 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?

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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)2x.

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)3x^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 1/3x^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 ba.

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 + 5x)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 + 5x)dx is equal to the integral from 0 to 10 of 50dx minus the integral from 0 to 10 of x^2 dx plus the integral from 0 to 10 of 5xdx. 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.

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