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# Definite integral of rational function

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)

## Video transcript

- [Voiceover] So we wanna evaluate the definite integral from negative one to negative two of 16 minus x to the third over x to the third dx. Now at first this might seem daunting, I have this rational expression, I have xs in the numerators and xs in the denominators, but we just have to remember, we just have to do some algebraic manipulation, and this is going to seem a lot more attractable. This is the same thing as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third, minus x to the third over x to the third dx. And now what is that going to be equal to? That is going be equal to the definite integral from negative one to negative two of, I could write this first term right over here, let me do this in a different color. I could write, I could write this as 16x to the negative three, x to the negative three, and this second one, we have minus x to the third over x to the third. Well x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx, so dx. And so what is this going to be equal to? Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds. So, let's see. The antiderivative of 16x to the negative three, we're just gonna do the power rule for derivatives in reverse. You can view this as the power rule of integration or the power rule of taking the antiderivative where what you do is you're gonna increase our exponent by one, so you're gonna go from negative three to negative two, and then you're gonna divide by that amount, by negative two. So it's going to be 16 divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight, x to the negative two. And then the antiderivative of negative one, well, that's just negative x, negative, negative x, negative x. And actually you might just know that, and then hey, if I take the derivative of negative x I get negative one, or if you view this as negative x to the zero power, because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power. And then you divide by one and so, I mean, you can view it as that right over there, but either way you get to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference. So we're going to evaluate that at negative two and then subtract from that this evaluated at negative one. And let me do those in two different colors so we can see what's going on. So we're gonna evaluate it at negative two and we're gonna evaluate it at negative one. So let's first evaluate it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative eight, negative eight times x to the negative two. So negative two the the negative two power minus negative two. And from that we're going to subtract it and evaluate it at negative one. So, it's going to be negative eight times negative one to the negative two power minus negative one. All right, so then what is this going to be? So, negative two the negative two. So negative two to the negative two is equal to one over negative two squared, which is equal to 1/4. So this is equal to positive 1/4, but then negative eight times positive 1/4 is going to be equal to negative two. And then we have negative two minus negative two, so that's negative two plus two. And so everything I've just done in this purplish color that is just going to be zero. And then if we look at what's going on in the orange, when you evaluate it at negative one, let's see, negative one to the negative two power, well, that's one over negative one squared. Well, this all is just going to be one. And so we're gonna have negative eight plus one, which is equal to negative seven. So all of this evaluates to negative seven. But remember, we're subtracting negative seven, and so this is going to result, we deserve a little bit of a drum roll, this is going to be equal to positive, positive seven. And obviously you don't have to write that positive out in front, I just wrote that just to emphasize that this is going to be a positive seven.
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