midterm2-solutions

midterm2-solutions - MATH 31B SECTION 2 SECOND MIDTERM...

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Unformatted text preview: MATH 31B SECTION 2 SECOND MIDTERM Please note: Show your work. Correct answers not accompanied by sufficent explanations will receive little or no credit. Please call one of the proctors if you have any questions about a problem. No calculators, computers, PDAs, cell phones, or other devices will be permitted. If you have a question about the grading or believe that a problem has been graded incorrectly, you must bring it to the attention of the professor within 2 weeks of the exam. #1 #2 #3 #4 #5 Total SID: Name: TA: Section(circle): Tuesday Thursday 1 MATH 31B SECTION 2 SECOND MIDTERM 2 Problem 1. Evaluate the definite integral π2 0 π2 0 ¢ £ ¦ ¥¤ 1 1 8 cos 4x dx π 16 ¢ ©¦ ¥¤ £ £ ¨¦ ¥¤ £ ¨¦ ¡ § ¤¦ 1 1 cos 2x 1 4 Hence the integral is the same as ¥¤ £ sin2 x cos2 x cos 2x 1 1 4 cos2 2x ¦ §¤ £ ¦ ¥¤ £ Solution. Using sin x 2 1 2 1 cos 2x and cos2 x 1 2 1 ¢ cos 2x , we get 12 sin 2x 4 1 1 8 cos 4x ¡ sin2 x cos2 x dx MATH 31B SECTION 2 SECOND MIDTERM 3 Problem 2. Find the limit Solution. Applying L’Hopital’s rule, we have Applying L’Hopital’s rule again gives   t 0 t 0 ¢ £ £ £ lim t2 tan t lim 2t sec2 t  t 0 0 1 ¢ lim t2 tan t  ¥  t lim 0 cos t sin t 1 t2   £ £  t 0 £ ¦ lim t ln sin t  t 0 t lim 0 ¢ ©¦ ln sin t 1t 0 lim t ln sin t ¤ ¤ MATH 31B SECTION 2 SECOND MIDTERM 4 Problem 3. Use a trigonometric substitution to evaluate the definite integral: 3 1 12 ¢ £ §§¥ ¡ !! ! "¦ ¥ ¥ ¤ 1 u u 2 12 ¤¡ u2 1 12 ¦ ¥ £ £ £ ¥ 12 1 du ¡ ¥ ¥£ £ Let u cos θ, du u2 1 1 u2 1 du 2 12 ¢ sin3 θ dθ cos2 θ 0 π 3 sin θ 1 cos2 θ dθ cos2 θ 0 sin θd θ, so that the integral is equal to π3 ¦ ¥¤ ¡ ¡ £ £ ¦ § ¤ 0 tan2 θ 1 3 £ sec2 θd θ 0 ¡ π3 tan3 θ π3 tan3 θ dθ sec θ       £ ¡ £ Solution. Let x is the same as tan θ, so that dx sec2 θd θ. Since 0 § ¤ 0 32 ¢ x 30 θ π 3. Thus the integral ¡¦ x3 x2 1   dx MATH 31B SECTION 2 SECOND MIDTERM 5 Solving for I gives that ¢ § &¦ ¥ ¤ £ I 1 x sin ln x 2 x cos ln x C ¢ ¥ § ¥ £ ¥ § £ I x sin ln x x cos ln x sin ln x dx x sin ln x x cos ln x £ ¥ %£ £ £ Integrating by parts again with u cos ln x, v dx, du 1 x sin ln x ¢ ¦ ¤ ¥ £ I x sin ln x cos ln x dx and v x, we get C I £ £ £ # $£ Solution. Let I du 1 cos ln x and v x sin ln x dx. We integrate by parts with u x. Hence ¦ ¤ Problem 4. Evaluate the indefinite integral sin ln x dx. sin ln x and dv dx. Thus £ ...
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