If you throw a ball in the air, the path that it takes is a polynomial. In this lesson, learn how to integrate these fantastic functions by putting together your knowledge of the fundamental theorem of calculus and your ability to differentiate polynomial functions.

## Distance Traveled

Okay, so if you’re traveling along at 30 miles per hour for 1 hour, you know that you will have gone 30 miles by the end of that hour. That’s not so bad. But let’s say that you have your velocity as a function of time, and it’s not a constant velocity. Say you’re zooming around traffic, slowing down, speeding up and slowing down. Is there a way to find out how far you’ve gone just given your velocity as a function of time?Sure! The distance that you’ve traveled is nothing more than the integral of your velocity as a function of time from time *a* to time *b*.

But how do we actually calculate this definite integral if your velocity’s not constant?

## Fundamental Theorem of Calculus

Well, we’re going to use the **fundamental theorem of calculus**, the **FToC**. Now, the fundamental theorem of calculus said that if you’re going to take the definite integral of *f(t)dt* from *t*=*a* to *t*=*b*, that that’s the same as the anti-derivative of *f(t)* evaluated at *b* minus the anti-derivative of *f(t)* evaluated at *a*. Well, we can write this shorthand as the anti-derivative of *t* from *a* to *b*. So this right here, this *F(t)* from *a* to *b*, is the same thing as this. This just says we’re going to evaluate this function at *b*, and we’re going to subtract from that this function evaluated at *a*.

If we have an indefinite integral, I know that I can write the indefinite integral of *f(x)dx* as the anti-derivative of *f(x)* plus some constant, some integration constant. It’s just a single number, although we don’t care what it is right now.

## Anti-Derivatives of Polynomials

So what happens if you’re trying to integrate a polynomial? Let’s say you’re integrating 1 – so this is like *x* to the 0 power.

The integral of *dx* is just equal to *x* plus your constant of integration. This makes sense, because if I take the derivative with respect to *x* of *x* + *C*, I get back 1, so *x* + *C* is the anti-derivative of 1. Said another way, 1 is the derivative of *x* + *C*.

What if you want to integrate *xdx*? Well, that’s just equal to (1/2)*x*^2 plus your constant of integration. Does this make sense? Well let’s check it. Let’s make sure that the derivative of this anti-derivative here equals our original function – in this case, *x*.

So the derivative, *d/dx*, of (1/2)*x*^2 + *C* is equal to (1/2) *d/dx*(*x*^2) because we pulled out the 1/2, and the derivative of *C* is just 0 because it’s the derivative of a constant. So (1/2) *d/dx*(*x*^2). I’m going to use my power rule for differentiating, and I get (1/2)2*x*.

My 2’s cancel, and I get back *x*. So if I take the derivative of (1/2)*x*^2 + *C*, I get *x*. Thus, (1/2)*x*^2 + *C* is an anti-derivative of *x*. So this works, too.

If I integrate *x*^2, what do I get? Well, what’s an anti-derivative of *x*^2? What about (1/3)*x*^3 + *C*? Again, I can take the derivative of (1/3)*x*^3 + *C*. Well, that’s (1/3) *d/dx*(*x*^3), which is (1/3)3*x*^2. Again using my power rule, the 3’s cancel, and I get back x^2. So yes, (1/3)*x*^3 + *C* is an anti-derivative of *x*^2. That is, the derivative of 1/3*x*^3 + *C* is *x*^2.Now you can generalize this to almost any order of polynomial. Let’s say you have *x*^*n*, and you want to integrate that.

As long as *n* does not equal -1, then the integral of *x*^*n* *dx* is (1/(*n*+1)) * (*x*^(*n*+1)) + *C*. You can think of this as being the opposite of the power rule.

## Example #1

Okay, so let’s do an example. Let’s say you’re at a stop sign, and as soon as you start moving forward, you hit ‘go’ on your timer. Your velocity increases *t*^2, where *t* is now your time. How far have you gone between time 0, when you were at that stop sign, and time 2, say 2 minutes in. Well, how far you’ve gone is the integral from *t*=0 to *t*=2 of *t*^2 *dt*.

How do we calculate this?

Well, let’s use the fundamental theorem of calculus. That says that the integral from *a* to *b* of *f(t)dt* is equal to the anti-derivative of *f(t)* evaluated at *b*, minus the anti-derivative of *f(t)* evaluated at *a*, which we can write shorthand as the anti-derivative evaluated at *b* – *a*.

So what is the anti-derivative of a polynomial? Well, the integral of *x*^*n* *dx*=(1/(*n*+1)) * (*x*^(*n*+1)) +*C*. In this case we’re looking at *t*^2, which is *n*=2, so the integral of *x*^2 *dx*=(1/3)*x*^3 + *C*. Okay, so let’s use this as our anti-derivative here. The integral from *t*=0 to 2, *t*^2 *dt*, is equal to this anti-derivative, (1/3)*t*^3 + *C*, evaluated from *a* to *b*. So let’s plug in *a* and *b*. Our integral is ((1/3)*b*^3 +*C*) – ((1/3)*a*^3 + *C*).

The *C*‘s cancel out, and we’re left with 1/3(*b*^3 – *a*^3). Now this brings up an important point. We chose the anti-derivative that included our constant here, but since we’re looking at a definite integral, we actually don’t need that constant. So let’s ignore it from here on out when we’re evaluating definite integrals.

Okay, so now we’ve got our integral as 1/3(*b*^3 – *a*^3). Let’s plug in our values for *a* and *b*. Well *a* is the left-hand side, the bottom of this range here, and *b* is the top of the range. So *a* is *t*=0, and *b* is *t*=2. So if I plug those in, *b*=2 and *a*=0, I get 1/3(2^3 – 0^3), which is just 8 / 3.

## Example #2

Let’s look at another example, one that we might have seen before.

Say you’re looking at the area of a new piece of property on Lake Heaviside. That area is given by the integral from 0 to 10 of (50 – *x*^2 + 5*x*)*dx*. Well, I can use my fundamental theorem and say that this is equal to the anti-derivative evaluated between 0 and 10. Well this doesn’t look like any integral I’ve seen before, so let’s use a property of integrals, and break this up into three different integrals.So, the integral from 0 to 10 of (50 – *x*^2 + 5*x*)*dx* is equal to the integral from 0 to 10 of 50*dx* minus the integral from 0 to 10 of *x*^2 *dx* plus the integral from 0 to 10 of 5*xdx*. I can rewrite this by pulling out these constant values.

Remember that’s another property of integrals. So my integral equals 50 times the integral from 0 to 10 of *dx* minus the integral from 0 to 10 of *x*^2 *dx* plus 5 times the integral from 0 to 10 of *xdx*. Let’s take a look at each one of these in turn. 50 times the integral from 0 to 10 of *dx* is equal to 50 times the anti-derivative of 1 from 0 to 10. I know that the integral of 1 is just equal to *x*, so let’s plug that in, and I get 50 times *x* evaluated from 0 to 10, which is 50(10 – 0), which is equal to 50(10), which is 500. Okay, so that’s my first term.

What about my second term? The integral from 0 to 10 of *x*^2 *dx* is equal to the anti-derivative evaluated from 0 to 10. Well, I know that if I’m integrating *x*^2, I’m going to get something like (1/3)*x*^3, because this is just a polynomial. I can take that, (1/3)*x*^3 and evaluate it from 0 to 10, and I get ((1/3)10^3) – ((1/3)0^3), and that’s equal to 1000 / 3. So now I’ve got my second term. Well let’s look at this last term: 5 times the integral from 0 to 10 of *xdx*. The integral of *xdx* is (1/2)*x*^2 + *C*, so my anti-derivative that I’m going to use is (1/2)*x*^2. And if I evaluate this at 10 and 0, I get 5((1/2)10^2) – ((1/2)0^2), which is 250.

And that’s my third term.

Okay, that’s all of my terms. If I take 500 – (1000/3) + 250, I get about 417, which is what we would have gotten if we had used a Riemann sum calculation across a whole lot of little rectangles.

## Lesson Summary

So let’s review the anti-derivatives of polynomials, that is, how to actually integrate polynomials. The integral of *dx* is equal to *x* plus a constant of integration. The integral of *x* times *dx* is equal to (1/2)*x*^2 plus some constant, *C*. This continues, and for all values of *n* not equal to -1, the integral of *x*^*n* *dx* is equal to (1/(*n*+1)) * (*x*^(*n*+1)) plus that constant of integration.

Using these, you can start to calculate all integrals that have polynomials in them.