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## Finding antiderivatives and indefinite integrals: basic rules and notation: definite integrals

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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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