Lecture 4.9.2: Antiderivatives

So we're going to continue with the idea of the antiderivative as we just previously have introduced. So just a reminder of the definition of the antiderivative. A function f, capital F, is called an antiderivative of lowercase f if the derivative of capital F is lowercase f. So in other words, an antiderivative is a function whose derivative is a given function. And what we saw is that because the derivative of a constant is always 0, then... every antiderivative in general form is going to have the form capital F of X plus C, where C is some arbitrary constant. So for any given function, without more information, we actually have an infinite number of possible antiderivatives. And so to take care of that, we're going to write the antiderivative in this form, capital F of X plus C, where C just stands to be a arbitrary constant. So We talked about the introduction of this notation. We call this the integral sign. This object here now is called an indefinite integral. And our interpretation of this object is the antiderivative. So when I see this symbol, this symbol is telling me to find the antiderivative of the expression it contains. So what we'll do here is we'll split this apart into two pieces. So this is going to be the same thing as the indefinite integral. We're going to have 3 over the root of 1 minus x squared, and then we're going to have 2 root 1 minus x squared over root 1 minus x squared. And so what you'll notice here is that these roots on the second term are going to drop out. And so what we can do is we can now look at this piece by piece. So if I look at this expression 3 over root 1 minus x squared, I want to think about something that I can take the derivative of that will give me this expression. So what might jump out to you is going to be 3 times the inverse sine of x. Now all I have right here is a 2 left over, and so the antiderivative of 2 with respect to x is just going to be 2x. Now what you might also notice here is we could just as easily say negative 3 arc cosine of x plus 2x plus c. So this will often be the case with antiderivatives. I often may have more than one way of expressing the antiderivative, in particular with these inverse trig functions. Both of these would be acceptable evaluations for this indefinite integral. In other words, both of these are antiderivatives for this expression. All right, another indefinite integral. I have the indefinite integral of t squared plus 1 times 2t minus 5. So again, I've got a product here. So what I want to do is I want to expand before I move forward. So we're going to FOIL this. And when we FOIL, what we're going to have is 2t cubed minus 5t squared plus 2t minus 5. And so what we can do now is we can now go term by term. We're effectively going to be reversing the power rule. So we're going to increase the power by 1, divide by our new power. And that's going to happen on the next term. We're going to increase the power by 1, divide by the new power. Same thing, 2t becomes 2t squared over 2, and then minus 5t, and then plus our constant. Okay, and so we could just simplify this. This is going to be 1 half t to the fourth power minus 5 thirds t cubed. Notice these 2's are going to cancel, so I'm going to have plus t squared minus 5t plus the constant. How about another? Here's the indefinite integral, cosine of x minus 1 over sine squared of x dx. Again, Our strategy here should be to simplify this, so let's split this into two terms. This is going to be cosine of x divided by sine squared of x minus 1 over sine squared of x. All right, and now cosine over sine squared, let's look at that. So sine squared is going to be 1 over sine times 1 over sine. So what you might notice here is that cosine over sine is cotangent. And this 1 over sine is going to be cosecant. So what I have here is cotangent x cosecant x. Now 1 over sine squared is cosecant. Cant squared of x. All right, so we've rewritten this expression into something that now looks a little bit more tractable. So those look like things that I know derivatives for, or those look like things that come out to derivatives for certain expressions. All right, so the derivative, the antiderivative of cotangent cosecant, that's going to be negative cosecant of x, right? Remember, the derivative of cosecant is negative cosecant cotangent. So I tack on a... a negative here to take care of the positive that's showing up. And then I'll minus cosecant squared. That's going to be minus cotangent of x. The derivative of cotangent is cosecant squared. And I should correct myself here this is going to be positive cotangent. The derivative of cotangent is negative cosecant squared. And then we'll tack on our constant of integration. Okay, so we rewrote this, and now we re-expressed this term, cosine over sine squared. We saw that that was cotangent cosecant. This is cosecant squared, and now it looks very obvious how I go to my antiderivative. So the derivative of cosecant is negative. negative cosecant cotangent, double negative gives me positive, derivative of cotangent is negative cosecant squared, and then the positive here makes sure that we do have the negative showing up in this expression. Alright, so this idea of the antiderivative now lets us think about differential equations. So we're really just going to dip our toes in the water here. Here's how I would define a differential equation. An equation involving an unknown function and one or more of its derivatives is called a differential equation. So the following here is a differential equation. 4x squared y double prime plus 12 x y prime plus 3y is equal to zero. All right, so this is a differential equation. It's an equation involving an unknown function. that unknown function is y and one or more of its derivatives. Now this is difficult to solve, so we're not going to look at things this complex, but we are going to be able to talk about this idea of a differential equation regarding some simple examples. So when we're solving a differential equation, the object is to find a function which satisfies the equation. So if I were to solve this differential equation, I'm looking for the function That's going to balance this equation. So the function y, its first derivative and its second derivative, when that's all found and plugged in, we should get a balancing of 0 equals 0. So this is a little bit different than solving an equation. When we solve an equation, in general, to this point, our result is a number. Now to solve a differential equation, our result is a function. So a little bit different idea of solving here. In general, when we solve an equation, we're looking for a number. When we're solving a differential equation, we're looking for a function. So we can talk about a very simple differential equation here. So the differential equation reads this way, 2y prime minus 3 is equal to the cosine of t. So let's rearrange this a little bit. Let's add that 3 over. So I have 2y prime equals the cosine of t plus 3. And then if we divide by 2, y prime is going to be 1 half cosine of t plus 3 over 2. All right, so to solve this equation, I want to find a function whose derivative is 1 half cosine t plus 3 halves. All right, so this is very simply going to be 1 half sine of t plus 3 halves t plus c. All right, so the function which satisfies this equation is going to be one half sine of t, so it's just the antiderivative. One half sine of t plus three halves t plus the constant. All right, so again we're just looking for the antiderivative, so a half sine t, the derivative will be a half cosine t, three halves t, the derivative will be three halves. Now notice here that we have this constant, right? We've got to include the constant. Without further information, I don't know what the constant is, so we'll have to just tack it on and leave it. But if I do have some more information, it's possible to solve for the constant. So if an initial condition is given, it's possible to solve for the constant. And a differential equation with an initial condition is referred to as an initial value problem. So here we're presented with the initial value problem. So e to the x times y prime minus 1. equals e to the 2x and we're given the condition that y of 0 must be equal to 3. So what we'd like to do is we'd like to solve this for the we'd like to solve this differential equation So we'd like to find the function y Who's which will satisfy this equation and then we're going to find the constant by using this piece of information here So let's isolate this expression So let's divide by e to the x so I have y minus 1 equals e to the 2x over e to the x And so we're going to have y prime is equal to, so e to the 2x divided by e to the x is going to be e to the 2x minus x. That's just going to be e to the x, and we'll have e to the x plus 1. So now we want to find the antiderivative. We want to find the function y, whose derivative is e to the x plus 1. So you should see here that that's going to be e to the x plus x plus c. Now we're told that y of 0 must be equal to 3. So plugging in 0, we're going to have e to the 0. plus zero plus the constant must equal three. So now we can solve for the constant e to the zero is one so we have one plus c is equal to three so of course c must be equal to two. What we have now is what we call a particular solution so this right here is called the general solution. Now what we have is the particular solution, e to the x plus x plus 2. And that's called the particular. All right, and we can check that y of 0 is going to be e to the 0 plus 0. That's 1 plus 0 plus 2. So y of 0 does indeed equal 3. All right, so without further information, we're stuck with just leaving the constant. But in an initial value problem, I have the ability to go ahead and find what my constant should be. All right, so we can now revisit the motion problems that we talked about earlier in the semester. So remember that if we have a position function, the velocity is the first derivative, the acceleration is the second derivative, or the acceleration is the derivative of velocity. So now we can turn that process backwards. So velocity can be recovered by finding the antiderivative of acceleration, and position can be recovered by finding the antiderivative of velocity. Right, so the velocity is the indefinite integral of acceleration. The position is the indefinite integral of the velocity. All right, so we took that chain forward. We went position, velocity, acceleration. Now we can go acceleration, velocity back to position. All right, so let's do that. Suppose that we've got an acceleration given to us. So a of t is negative 32 feet per second squared. So let's find the velocity and position functions given these conditions that the velocity at time zero is two feet per second. And the position at time zero is 10 feet. So in other words, what I have here is I have an initial value problem. So I'm going to be able to start with a function. That function is the second derivative. I'm going to be able to work my way up towards the function that we began with. That would be the position function. So we're going to take this process backwards. So the velocity is going to be the indefinite integral of acceleration. So that's going to be the indefinite integral of negative 32 dt. And so the antiderivative is going to be negative 32t plus the constant. So the initial condition says that the velocity at time 0 must equal 2. So the velocity at time 0, that's going to be negative 32 times 0, plus the constant, that must equal 2. So what we're going to find here is negative 32 times 0 is equal to 0. So 0 plus c equals 2 is our equation. So that's going to force c to be equal to 2. Now we can take the velocity function. The velocity function is now going to be negative 32t plus 2. And we can recover the position. So the position function is going to be the indefinite integral of velocity. That's going to be negative 32t plus the number that we found, 2. All right, so now our antiderivative, so negative 32t, increase the power by 1, divide by the new power, antiderivative of 2t, excuse me, the antiderivative of 2 is going to be 2t. So what you're going to find here is that this is going to be negative 16t squared. And just to eliminate any confusion, let's call this c1. They're different constants, but we're recycling the label. So we'll just call this C2. So I get negative 16t squared plus 2t plus C2. So what am I told? I'm told that S of 0 must be equal to 10. So what we're going to have is that S of 0 Right, plugging in zero here, plugging in zero here, those are going to go away. So we're going to get S of zero is equal to C2, and C2 must be equal to 10. And so there is our constant C2. All right, so we can now recover the velocity function. The velocity function is going to be negative 32t. plus 2, that's going to be feet per second. And then the position function is going to be negative 16t squared plus 2t plus 10 feet.

So in other words, an antiderivative is a function whose derivative is a given function. And what we saw is that because the derivative of a constant is always 0, then... every antiderivative in general form is going to have the form capital F of X plus C, where C is some arbitrary constant.

So for any given function, without more information, we actually have an infinite number of possible antiderivatives. And so to take care of that, we're going to write the antiderivative in this form, capital F of X plus C, where C just stands to be a arbitrary constant. So We talked about the introduction of this notation. We call this the integral sign.

This object here now is called an indefinite integral. And our interpretation of this object is the antiderivative. So when I see this symbol, this symbol is telling me to find the antiderivative of the expression it contains. So what we'll do here is we'll split this apart into two pieces. So this is going to be the same thing as the indefinite integral.

We're going to have 3 over the root of 1 minus x squared, and then we're going to have 2 root 1 minus x squared over root 1 minus x squared. And so what you'll notice here is that these roots on the second term are going to drop out. And so what we can do is we can now look at this piece by piece.

So if I look at this expression 3 over root 1 minus x squared, I want to think about something that I can take the derivative of that will give me this expression. So what might jump out to you is going to be 3 times the inverse sine of x. Now all I have right here is a 2 left over, and so the antiderivative of 2 with respect to x is just going to be 2x. Now what you might also notice here is we could just as easily say negative 3 arc cosine of x plus 2x plus c. So this will often be the case with antiderivatives.

I often may have more than one way of expressing the antiderivative, in particular with these inverse trig functions. Both of these would be acceptable evaluations for this indefinite integral. In other words, both of these are antiderivatives for this expression.

All right, another indefinite integral. I have the indefinite integral of t squared plus 1 times 2t minus 5. So again, I've got a product here. So what I want to do is I want to expand before I move forward. So we're going to FOIL this.

And when we FOIL, what we're going to have is 2t cubed minus 5t squared plus 2t minus 5. And so what we can do now is we can now go term by term. We're effectively going to be reversing the power rule. So we're going to increase the power by 1, divide by our new power.

And that's going to happen on the next term. We're going to increase the power by 1, divide by the new power. Same thing, 2t becomes 2t squared over 2, and then minus 5t, and then plus our constant.

Okay, and so we could just simplify this. This is going to be 1 half t to the fourth power minus 5 thirds t cubed. Notice these 2's are going to cancel, so I'm going to have plus t squared minus 5t plus the constant. How about another?

Here's the indefinite integral, cosine of x minus 1 over sine squared of x dx. Again, Our strategy here should be to simplify this, so let's split this into two terms. This is going to be cosine of x divided by sine squared of x minus 1 over sine squared of x. All right, and now cosine over sine squared, let's look at that. So sine squared is going to be 1 over sine times 1 over sine.

So what you might notice here is that cosine over sine is cotangent. And this 1 over sine is going to be cosecant. So what I have here is cotangent x cosecant x.

Now 1 over sine squared is cosecant. Cant squared of x. All right, so we've rewritten this expression into something that now looks a little bit more tractable. So those look like things that I know derivatives for, or those look like things that come out to derivatives for certain expressions. All right, so the derivative, the antiderivative of cotangent cosecant, that's going to be negative cosecant of x, right?

Remember, the derivative of cosecant is negative cosecant cotangent. So I tack on a... a negative here to take care of the positive that's showing up.

And then I'll minus cosecant squared. That's going to be minus cotangent of x. The derivative of cotangent is cosecant squared.

And I should correct myself here this is going to be positive cotangent. The derivative of cotangent is negative cosecant squared. And then we'll tack on our constant of integration.

Okay, so we rewrote this, and now we re-expressed this term, cosine over sine squared. We saw that that was cotangent cosecant. This is cosecant squared, and now it looks very obvious how I go to my antiderivative.

So the derivative of cosecant is negative. negative cosecant cotangent, double negative gives me positive, derivative of cotangent is negative cosecant squared, and then the positive here makes sure that we do have the negative showing up in this expression. Alright, so this idea of the antiderivative now lets us think about differential equations. So we're really just going to dip our toes in the water here. Here's how I would define a differential equation.

An equation involving an unknown function and one or more of its derivatives is called a differential equation. So the following here is a differential equation. 4x squared y double prime plus 12 x y prime plus 3y is equal to zero.

All right, so this is a differential equation. It's an equation involving an unknown function. that unknown function is y and one or more of its derivatives.

Now this is difficult to solve, so we're not going to look at things this complex, but we are going to be able to talk about this idea of a differential equation regarding some simple examples. So when we're solving a differential equation, the object is to find a function which satisfies the equation. So if I were to solve this differential equation, I'm looking for the function That's going to balance this equation. So the function y, its first derivative and its second derivative, when that's all found and plugged in, we should get a balancing of 0 equals 0. So this is a little bit different than solving an equation.

When we solve an equation, in general, to this point, our result is a number. Now to solve a differential equation, our result is a function. So a little bit different idea of solving here.

In general, when we solve an equation, we're looking for a number. When we're solving a differential equation, we're looking for a function. So we can talk about a very simple differential equation here.

So the differential equation reads this way, 2y prime minus 3 is equal to the cosine of t. So let's rearrange this a little bit. Let's add that 3 over. So I have 2y prime equals the cosine of t plus 3. And then if we divide by 2, y prime is going to be 1 half cosine of t plus 3 over 2. All right, so to solve this equation, I want to find a function whose derivative is 1 half cosine t plus 3 halves.

All right, so this is very simply going to be 1 half sine of t plus 3 halves t plus c. All right, so the function which satisfies this equation is going to be one half sine of t, so it's just the antiderivative. One half sine of t plus three halves t plus the constant.

All right, so again we're just looking for the antiderivative, so a half sine t, the derivative will be a half cosine t, three halves t, the derivative will be three halves. Now notice here that we have this constant, right? We've got to include the constant. Without further information, I don't know what the constant is, so we'll have to just tack it on and leave it. But if I do have some more information, it's possible to solve for the constant.

So if an initial condition is given, it's possible to solve for the constant. And a differential equation with an initial condition is referred to as an initial value problem. So here we're presented with the initial value problem. So e to the x times y prime minus 1. equals e to the 2x and we're given the condition that y of 0 must be equal to 3. So what we'd like to do is we'd like to solve this for the we'd like to solve this differential equation So we'd like to find the function y Who's which will satisfy this equation and then we're going to find the constant by using this piece of information here So let's isolate this expression So let's divide by e to the x so I have y minus 1 equals e to the 2x over e to the x And so we're going to have y prime is equal to, so e to the 2x divided by e to the x is going to be e to the 2x minus x. That's just going to be e to the x, and we'll have e to the x plus 1. So now we want to find the antiderivative.

We want to find the function y, whose derivative is e to the x plus 1. So you should see here that that's going to be e to the x plus x plus c. Now we're told that y of 0 must be equal to 3. So plugging in 0, we're going to have e to the 0. plus zero plus the constant must equal three. So now we can solve for the constant e to the zero is one so we have one plus c is equal to three so of course c must be equal to two.

What we have now is what we call a particular solution so this right here is called the general solution. Now what we have is the particular solution, e to the x plus x plus 2. And that's called the particular. All right, and we can check that y of 0 is going to be e to the 0 plus 0. That's 1 plus 0 plus 2. So y of 0 does indeed equal 3. All right, so without further information, we're stuck with just leaving the constant.

But in an initial value problem, I have the ability to go ahead and find what my constant should be. All right, so we can now revisit the motion problems that we talked about earlier in the semester. So remember that if we have a position function, the velocity is the first derivative, the acceleration is the second derivative, or the acceleration is the derivative of velocity. So now we can turn that process backwards. So velocity can be recovered by finding the antiderivative of acceleration, and position can be recovered by finding the antiderivative of velocity.

Right, so the velocity is the indefinite integral of acceleration. The position is the indefinite integral of the velocity. All right, so we took that chain forward. We went position, velocity, acceleration. Now we can go acceleration, velocity back to position.

All right, so let's do that. Suppose that we've got an acceleration given to us. So a of t is negative 32 feet per second squared.

So let's find the velocity and position functions given these conditions that the velocity at time zero is two feet per second. And the position at time zero is 10 feet. So in other words, what I have here is I have an initial value problem.

So I'm going to be able to start with a function. That function is the second derivative. I'm going to be able to work my way up towards the function that we began with. That would be the position function.

So we're going to take this process backwards. So the velocity is going to be the indefinite integral of acceleration. So that's going to be the indefinite integral of negative 32 dt.

And so the antiderivative is going to be negative 32t plus the constant. So the initial condition says that the velocity at time 0 must equal 2. So the velocity at time 0, that's going to be negative 32 times 0, plus the constant, that must equal 2. So what we're going to find here is negative 32 times 0 is equal to 0. So 0 plus c equals 2 is our equation. So that's going to force c to be equal to 2. Now we can take the velocity function. The velocity function is now going to be negative 32t plus 2. And we can recover the position.

So the position function is going to be the indefinite integral of velocity. That's going to be negative 32t plus the number that we found, 2. All right, so now our antiderivative, so negative 32t, increase the power by 1, divide by the new power, antiderivative of 2t, excuse me, the antiderivative of 2 is going to be 2t. So what you're going to find here is that this is going to be negative 16t squared.

And just to eliminate any confusion, let's call this c1. They're different constants, but we're recycling the label. So we'll just call this C2. So I get negative 16t squared plus 2t plus C2.

So what am I told? I'm told that S of 0 must be equal to 10. So what we're going to have is that S of 0 Right, plugging in zero here, plugging in zero here, those are going to go away. So we're going to get S of zero is equal to C2, and C2 must be equal to 10. And so there is our constant C2.

All right, so we can now recover the velocity function. The velocity function is going to be negative 32t. plus 2, that's going to be feet per second. And then the position function is going to be negative 16t squared plus 2t plus 10 feet.

Transcript for:Lecture 4.9.2: Antiderivatives

Transcript for:
Lecture 4.9.2: Antiderivatives