BAM!!! Mr. Tarrou. In this video we are going to continue our work of integration using
substitution, or u substitution, or changing the variable of integration. Now we have had
a video of doing indefinite integration. We are going to move on to now working with the
First Fundamental Theorem of Calculus along with this substitution to find definite integrals.
So change of variable, if the function u = g(x)... This again us looking normally at that inner
function, finding that inner function and letting it equal u. Then doing all the substitutions
to eliminate all references of the original variable in your integration expression changing
it to that variable of u... has a continuous derivative on the closed interval of [a,b]
and f is continuous on the range of function g, then the definite integral of f(g(x)) times
g prime of x dx from a to b is equal to the definite integral of f(u)du... But note here
the lower bound instead of being a is g(a) and the upper bound of our definite integral
is not g(b). So with indefinite integration by substitution we did a u substitution, we
went through the indefinite integration process, then at the end we had to take the value or
expression that we set u equal to and plug it back into our final answer. We need to
have an answer in terms of the same variable that the integral started off with. But with
definite integrals we are looking for a numerical value. So we can just change that lower and
upper limit of our definite integral by taking that inner function g, that is equal to u,
and plugging in a our original lower limit and plugging in b our original upper limit
of the definite integral. And that way we can just evaluate this definite integral,
get that numerical value, and that is it. We don't have to take whatever u is set equal
to and plug it back into our answer and get it back into say terms of x. We can just leave
it in terms of u, find the definite integral, and we are done. Here is what it is going
to look like. Our first example here is we want the definite integral of x times (x^2+2)
to the fifth power dx from negative three to three. Ok, so we are going to look at this
expression. We are going to find the inner function which looks like it is x^2+2 because
it is inside that other math function of raising it to the fifth power. So we are going to
let U equal x squared plus two. Then du is going to be equal to 2x dx. I also see a lot
of clouds rolled in which messed up my white balance a little bit, so let me pause the
tape and fix that white balance and we will get back to my example. I look a little bit
less orange now. So what we can do with this... Oops, I forgot the dx. Ok. Inside our definite
integral we have x dx and we don't have a 2. I am going to move this 2 to the other
side and have x dx. Now with this U substitution we are going to take x^2+2 out and plug in
u. With our x dx, we can plug in du/2. We have enough information here with this equation
and this equation to eliminate all the references of the variable of x and change it to the
variable of u in our definite integral. But if we don't remember to change the lower and
upper limit on our definite integral we are going to of course going to get the wrong
answer. How are we going to change those? Well, we know that the lower limit is x equals
-3. So we are going to take this u substitution and find that lower limit. U is going to be
equal to (-3)^2+2. See I am just taking this lower value of x that was defined by our definite
integral and plugging it into our expression. Well -3 squared, keep that negative wrapped
in the parenthesis, is 9, and 9+2 is equal to 11. Now we need to find our upper limit.
This again was our lower limit. Our upper limit is, take the positive value of 3 and
u is equal to (3)^2+2. Now because when you raise a negative number to an even power you
cancel out the negative sign, we are going to get the same value of 11 which is a little
bit weird. Let's see what happens. We have the definite integral, now we are taking all
references of x out and using reference of U. So it is going to be... let's see here...
x dx is again du/2. Instead of writing x^2+2 to the fifth, we are just going to write u
to the fifth power. Then we have a new lower limit of 11 and an upper limit as well of
11. Ok. So a couple of things here. When we looked at our original lower and upper limit
of our definite integral... Pardon me. Sorry for the interruption. I had to get my little
puppy calmed down. Ok, so clearly we had a lower and upper limit that were different
of course... negative three and positive three. Now our new lower and upper limits are actually
the same value. So let's not forget with a definite integral, one of the properties that
we learned of definite integrals, is when a and b are the same... or now g(a) and g(b)...
when our lower and upper limits are equal, we are basically... um... at least if you
are using definite integrals for area... That is not the only thing it can represent...
but when you represent it as an area, the lower and upper limit bound are the same.
You are squishing that area between the lower and upper bound. It is closed together. There
is no area there. So the value of this definite integral because these lower and upper limits
are the same is going to be equal to zero. I don't even need to walk through the First
Fundamental Theorem of Calculus to calculate this. Those are equal, so this is equal to
zero. That is because this original function turns out to be, when you multiply it out,
turns out to be an odd function. And my lower and upper limit values are opposites. So the
area, so this is symmetric to the origin... So we have some area below the x axis and
some area above the x axis. Odd functions are symmetric to the origin. Our lower and
upper limits are balanced to that origin, so they are just cancelling each other out
and getting a value of zero. The same answer that we got without, you know, discussing
that idea of odd functions and their symmetry to the origin and the fact that the lower
and upper limits are simply opposites. So we are going to get that value of zero. Let's
take a look at one more example. For our second and last example we have the definite integral
of (x-3) times the square root of 6-x dx from the lower limit of 2 and upper limit on the
definite integral of 5. Again we have a somewhat complicated definite integral here. We are
going to come off to the side and set up our U substitution. We are going to let U equal...
now we have an x-3 but it is not inside a math function like the 6-x underneath that
square root symbol. So we are going to let U equal 6-x. That is going to make du equal
to negative... let's see here... that is a constant so that derivative is zero.. the
derivative of -x is -1 dx. Ok. Now let's go ahead and take this negative one to the other
side and write negative du is equal to dx. With this much here we have enough information
to get rid of this reference to x and plug in U. We have dx which is equal to negative
one du. But I don't have enough information here with my U substitution yet to be able
to get rid of that x-3. So we are going to again like we learned with indefinite integration
manipulate this equation and allow it to do two different types of substitution. We are
going to solve it for... Well I am going to solve it for x just to show all the steps.
But ultimately I want to change this side of the equation to look like x-3 so I have
that direct substitution. So we are going to subtract both sides by 6. We get u-6 is
equal to negative x. Then we are going to divide everything by negative one. That is
going to change this sign so we have x. It is going to change this negative six to positive
six when we divide by negative one. So 6, and then u divided by -1 is of course is negative
u. Well I am almost there. I have the x, but not the minus 3. So now I am going to introduce
that part to the equation, and subtract both sides by 3. I get 6-3 is equal to 3 minus
u is equal to x-3. I guess I should pause the video and get that phone. Ok. So what
do we have here? I have enough information now to get rid of all my references of x in
my definite integral but we have not changed the lower and upper limits yet. So my lower
limit... Let's pick a different color here... is.. ok my lower limit is now going to become
g(a). So we are going to take the function g which was that inner function that is our
U. We are going to take the 2 and plug it into what we set U equal to. So L, my lower
limit... my g(a), is going to be equal to... clean that up a little bit... we have 6 minus
2 which is equal to 4. My Upper limit of the integration is going to be g(b). That is going
to be 6 minus my original value of b which is 5 and we get a value of 1. So a cautionary
note here, my lower limit in my original definite integral is a small number like we generally
see. We don't to have it this way, but it is like this often. My upper limit is a larger
value of 5. Now my new lower limit after I finish this U substitution is going to be
4 and my upper limit is going to be 1. A little bit backwards. So don't accidentally put that
1 in the bottom of your definite integral because it is the value of g(b). It is your
upper limit if you will. So we have enough information, let's get this substitution done.
We have the definite integral... what was 2 is now equal to 4 and going to 1... x-3
is 3-u... that time the square root of 6-x, or U... so the square root of U du. Now don't
forget that dx was equal to negative du. So if I really want to show all of my steps,
I need to show that this is getting multiplied by negative one times du. Then we are going
to move that -1 out in front of the definite integral. You can pull out those constants
remember. So again, we times du with the negative out front. Now at this point hopefully with
your new knowledge of the First Fundamental Theorem of Calculus, and now that we have
completed the U substitution, you should be able to finish this. So I am going to step
off, reveal the answer step by step. Maybe you can pause the video and try it on your
own and make sure that you are getting these answers right. Do not forget that this is
going to come out to be a numerical value. So when we are done, we are done. We don't
need to take the 6-x and plug it back in. It is already to be finished. So let's see
what the answer looks like. BAM! So I hope this is what you got. We needed to distribute
this square root of u through the parenthesis. I went ahead and changed that to a U to the
1/2 power so I could add the exponents when I multiplied like bases. When you integrate
and you are dividing by that fraction, because we had to raise those exponents by 1, let's
not forget that dividing by a fraction is the same as multiplying by the reciprocal.
That is how I got this line here, just continued through the First Fundament Theorem of Calculus
until we got that definite integral value of 8/5. So, I am Mr. Tarrou. BAM! Time to
Go Do Your Homework:D