Definite integral involving natural log | AP Calculus AB | Khan Academy

Let's now take the definite integral from 2 to 4 of (6 + x^2) over (x^3) dx. At first, this might seem pretty daunting. I have this rational expression, but if we just rewrite this, it might jump out at you how this could be a little bit simpler.

So, this is equal to the integral from 2 to 4 of (\frac{6}{x^3} + \frac{x^2}{x^3}) dx. I just separated this numerator out; I just divided each of those terms by (x^3). This I could rewrite this as equal to the integral from 2 to 4 of (6x^{-3}), that's that first term there, and (\frac{x^2}{x^3}), well that is going to be (\frac{1}{x}), so plus (\frac{1}{x}) dx.

Now, this is going to be equal to, let's take the anti-derivative of the different parts, and we're going to evaluate that at four and we're going to evaluate that at two. We're going to find the difference between this express the anti-derivative evaluated at four and at two.

Now, what is the anti-derivative of (6x^{-3})? Well, here once again we can just use, we could use the power rule for taking the anti-derivative, or it's the reverse of the derivative power rule. We know that if we're taking the integral of (x^n) dx, the anti-derivative of that is going to be (\frac{x^{n+1}}{n+1}), and if we were just taking an indefinite integral, there would be some plus C.

The reason why we don't put the plus C's here is when you evaluated both bounds of integration, the C would cancel out regardless of what it is. So, we don't really think about the C much when we're taking definite integrals. But let's apply that to (6x^{-3}).

So, it's going to be, we're going to take (x^{-3 + 1}) so it's (x^{-2}), and so we're going to divide by -2 as well. And, of course, we had that six out front from the get-go, so that's the anti-derivative of (6x^{-3}).

And what's the anti-derivative of (\frac{1}{x})? You might be tempted to use the same idea right over here. You might be tempted to say, "All right, well the anti-derivative of (x^1), which is the same thing as (\frac{1}{x}), would be equal to (\frac{x^{-1 + 1}}{-1 + 1})." But what is (1 + 1)? It is zero, so this doesn't fit this property right over here. But lucky for us, there is another property.

We went the other way when we were first taking derivatives of natural log functions. The anti-derivative of (\frac{1}{x}) or (x^1) is equal to sometimes you'll see it written as (\ln(x) + C), and sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value of (\ln(\lvert x \rvert)).

This is useful because this is defined for negative values, not just positive values. The natural log of (x) is only defined for positive values of (x), but when you take the absolute value, now it could be negative or positive values of (x) and it works; the derivative of this is indeed (\frac{1}{x}).

Now, it's not so relevant here because our bounds of integration are both positive, but if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is a natural log of the absolute value of (x). So this we could say is plus the natural log of the absolute value of (x).

It's not a bad habit to do it, and if everything's positive, well, the absolute value of (x) is equal to (x). And so what is this going to be equal to? This is equal to, let's evaluate everything at four.

And actually, before I even evaluate at four, what's (6 \cdot (-2))? That's (-3). So if we evaluated at 4, it's going to be (-\frac{3}{4^2}); (4^{-2}) is (\frac{1}{4^2}) and then plus the natural log of, we could say the absolute value of four, but the absolute value of four is just four, so the natural log of four.

And from that, we're going to subtract everything evaluated at two. So let's do that. If we evaluated at two, it's going to be (-\frac{3}{2^2}); (2^{-2}) is (\frac{1}{2^2}), plus the natural log of the absolute value of positive two is once again just two.

And so what does this give us? So let's try to simplify it a little bit. So this is (-\frac{3}{16}), doing that same color. So this is going to be equal to (-\frac{3}{16}) plus natural log of (4), and then this right over here is (-\frac{3}{4}); (-\frac{3}{4}), do that same color, this right over here is (-\frac{3}{4}), but we have this negative sign out front that we're going to have to distribute.

So the negative of (-\frac{3}{4}) is (+\frac{3}{4}) plus (\frac{3}{4}), and then we're going to subtract, remember we're distributing this negative sign to the natural log; the natural log of (2).

And what is this equal to? Alright, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve the natural log and let's see if we have a common number.

(-\frac{3}{16}) is the same thing as, we multiply the numerator and denominator by 4, that is (-\frac{12}{16}). So, you have (-\frac{12}{16} - \frac{3}{16} + \frac{12}{16}) will give you (-\frac{3}{16}). And then we're going to have the ones that do involve the natural log, (\ln(4) - \ln(2)).

So we could write this as plus the natural log of (4 - 2). You might remember from your logarithm properties that this, over here, is the same thing as the natural log of (\frac{4}{2}). This comes straight out of your logarithm properties, and so this is going to be the natural log of (2).

So we deserve a little bit of a drum roll now. This is all going to be equal to, this is going to be equal to (-\frac{3}{16} + \ln(2)). And we are done.