if we get some nice equations that will allow
us to do some substitution here.
So let U equal x^5, then du is going to be equal to... bring this power down and subtract
by one... so 5x^4 dx.
Ok well we are really close.
We have an x^4dx, we have x^4dx and there is a couple of ways to do this.
I can put a 1/5 out front and have a multiplication of five in here so that 1/5 and that 5 are
really just a multiplication of one making this indefinite integral look a little bit
different.
But I like to work with these equations instead.
So I am going to divide both sides by five and get 1/5 du is equal to x^4dx.
Now we have this equations some very nice substitutions to get rid of all of our references
of x.
So the indefinite integral of e^u.. x to the fourth dx... we have the du and I am going
to put that on the right like I have been doing with all of my videos...
then put that constant multiplication of 1/5 out in front of the integrand.
Ok, so now just like we have in this rule here the indefinite integral of e^u is e^u,
the something, plus c.
So we have 1/5 e^u + c.
Now... excuse me, plus c. [yelling] CHALK DOWN!
Now ok, we have done our integration.
Lets' get rid of that u because of course our original problems had x's in it.
So we have 1/5 e to the x^5 power plus c.
That is the end of our first example.
Pretty straight forward with that.
Let's get to the next one or two, whatever I put up here.
For our next example, it looks quite a bit more complicated with the division here or
the ratio.
But let's go ahead and try the U substitution with that variable in the exponent of e again
and see if this is really that much more complicated then the last problem or is just more of the
same.
Looks a lot different.
We are going to let U equal 1 over x to the third, which if you are like me you might
want to write that as x to the negative third power.
Then du is going to be equal to.... well I am going to bring this exponent down and have
-3x... and of course reduce that by one... so -3x^-4 dx.
A lot of times when we take derivatives, we are used to of course the exponents counting
down by one.
I see an exponent of three and I see an exponent of four, so we think maybe the U substitution
will be that x to the fourth in the denominator.
But with that numerator's x being in its own denominator we can see that we are dealing
with negatives and they may act a little differently than we might expect.
Because we are really pretty close here.
We can rewrite this getting rid of the negative exponent as du is -3/x^4 dx.
I hope that air conditioner is not too distracting.
See my numerator has a coefficient of positive three and my numerator over here that I am
going to do the substitution with is simply negative three.
The denominator is x to the fourth.
I have an x to the fourth in the denominator here.
So all I have to do for a nice substitution is to get this negative to the left side of
the equation.
Ok, now I don't want to have to talk over the air condition if that is distracting for
the rest of this problem.
Let me go turn that off.
Ok, that is much better.
So we have e to the u here, so we have the indefinite integral of e^u.
I am kind of wanting to write that three right, but three over x to the fourth dx, 3/x^4 dx
is -du.
So really I kind of made this a bit too big right.
It is just...
Just bring that negative out front, e^u du.
Ok so this is working out pretty much as easily as the last one.
We have negative e to the u plus c.
I am kind of thinking here a little bit.
I didn't do this with the last problem, but maybe I should just remind you one more time.
It is negative whatever you have from this indefinite integral.
Really we should be thing about that negative having to get multiplied through the parenthesis.
But of course c is just an unknown constant.
If I multiply by a negative, I am just chasing the sign of an unknown constant.
So we don't have to show that sign change on the c.
We can just leave it as plus c.
That is why I am just simply writing e to the u, but I am going to put that back in,
u was 1/x^3 plus c.
That is the answer for this example.
Let's get the third one up here.
It will be a definite integral.
nanananana...
The definite integral of 2 times e to the x over 2 plus e to the x dx where the lower
limit is zero to three.
Ok.
So what are we going to do for this U substitution.
Well, you know, in the numerator we have just e^x.
So that is really really basic.
I think that is going to be too simple for our U substitution to work.
I am going to focus on the denominator.
We are going to let u equal 2+e^u... trying that again... let u equal 2+e^x.
Then du is going to be equal to... the derivative of a constant is zero... so just e to the
x.
Second piece of chalk I have broken.
e^x dx.
Ok.
I don't have an e^x dx, I have a 2.
So I can either move this 2 out in front of the integral symbol.
But will all of my examples of U substitution and integration, I have been making these
equations match what I have in my integrand.
So what I am going to do instead, and there is again I just gave you an alternative way
of showing this, I am going to multiply both sides of this equation by 2.
I get 2du is equal to 2e^x dx.
Now we have direct substitutions to get rid of all the references of x... almost.
So we have 2e^xdx is 2 time, and then we are going to have some kind of definite integral,
du.
So I have just floated that 2 out front of our integral.
Ok the numerator has all been taken care of with that substitution so the numerator is
equal to 1.
The denominator of 2+e^x is u.
Now I am pausing because, and I kind of gave you a hand wiggle there like... kind of sort
of...
I have not really...
If I put in my lower and upper limits of 0 and 3, those are in terms of x.
So if I still want to write that lower limit of zero and that upper limit of three, I am
going to put an asterix.. or a little star next to them to remind myself those are for
when this expression is in terms of x.
So I can still do the integration, but then I will have to undo the u substitution to
get my problem into terms of... back into that variable of x or referencing x.
Hopefully I am reviewing so hopefully you already you understand this by now, but just
in case you forgot.
So to save myself from having to do that, I am going to write my lower and upper limits
in terms of U. I am going to go ahead and say for my upper limit U is equal to 2 plus
e to the, my upper limit is 3.
My lower limit is U is equal to 2 plus e^0.
Anything to the zero power is 1, so 2+1 is 3.
So my lower limit is going to be 3.
My spacing is not really great here.
I should have maybe set my lower and upper limits first.
So I am doing the definite integral of 1/u du from 3 to 2+e^3.
You cannot forget all of those integration rules.
So the integral of 1/u du is the natural log of the absolute value of u.
So we have 2 times the definite integral from 3 to 2+e^3 of the natural log...
Excuse me...
I am doing the integral right now.
Let's change that.
Natural log of absolute value of u with an upper limit of 2+e^3 and a lower limit of
3.
Ok.
So let's continue on evaluating this definite integral.
We have 2 times the natural log of 2+e^3 minus the natural log of 3.
You might notice that I got rid of this absolute value sign as I wrote each of these natural
logs.
It would not necessarily be wrong to put them in, but 2 which is positive plus a positive
base of e which is approximate 2.718 raised to a power... this expression cannot be negative.
So the absolute value signs or symbol there is not necessary and of course 3 is positive
as well.
Ok.
I am not going to write this as a decimal, I am not going to approximate this solution
so how can we clean this up a little bit.
Of course when you have the subtraction of two logs with the same base, these are both
natural logs, I can write this as 2 times the natural log... just writing one natural
log function compressing this here... we have 2+e^3 over 3.
I like to wrap the expression that has the log function applied to it in parenthesis.
So that is the value of my answer in exact form.
If I wanted an approximated form, or an approximated value, of course we could put this in our
calculator and get some decimal rounded off to what ever.
But that is the value, that is the answer in exact form.
One last example.
I was going to say BAM!
but I am so sad this is our last example.
That means the video is almost over:(
Use integration to find a particular solution to the differential equation.
[pretending to cry] dy/dx is equal to e^x+e^-x over 2e^x-2e^-x squared.
The particular solution is that I want a function that goes through this point of (1,2).
Well dy/dx just means the derivative of y with respect to x.
So if I want to solve this, I need to basically integrate both sides.
So y is going to be equal to the indefinite integral of this.
So y is equal to the indefinite integral of e^x+e^-x over... you can read the problem
dx.
So we are going to look for a U substitution, we are going to do the indefinite integration
and have the plus c of course because it is not definite integration like the last problem,
and then we are going to use this point to find the particular solution.
Remember when we have the plus c in there, that is your family of curves, or your family
of solutions and we want that particular one that goes through the point of (1,2).
At this point I have done three U substitutions for you, or with you.
So now I want to just step off, reveal the answer one step at a time so you can pause
the video and try and finish this on your own.
If you don't get it just keep watching the video, or pause it and give it a shot.
So with a lot more going on in the denominator plus we have that exponent of 2 in the denominator,
I went ahead and decided to try and let that inside... that very complicated base of that
power of 2 be my U.
So we let U equal 2e^x-2e^-x.
Found that du was basically very very similar because of course the derivative of e^x is
e^x except we have this exponent here of negative one x.
That u is negative one x, then du.... sort of like another u substitution is equal to
negative one.
So negative one times the negative that was there changed that sign to a plus.
And we had du is equal to well... as you can read here.
But my numerator was e^x+e^-x.
We did not have these coefficients of 2, so I divided both sides by 2... or factored out
that 2 and divided it over and had 1/2 du is equal to e^x+e^-x dx.
Now we were able to use our U substitutions to get rid of all of our references of x.
Do our indefinite integration, and then when we got done cleaning that up we went ahead
and applied that particular solution where the function had to go through the point of
(1,2).
We can see here that when you make up your own problems, sometimes they do not come out
to be very neat like most of the problems in the textbook I am going to be teaching
out of.
So we ended up with a C value that is approximately 2.1064.
Our final solution, our particular solution is right here as we can see.
You can read it:)
So I am Mr. Tarrou.
BAM!!!
Go Do Your Homework:)