So we're ready to begin to connect this idea of integration back to this topic that we started with, the topic of the antiderivative. So as a reminder, here is the definition that we've previously introduced, the definition for an antiderivative. So now it's time to see how antiderivatives are going to connect back to this topic that we've started to introduce, this topic of what we're calling integration now, right?
Finding the limit. of a Riemann sum. So what we'll say is that if lowercase f is continuous on a closed interval, then for every number inside of that interval, the definite integral here, the definite integral from a to x, right, x is just some number between a and b, if that function is continuous, this definite integral is again going to be a number when we select that value of x in between a and b. And so what we're going to do now is we're going to define a function, a function called capital F, on that same interval and that function is going to be exactly this definite integral from a to x. Right, so we will pick a number, plug that number into the function, and out comes a number, right, an x value, and this will be our y value for every x that we select inside the interval.
So the first fundamental theorem of calculus makes an important connection. between the topic of the antiderivative and this function right here. So this is called the first fundamental theorem of calculus or the fundamental theorem of calculus part one. So what the theorem says is this. If lowercase f is a continuous function on the interval a, b, then the function called capital F defined this way, defined as the definite integral from a to x of f of t dt, that function is continuous on the same interval.
It's differentiable on the same interval, and its derivative, capital F prime of x, turns out to be just the original function, f of x. So when I take the derivative of this definite integral defined as this function, what happens is the two sort of wash each other out, and we're just left with f of x. So in other words, what we're seeing is, we're seeing exactly the definition of the antiderivative capital F prime of x is equal to lowercase f of x. And so what this establishes is this function here is an antiderivative of our function lowercase f.
So what we've established now is that differentiation and integration are essentially inverse operations, much like multiplication and division. And our knowledge that this is now representing an antiderivative will lead to an important result that we'll see in the next section. So what we're seeing here is that if I take the derivative of an integral, the two operations are sort of washing each other out.
So if I take the derivative with respect to x of the definite integral from a to x of e to the t cosine of t dt, so effectively the x just jumps in and gives us e to the x times the cosine of x. Likewise, if I move the derivative to the inside and then apply the integral, we have sort of the same effect. They're going to wash each other out. And they're identical up to a constant in this case.
So the first fundamental theorem of calculus establishes that this function is an antiderivative. And now we can say that the operations of differentiation and integration are effectively inverses of each other, like multiplication and division. So now I know how to observe this function.
This function is just an antiderivative. And since I'm taking the derivative of an antiderivative, I know the result I should have. I should have this capital F prime equals lowercase f.
We can now treat this function like we've treated functions all semester. We can do things. that we've done to other functions. For instance, we can find local extrema on an interval.
So let's look at this function. The function called capital F of x equals the definite integral from 0 to x of the cosine of t dt. So by the first fundamental theorem we know that capital F prime of x is just going to be the cosine of x.
So our derivative is just the cosine of x. How do I find local extrema? I find critical points. So our critical points are going to be where the cosine of x is equal to 0 and that's going to happen at pi over 2 and 3 pi over 2 which is within 0 to 2 pi.
So now I'm going to do what we did before. Let's look at the behavior of the derivative around those critical points. So we've got a critical point at pi over 2, a critical point at 3 pi over 2. So we'll just select some test points.
So maybe pi over 4. Maybe 3 pi over 4. And maybe 5 pi over 4. Let me correct this. This should be 7 pi over 4. So what do I know about the cosine? Our derivative, cosine of pi over 4, we know is root 2 over 2, so we're positive. Cosine 3 pi over 4, we're in the second quadrant. Cosines in the second quadrant are negative.
negative and 7 pi over 4 cosines in the fourth quadrant are going to be back positive. So what is this showing us? We're transitioning from positive to negative at pi over 2, so that's going to be a local maximum.
And we're transitioning from negative to positive at 3 pi over 2, that's going to be a local minimum. Alright, so what I know here, we don't quite know how to evaluate this function. So this function takes on a local max at pi over 4. and then a local min at three excuse me at pi over two not our test point our critical point and a local minimum at three pi over two so again this is now just a different kind of function that we can now discuss things about So what do I know?
I know that the y value of this function is going to be a local maximum at pi over 2, and it'll be a local minimum at 3 pi over 2. We can continue with the same ideas. Let's think about the intervals of increase and decrease and the intervals of concavity for this function. Capital F of x equals the definite integral from 1 to x. t squared minus 7t plus 10 dt.
Again, this is just a function, so I pick the x, plug it in, out comes a y value. All right, we don't quite know how to find the y value yet, but we can say some things about this function. So the derivative capital F of x, of capital F of x, is going to be capital F prime, and that's going to be x squared minus 7x plus 10. And we'll need the second derivative, so let's go ahead and find it.
That's going to be 2x minus 7. So for our intervals of increase and decrease, we'll find the critical points. We'll set the derivative equal to 0. And so our derivative here is going to factor into x minus 5 and x minus 2. So we've got critical points at x equals 2 and x equals 5. So let's look at the behavior of the derivative around those critical points. So here's 2 and 5. Let's pick our test points.
Maybe we'll pick 6, maybe 3 and 0. So 0 goes into the derivative. Out is going to come 10. That'll be positive. 3 is going to go into the derivative.
What's going to come out? That'll be 3 squared. 9 minus 7 times 3. So 9 minus 21. That'll be negative 12. Negative 12 plus 10 will be negative 2. So we're negative in this region.
And then plugging in 6. We'll have 6 squared which is 36. 36 minus 7 times 6 that'll be 36 minus 42 that'll be negative 6. Negative 6 and 10 is going to be positive 4. Alright so we are seeing that this function increases On the interval negative infinity to 2 and then the interval 5 to infinity. And we're going to decrease just on that little segment in between. And that'll be the interval, excuse me, 2 to 5. All right, now for concavity, we'll look at the second.
Second derivative equal to zero is going to give me a potential inflection point at seven halves. So again, we'll just look at the behavior around that point for the second derivative. So seven halves will go above and below. Seven halves is 3.5, so let's pick maybe zero and maybe five.
So our second derivative was 2x minus 7. 0 goes in, negative 7 comes out. Second derivative at 5, that would be 10 minus 7, which is going to be positive. So we're going to see here that we are concave up on the interval 7 halves to infinity.
And we'll be concave down otherwise. Concave down on negative infinity to 7 halves. All right, so with this connection that the first fundamental theorem makes, this may look like a complicated function, but we know how to find the derivative, and from there we can do things that we've already done this semester. we can even further we can find the equation of a tangent line for this kind of function. So for the tangent line I need the derivative.
So the derivative by the first fundamental theorem is going to be 3x squared minus x. So the slope at the point 2 is going to be capital F prime of 2. So that's going to be 3 times 2 squared minus 2. So our slope is going to be 12 minus 2 which is 10. And now for our point, well, our point will be f of 2. And that's going to be, notice here, that's going to be the integral from 2 to 2. We're plugging in 2. And if I integrate from 2 to 2, remember our property, if I have the interval being width 0, if I integrate from 2 and stop at 2, I don't go anywhere. So I know f of 2 is going to be 0. So our tangent line is now very simple. That's our y minus our y-coordinate of 0 equals our slope times x minus our x-coordinate of 2. So that's going to be 10x minus 20. So we should generalize this first fundamental theorem to account for the fact that perhaps we might see something besides a plane x in this upper limit of integration.
So this is somewhat of a generalization of the chain rule. The idea is very simple. There's some technical details here that I don't want to spend much time on.
But if I have some larger function besides a plane x in the upper limit of integration, the idea for the derivative is still the same. So what's going to happen, that function is going to jump into our integrand. So that'll be this f of u of x.
So that function will jump in. And we want to finish things off by multiplying by the derivative of that object that we plugged in. So this reads a lot like the chain rule.
So capital F prime of x is going to be f of u of x times u prime. All right, so notice here this lower limit of integration is really irrelevant. The only thing relevant is the function in the upper limit. So let's generalize this now. And I should say one thing.
We're using the variable x, which means that variable is spent. And so when I go to my integrand, I have to use a different variable other than x. So that's why you're seeing t here.
Here I've selected z. But this variable... corresponds to the input of the function.
And the only places that we're putting inputs are going to be in these limits of integration. So we need to use another variable here. So we're going to have the same idea for the derivative.
And now instead of just x taking the place, this function x cubed is going to jump in. So the derivative capital F prime of x is going to be 2x cubed. And then minus x cubed squared. And then that has to be multiplied by the derivative of what was plugged in.
And that's going to be 3x squared. Alright, and then we can simplify this. This will be, we could distribute, x cubed squared is going to be x to the sixth power, but I'm just going to leave it in that form.
This is the expression for the derivative of our function. Alright, how about the derivative of capital F of x is equal to the integral from 1 to the natural log of x, e to the t dt. So again, Again, That function in the upper limit jumps in.
That's going to be e natural log x. And then we have to multiply by the derivative of what was plugged in. The derivative of ln is 1 over x. Now you should notice here that e natural log x simplifies to just x. So we have x times 1 over x, which just is equal to 1. All right, so what does that tell us about this function?
Our slope here is a, excuse me, our derivative is a constant. When the derivative is a constant, that means our function was linear. So if I was to graph this function, I would see, I should see that it has the shape of a line. So lastly, here is a function capital F of x equals the definite integral from the tangent of x to 0 of 1 over 1 plus k squared dk.
Now again, let's look at the definition here. The definition. has the upper limit of integration being the function and the lower limit being the constant. So notice that's backwards here. The function is in the lower limit, the constant's in the upper limit.
So no big deal. We can just move on. move those around.
We can write this as the integral from 0 to the tangent of x of 1 over 1 plus k squared dk. But remember, the interchange of those limits requires us to introduce a negative. And so now the derivative is just going to be capital F prime of x. So we'll bring that negative along.
The tangent is going to jump in for the variable. That's going to be 1 over 1 plus tangent squared of x and then times the derivative of the tangent which is the secant squared. Alright so we could express our derivative as just the negative secant squared.
over 1 plus the tangent squared. And we have an identity here that we could simplify this, but I'm going to leave this as just negative secant squared of x over 1 plus the tangent squared. Alright, so the name first fundamental theorem suggests that there is a second fundamental theorem or a second part of this fundamental theorem. And so let's begin to look at where the first fundamental theorem will take us.
So what I want to do here is I want to use the first fundamental theorem to evaluate the definite integral from 2 to 5 of e to the t plus 2t dt. So here's how we'll approach this. Let's define a function. We're going to define a function called capital F of x.
We're going to set that equal to the definite integral from 2 to x of e to the t plus 2t dt. Now what we have by the first fundamental theorem, we know the derivative of this function. So capital F prime of x is going to be equal to e to the x plus 2x.
And since we know that this function represents an antiderivative, we can go ahead and just find the antiderivative of this function. So our function capital F of x should read, in terms of its antiderivative, e to the x plus x squared plus c. So let's solve for the constant c.
How can we do that? So we can observe here that f of 2 is going to be the integral from 2 to 2. For e to the t plus 2t dt, and by properties of integrals, that should be equal to 0. And so we can solve for c now. f of 2 is going to be e squared plus 2 squared plus c, and that should be equal to 0. and that's going to give us our constant.
Our constant is going to be negative e squared minus 4. So now we can put this back together. Capital F of x is defined to be the integral from 2 to x. We have e to the t plus 2t dt.
And now we know that's equal to e to the x plus x squared plus c. And we have solved for c. So this is going to be equal to e to the x plus x squared and then add in our constant. We'll write this as negative e squared plus 4. And so we're ready to find this original Original question now. So the integral from 2 to 5 of e to the t plus 2t dt, well that's going to be equal to f of 5 as we have defined this function.
And so now we know that this function is equal to e to the x plus x squared. minus e squared plus 4. So we're just going to plug in the 5. So we'll have e to the fifth plus 5 squared and then minus e squared. Let's write this 4 as 2 squared. So what is a pattern that is jumping out here?
This definite integral turns out to be equal to its antiderivative with 5 plugged in, subtract the antiderivative with 5, with two plugged in. So in other words, this definite integral turns out to be equal to F of five, capital F of five minus capital F of two. So this suggests that we now have a way of evaluating a definite integral.
Again, remember how it's defined. It's defined as the limit of the Riemann sum. All right, remember all of that.
So being defined as the limit of a Riemann sum, that's very difficult to evaluate in general. But it seems now that we have a way around this. It seems that this definite integral, the definite integral from 2 to 5, is just the antiderivative.
of our integrand, we'll plug in the upper limit, and we're going to plug in the lower limit. And that's exactly what the second fundamental theorem, or the second part of the fundamental theorem of calculus, will establish. So our method going forward for finding the definite integral is going to tie back to the antiderivative. Once I have the antiderivative, it's a matter of just plugging in and subtracting two numbers.
So all of that limit will turn out to boil down to just a very simple calculation through the antiderivative. And that's the second fundamental theorem which we will introduce in the next part of this section.