Chapter 7 pre-Test: Antidiffernetiation
The multiple choice were pretty much calculator work. I will just list the answers our group got.
#1) If the definite integral 1S3 (x2-1)dx is approximately using the Midpoint Rule with n=4, the error is:
#2) If -1Sk (3x2-2x)dx=6, then k=?
#3) If 0Sk (sec2x)/(1+tanx)dx=ln2, and k>0, then the value of k is:
#4) Let R be the region in the first quadrant enclosed by the graph of y=xcosx, x=0, y=0, and x=k for 0<(3.14/2). The area of R, in terms of k is:
This is where all of us (as a class) just became clueless. We looked at it for a long time and read it over again, but we were not 100% or even 90% sure what to do or where to even begin. We're not sure what that -2 is all about and if it's supposed to be in that missing spot in the table.
#1) Let f be a differentiable function such that f " is continous and f and f ' have the values given in the table below.
Use the information in the table to evaluate:
a) 0S2 xf ' (x2)dx
b)1S3 xf ' (x)dx
This was the answer our group got (aka manny figured it out :) ).
Well that's it for that class. As a class we are really confused about this section, we feel we're not ready for that test on tuesday. We just need a huge review Mr K., a HUGE review.
Oh the scribe for next class is ANH :)