Definite integral of rational function | AP Calculus AB | Khan Academy

So we want to evaluate the definite integral from -1 to 2 from 1 to -2 of 16 - x³ over x³ dx.

Now, at first, this might seem daunting. I have this rational expression; I have x's in the numerator and x's in the denominator, but we just have to remember we just have to do some algebraic manipulation, and this is going to seem a lot more tractable.

This is the same thing as a definite integral from -1 to -2 of 16 over x³ - x³ over x³ - x³ over x³ dx.

And now, what is that going to be equal to? That is going to be equal to the definite integral from -1 to -2 of... I could write this first term right over here. Let me do this in a different color. I could write this as 16x^(-3) - x³ over x³ well, x³ is just over x³ over x³.

Well, x³ over x³ is just going to be equal to 1, so this is going to be minus 1 dx.

So what is this going to be equal to? Well, let's take the anti-derivative of each of these parts, and then we're going to have to evaluate them at the different bounds.

So let's see, the anti-derivative of 16x^(-3), we're just going to do the power rule for derivatives in reverse. You could view this as the power rule of integration or the power rule of taking the anti-derivative, where what you do is you're going to increase our exponent by one. So you go from -3 to -2, and then you're going to divide by that amount by -2.

So it's going to be 16 / -2 * x^(-2). All I did is I increased the exponent, and I divided by that amount, so that's the anti-derivative here. And 16 / -2 that is just 8, so we have 8x^(-2).

And then the anti-derivative of -1, well, that's just -x. Negative negative x gives you +x. Actually, you might just know that!

And hey, if I take the derivative of x, I get 1. Or if you viewed this as x^0 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 divide by one.

So, I mean, you could 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 -2 and then subtract from that this evaluated at -1. And let me do those in two different colors just so we can see what's going on.

So we're going to evaluate it at -2, and we're going to evaluate it at -1.

So let's first evaluate it at -2. This is going to be equal to... when you evaluate it at -2, it's going to be -8 * (-2)^(-2) - (-2).

And from that, we're going to subtract what we evaluated at -1. So it's going to be 8 * (-1)^(-2) - (-1).

Alright, so what is this going to be? So -2 to the -2... so -2 to the -2 is equal to 1 over (-2)², which is equal to 1/4.

So this is equal to positive 1/4, but then -8 * positive 1/4 is going to be equal to -2.

And then we have -2 - (-2), so that's -2 + 2, 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 we evaluate at -1, let's see, -1 to the -2 power... well, that's 1 over (-1)².

Well, this is all just going to be 1, and so we're going to have -8 + 1, which is equal to -7.

So all of this evaluates to -7, but remember we're subtracting -7. So this is going to result... we deserve a little bit of a drum roll.

This is going to be equal to positive 7. And obviously, we don't have to write that positive out front. I just wrote that just to emphasize that this is going to be a positive 7.