BAM! Mr. Tarrou. In this video we are going to introduce the Second Fundamental Theorem
of Calculus. Yes, there is two. If f is continuous on an open interval I containing a, then for
every x in the interval the derivative with respect to x of the definite integral of f
of t dt from a to x is equal to f(x). So, let's see here. It seems like if we have a
definite integral whose lower bound is a constant and whose upper bound is the variable x, so
it is not a constant like most of the time when we do definite integrals...it is just
that variable x, we can see here there is a lower case f and and another lower case
f. So, it looks we can just take t out and plug in x and get our answer. What we are
going to do to show that this actually works is use three of our four examples from my
previous video Definite Integrals as a Function of x. Why did we get definite integrals that
were functions? That is because we did definite integrals that had an upper bound or an upper
and lower bound which were not just simple constants, they were algebraic expressions.
Through that process we got an anti-derivative, or capital F in terms of x. So, we should
be able to take the derivative with respect to x of those examples or those answers that
we got from those examples in the previous video and see that the derivative of those
F(x)'s...those capital F of x's...those antiderivatives... is equal to f(x). So, we can kind of cut out
this middle work here of doing the definite integral and then finding the derivative,
because we can see oh..we can just take the t and plug in x. Well, let's see if that actually
works. Of course it is going to. In the previous video we talked about finding the definite
integral of 10 over the square root of t dt from 4 to x. When we did that, we got F(x)
is equal to 20 square root of x minus 40. Now again notice, we are going to a capital
F so it is the antiderivative in terms of x. Ok, so we should be able to...let's see
here...F prime of x...the derivative of F(x) is equal to the derivative with respect to
x of 20 times the square root of x minus 40. Now let's not forget this is x to the 1/2
power, so taking the derivative of 20 square root of x is going to give us 20, we are going
to take that power of 1/2 and bring it down, and then reduce the power by one. Then of
course minus 40 is the constant and the derivative of a constant is always equal to zero. Well,
1/2 of 20 is equal to 10. x to the negative 1/2 power, that is the same as 1 over x to
the 1/2 power so that is going to be 1 over the square root of x. It is going to be 10
times...uh...let's just say 1 over the square root of x just to show our work. Then it is
10 over the square root of x. You see what we have here? 10 over the square root of x
was in that same pattern of 10 over the square root of t. Is the final function the was the
function was written in the definite integral? We are just taking the t out and replacing
it with x. So, the derivative with respect to x of the definite integral from 4 to x
of 10 over the square root of t dt is equal to 10 over the square root of x. It is the
same function, just with t taken out and x put in it's place. So, here we have F(x) is
equal to negative cotangent of x plus 1. So, the derivative of F(x) is equal to the derivative
with respect to x of negative cotangent of x plus 1. Well, the derivative of the cotangent
of x is equal to negative cosecant squared of x. Again we have a constant in here of
1, so it's derivative is going to be equal to zero. The negative that is in front of
cotangent along with the negative that comes in with doing the derivative of the cotangent
of x are going to cancel out and we end up with just cosecant squared of x. Again, because
our lower bound is a constant and our upper bound is just a basic variable of x, we can
take this expression that is in our definite integral and just say that f(x) is the same...just
take out the t and replace it with x and we get cosecant squared of x. Now, things get
a little bit trickier when your lower bound is not a constant and your upper bound is
not just a little basic x. So, let's look at our last example and see how we are going
to deal with extra difficulties. In our last example the use of the Second Fundamental
Theorem of Calculus is not going to be as straight forward as our previous two examples.
Why? Because our definite integral does not have a lower bound which is a constant, nor
is the upper bound just a variable of x. So, we learned in our previous video Definite
Integrals as a Function that this definite integral of 3t^2-1 dt from x to x+3 is equal
to capital F of x, that antiderivative in terms of x, is equal to 9x^2 plus 27x plus
24. Ok, so that means that the derivative with respect to x of this definite integral
is equal to, let's see here F'(x) or the derivative with respect to x of this expression here,
is going to be...bring down the 2 and 2 times 9 is 18x...and the 27x just simply becomes
27. So, we can see that we are not just taking t out and replacing them with x. There is
a lot more going on. So, I am going to step off and reveal the solution to this problem
and it is going to deal with a couple of properties that we have learned in the past to get this
done. Let's not talk, and just see that example. Ok, so I have written out the solution. Let's
figure out what the heck we did. We had to use two properties of definite integrals,
the Additive Interval Property and this property here where I am switching the upper and lower
limit of our integral which is going to give us an opposite answer. Then we finally finished
off with the chain rule so let's talk about. For the Second Fundamental Theorem of Calculus
to work our lower bound must be a constant. Well, my upper bound is x+3 on my definite
integral and my lower bound is x, I have no constant so I need to introduce that. For
convenience sake we are going to introduce the constant of zero. Ok, well here I have
a definite integral going from x to zero and I need that constant to be my lower bound
and not my upper bound. By the way, I have split this one definite integral into two
and the property that allows me to do that is the Additive Interval Property. If you
have trouble thinking that zero which I have introduced, I have gone from x to zero and
from zero to x+3, and this would act as my variable c in that property of the Additive
Interval Property. When we talked about that extra break that we are putting in, this value
of c does not need to be in-between a and b for the Additive Interval Property to work.
So, if it turns out that zero is not in-between x and x+3, it is ok. It will still work out.
Ok, so I need that lower bound to be constant so I am going to take that lower bound here
and that upper bound of my definite integral and swap them. By doing that I am going to
get the opposite answer from my definite integral. That is why this negative is here. That is
another property of those definite integrals. And I have in my second term here I have a
lower bound that is a constant but my upper bound is an algebraic expression of x+3. I
need that to be a basic variable of x if you remember from that property...or that Second
Fundamental Theorem of Calculus. So, I am going to a u substitution and let u be equal
to x+3. That is going to allow me to rewrite this definite integral with that basic variable
of x or in this u as my upper bound. Well, that means that I need to switch from the
derivative with respect of x to the derivative with respect to u and then go through that
chain rule. The derivative of F(x) or F'(x) is equal to dF/du and we are going to work
through that variable of u to ultimately get to that next step of multiplying by du/dx,
that chain rule. Ok, so constant and x, just take out the t and plug in x. When we work
out that derivative with respect of x of that definite integral. Well, here we have the
derivative with respect to u. Here is an upper bound of u, so we are going to just take that
t out and put in u which we have here, but we also now need to include that du/dx part
that we have at the second part of our chain rule. Well now that substitution allows us
to directly apply the Second Fundamental Theorem of Calculus. We need to get u and plug that
back in, so u comes out and x+3 goes back into it's place. But, to take care of that
du/dx, plugging in back that x+3 we get the derivative with respect to x and that du comes
out and I am putting the substitution back in of x+3. Well, I have kind of made my example
a little bit easier by letting my upper limit bound of my original definite integral be
a single degree or have a degree of one, so when we take the derivative of x+3 it does
come out to be a value of 1. That is where this is coming from, the derivative of x+3
is equal to 1. This is not going to happen. If my upper bound happened to be x squared,
then this would be a 2x here. At this point... So, just be aware of that. At this point we
are just going to finish simplifying all of this up. We get 18x+27. Now again when we
did Definite Integrals as a Function, we saw that we got F(x) was equal to... What was
that again? Let's see (it is right behind me!!!) 9x^2 plus 27x plus 24. That is what
F(x) and when we took the derivative of that, we of course got the 18x+27 that we were hoping
to arrive at. So, we could have gotten that same answer F'(x) by not going through the
Second Fundamental Theorem of Calculus. However, if you need to use the Second Fundamental
Theorem of Calculus this is again what it looks like. This is the end of our last example.
So, I am Mr. Tarrou. BAM!!! Go Do Your Homework:D