Question

In this example from Session 54 (end of the first video excerpt), the lecturer switches the order of integration and a negative sign at the front. But then he says that the integrand is a positive area since x < 1. I don't understand this. Isn't it a positive area if x > 1, and then the negative sign makes the result negative? Is x being defined differently in the two expressions?

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-second-fundamental-theorem-areas-volumes/session-54-the-second-fundamental-theorem-and-ln-x/

Answer 1

I have found it useful sometimes when trying to understand things, to use real numbers.
For a start, when Professor Jerison puts the question: Why is L(x) < 0 for x < 1 he is only referring to 'x' as being less than one.
Lets say x = 1/2, making x < 1, we have
then \[\int\limits_{1}^{1/2}\frac{dt}{t} = \ln(1/2) - \ln(1) = \ln(1/2) - 0 = -0.693147181\]
now when we change the order of integration we have
\[-\int\limits_{1/2}^{1}\frac{dt}{t} = \ln(1)-\ln(1/2) = 0 - \ln(1/2) = -(-0.693147181)\]
This equals -0.693147181, the same as above, what you will note is that
when he says \[-\int\limits_{x}^{1}[\frac{dt}{t}]\]
the part in the box is a positive quantity he is referring to the above -(-0.693147181).
So two points really, he is always referring to AS less than one (in this instance) and second point you may find it useful as I do to put real numbers into things to help with your understanding.

Answer 2

The formatting didn't come out very well above so ill redo it here.
The part in the box is a positive quantity above it should read
-(-0.693147181)