MATH 21B Midterm: MATH 21B Harvard 21b Spring 15Solution6
Document Summary
Solutions to the first exam for mathematics 1b. March 4, 2003: (a) change of variables u = x2: The last limit equals in nity, so the integral diverges: (b) change of variables u = sin x (notice sin( /2) = 1) : dx =z 1. This is the integral of an odd function, but we cannot conclude the integral is zero because it is an improper integral. We have to break it down into to pieces: 1 u du = lim t 0 ln|t| ln| 1| = , so the original integral diverges as well. (c) the volume of rotation equals (using vertical slices): After a change of variables u = ln x, the integral becomes: 1 x(ln x)2 dx so the integral converges to . The amount of capsaicin on the ith ring (density in mg/ cm2) (area of ring in cm2) e x2/100 x dx. The integral can be done by simple substitution: let u = x2/100.
