Gavin's Calc II Class Clarification: Sep 20

Excellent question, answered (again) belatedly. Let's think about what we mean by int_a^b(1/x)dx. The graph of f(x)=1/x is shown below, with two possible areas shaded. (Ignore the dashed curve for the moment.) Let's think about the area given by the right-most shaded region. This is, ignoring for the moment the issue of the absolute value (because all values of x are positive), int_0.5^1.5(1/x)dx = ln(1.5) - ln(.5)
= ln(1.5/.5) = ln(3) (where we have used in the last line fun rules of logs to intrigue the intrigue-able). Clearly the left-most shaded region must have area equal to the negative of this, namely, -ln(3). How could we find this?

Notice that the area we want is the negative of the area under the graph of 1/(-x), the curve shown as the dashed curve above. This would be

-int_-1.5^-0.5(1/(-x))dx = int_-1.5^-0.5(-1)(1/(-x))dx. Next notice that (d/dx)(ln(-x))=(1/(-x))(-1), which is the integrand in this equation. Thus, int_-1.5^-0.5-(1/(-x))dx = ln(-x)|_-1.5^-0.5
= ln(0.5) - ln(1.5) = ln(0.5/1.5)
= ln(1/3) = ln(3^-1) = -ln(3)

Of course, this is a bit sneaky, because we rewrote the integrand taking advantage of the fact that we knew we had to worry about what happened when x<0. However, what we showed when we took the derivative of ln(-x) is that (for x<0) ln(-x) is in fact an antiderivative of 1/x. (Why is this? We showed that it is an antiderivative of (-1)(1/(-x)), which is the same thing.) Of course, for x<0,

-x = |x|, so
ln(-x) = ln(|x|), while if x>0 we can certainly also say ln(x) = ln(|x|). Therefore we've confirmed that the antiderivative of 1/x is ln(|x|) whether x is positive or negative. Further, because ln(x) isn't defined for x<0, we have to use the absolute value for this case.

---

We can also do this integral through a sneaky substitution: The integral is

int_-1.5^-0.5(1/x)dx. Because we know that we might run into trouble with x<0 (this is 20/20 hindsight), let's make the substitution w=-x to see if that dodges the problem. With this w, dw=-dx, so the integral becomes int_1.5^0.5(1/(-w))(-dw) = int_1.5^0.5(1/w)dw. (be sure you see why the limits on the integral changed). To quickly finish the calulation, we can reverse the limits of integration and get -int_0.5^1.5(1/w)dw = -(ln(1.5) - ln(0.5)) = -ln(3).

All around, a really neat problem!