MATH 3B Midterm: Math 3B Exam 5
Document Summary
Practice for midterm ii (solutions: find z e. 0 ln x dx. (hint: l"hospital"s rule might be useful. ) Before we start we note that ln x has a vertical asymptote at x = 0 and so this is an improper integral. First we nd the inde nite integral of ln x, using integration by parts we have. 1 x dx = x ln x x + c. | {z } u=ln x x dx du= 1 dv=dx v=x. 0 ln x dx = lim c ln x dx(cid:19) c 0+(cid:18)z e c 0+(cid:18)(x ln x x)(cid:12)(cid:12)(cid:12)(cid:12) c 0+(cid:18)(e ln e e) (c ln c c)(cid:19) c(cid:19) e. = lim c 0+ c ln c ln c. The last part coming from l"hospital"s rule (it turned out that it was useful): (a) find z dx e2x 1 by substituting u = ex.
