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MATH 2202 Exam 2 Solutions: Volume Calculation, L'Hopital's Rule, Improper Integrals - Pro, Exams of Calculus

The solutions to exam 2 of math 2202, which includes problems on finding the volume of a solid by revolving a shaded region about the x-axis, applying l'hopital's rule to determine limits, and evaluating improper integrals. Detailed solutions with correct notation and explanations are provided.

Typology: Exams

2010/2011

Uploaded on 06/03/2011

koofers-user-0rt
koofers-user-0rt 🇺🇸

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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 L’Hopital’s 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 L’Hopital’s 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 L’Hopital’s Rule to show that lim x!1+ x 1 x1 = 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 L’Hopital’s Rule, we let y = x 1 x1 and then note that ln (y) = ln  x 1 x1  = 1 x 1 ln (x) = ln (x) x 1 . Since limx!1+ ln (x) = 0 and limx!1+ (x 1) = 0, we can use L’Hopital’s 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
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