Download MATH 2202 Exam 2 Solutions: Volume Calculation, L'Hopital's Rule, Improper Integrals - Pro and more Exams Calculus in PDF only on Docsity! MATH 2202 Exam 2 (Version 1) Solutions October 15, 2010 S. F. Ellermeyer Name Instructions. This exam contains 5 problems. In giving solutions to these problems, you must be detailed, use correct notation, and present the solution in an organized way. 1. Find the volume of the solid obtained by revolving the shaded region pictured below about the x axis. All details of your work must be shown including setting up the correct integral that gives the volume evaluation of the integral. You may use either the disk method or the shell method. Solution: By using the disk method, we see that the volume is V = Z 3 0 p 9 x2 2 dx = Z 3 0 9 x2 dx = 9x 1 3 x3 x=3 x=0 = [(27 9) 0] = 18. 2. Use LHopitals Rule to show that lim x! 3 3x+ sin x+ 3 = 3. You must be detailed (and organized) in your explanation in order to get full credit for this. Make sure to use correct notation and to write some explanatory sentences. 1 Solution: Since lim x! 3 (3x+ ) = 0 and lim x! 3 sin x+ 3 = sin (0) = 0, we see that LHopitals Rule is applicable. (We have a 0=0 indeterminate form.) Thus we have lim x! 3 3x+ sin x+ 3 = lim x! 3 3 cos x+ 3 = 3 cos (0) = 3. 3. Use LHopitals Rule to show that lim x!1+ x 1 x 1 = e. You must be detailed (and organized) in your explanation in order to get full credit for this. Make sure to use correct notation and to write some explanatory sentences. Solution: Since lim x!1+ x = 1 and lim x!1+ 1 x 1 =1, we see that we have a 11 indeterminate form. To be able to use LHopitals Rule, we let y = x 1 x 1 and then note that ln (y) = ln x 1 x 1 = 1 x 1 ln (x) = ln (x) x 1 . Since limx!1+ ln (x) = 0 and limx!1+ (x 1) = 0, we can use LHopitals Rule to see that lim x!1+ ln (x) x 1 = limx!1+ 1 x 1 = 1. Thus limx!1+ ln (y) = 1 which means that limx!1+ y = e. 4. Consider the integral Z 4 0 1p 4 x dx. (a) Explain why this is an improper integral. (Write in complete sentences.) (b) Show, in detail, how to evaluate this integral. (Your calculations must involve a limit of something. In other words, you must use the correct de
nition of the improper integral.) 2