Area Under 3 Curves PtII

1. Feb 24, 2006

h6872

hi! Thanks for the advice, I managed to find the intersection points, but I seem to have run into a problem.
I set up the integral as follows:

(4-x^2) (for 1 to 2) + (4 - 1/x) (for 1/4 to 1)
this worked out to
((8-8/3) - (4 - 1/3)) + (4 - (1 - ln1/4))
(16/3 - 11/3) + (3 + ln1/4)
5/3 + 3 + ln 1/4
= 14/3 + ln1/4

but the answer at the back says that the correct response is
14/3 - ln4... What did I do wrong?
Thanks so much,
heather

2. Feb 24, 2006

assyrian_77

The (4-x^2) integral gives you the total area in that interval. However, the 1/x curve "intersects" the area. Therefore, you must subtract that part from the first integral. If you plot the curves, you will see what I am talking about.

3. Feb 24, 2006

assyrian_77

By the way, there is no need to start a second thread about the same problem. You could have just as well continued in your first thread.