What does an anti-derivative have to do with a derivative? Is a definite integral a self-confident version of an indefinite integral? Learn how to define these in this lesson.
The fundamental theorem of calculus says that the integral from a to b of f(x)dx equals the antiderivative of f(x) evaluated at b minus the antiderivative of f(x) evaluated at a, or F(b) – F(a).
Here, I’m calling F(x) an antiderivative of f(x). So dF/dx=f(x); that’s the antiderivative.Let’s say that f(x)=1, and you’re trying to find an antiderivative of f(x). That means that you’re trying to find a function for which, if you differentiate, you’ll get back 1. Let’s try the function x. If I take the derivative of x with respect to x, I get 1.
So, x is an antiderivative of 1. But let’s say I take F(x)= x + 3. If I take the derivative of x + 3, I still get back 1, so x + 3 is also another antiderivative of 1. It’s the same thing if I take F(x) = x – 47/pi. The derivative of x – 47/pi is still just 1. In fact, I could take any function that looks like F(x) = x + C (C being a constant), because the derivative of x + C is the derivative of x plus the derivative of C. That just equals 1 + 0, or 1.
How could this be true?
Let’s look at an example. Let’s say I want to know the integral from a to b of 1dx.
So, this is the area under the curve y=1 between x=a and x=b. According the fundamental theorem, this equals F(b) – F(a). But we just found four different antiderivatives. Let’s try one; let’s say we use F(x)=x.
According to the fundamental theorem, the integral from a to b of 1dx = b – a. So my derivative equals b – a. This makes a lot of sense if I think of it on a graph. I’m just finding the area under a straight line, so that’s really just going to be a rectangle. The height of my rectangle is 1, and the width is b – a. So the area is 1(b – a) or b – a.
This first antiderivative, where we just used x as the antiderivative, gives us exactly what we’d expect.But what if we use the antiderivative x – 47/pi? The left-hand side is still going to be the same; we’re still finding the integral from a to b of 1dx. On the right-hand side, we’re now going to use the function x – 47/pi.
If I evaluate this at x=b, I get b – 47/pi. I’m going to subtract from that my antiderivative at x=a, which is a – 47/pi. If I expand this term, I get b – 47/pi – a + 47/pi. I can actually cancel this – 47/pi + 47/pi, and again, I get b – a. In fact, I can use any antiderivative in the fundamental theorem, but it makes the most sense to use ones without any constants.
I could use, for the case of f(x)=1, any antiderivative that is F(x)= x + C. It makes the most sense, and I don’t have to do any simplification, if I choose my constant to be 0.So what’s the point? Consider your velocity as a function of time, and time goes from t=a to t=b.
If I integrate velocity from a to b, then I get the area under the curve from a to b. That is, the distance I’ve traveled from time a to time b. Does that tell me where I am at any given point in time? No, it just tells me how far I’ve gone in some amount of time. In order to know where I am, I have to know where I started. Just because I’ve gone 30 miles doesn’t mean I’m in Las Vegas, it might mean I’m in Kalamazoo.
It all depends on where I started.
Indefinite Integrals as Antiderivatives
Let’s make this a little more concrete. Let’s define the indefinite integral. The indefinite integral is the integral of the integrand, f(x)dx where x is a variable of integration.
Here I don’t have lower and upper limits. The indefinite integral of f(x)dx equals an antiderivative of x plus some constant of integration, or F(x) + C . That is, if I take the derivative of the right-hand side, I’m going to get back f(x).What does all of this matter? Let’s say my velocity is 30 mph. This is dx/dt, my change in position over time. I can take this function v=dx/dt, multiply both sides by dt and integrate.
I’m not integrating over some set period of time, because maybe I don’t know how long I’m going to be on the road. I just know that I’ll always be going 30 mph. So I’m going to take the indefinite integral of both sides of the equation. Let’s plug in 30 for my velocity, and I get the indefinite integral of 30dt = 30t + C. This is an antiderivative of 30; if I take the derivative of this, I will get back 30. The integral of dx is just x + C, because I can take the derivative of x + C and get 1. My equation becomes 30t + C = x.
Why did I do this? Now I have my position, x, at any given point in time just given my velocity. What about this C? That’s an unknown; I just called it a constant of integration. Remember, if I just know my velocity as a function of time, I need to know where I started to know where I am, how close I am to Las Vegas.
Let’s say that at t=1 hour, I was 60 miles from home. Let’s plug that into my equation, 30t + C = x. x is 60 and t=1, so 60 = 30 + C. If I solve this, I get my constant of integration as being 30. So I can write my equation for a position as a function of time as x = 30t + 30.
Now, given a single point and my velocity, I know exactly where I am at any given point in time. By using an indefinite integral rather than a definite integral, I found my position for any time, rather than where I am at a certain point in time. There’s a slight difference here; this is just a bit broader.
So let’s review. The fundamental theorem of calculus says that the integral from a to b of f(x)dx = F(b) – F(a). Similarly, the indefinite integral of f(x)dx = F(x) + C.