Indefinite Integrals as Anti Derivatives

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  • 0:02 Fundamental Theorem
  • 1:55 Antiderivatives
  • 5:50 Indefinite Integrals
  • 9:11 Lesson Summary
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Lesson Transcript
Instructor: Erin Monagan

Erin has been writing and editing for several years and has a master's degree in fiction writing.

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.

Fundamental Theorem

Taking the derivative of these functions gives you back 1
Antiderivative List

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?


The integral from a to b of 1dx equals F(b) - F(a)
Antiderivative Graph Example

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.

The - 47/pi + 47/pi can be canceled out in this problem
Antiderivative Example Problem

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