midterm2-practice-solutions

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

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Unformatted text preview: MATH 31B SECTION 2 SECOND PRACTICE MIDTERM Problem 1. Evaluate the definite integral π2 0 Hence π2 0 0 1 ¢ π 8 ¦ ¥ ¥ ¤ ¥ £ £ sin2 x cos 2x dx π2 1 cos 2x 2 1 4 ¢ ¨¦ § ¤ ¥ ¡ £ sin2 x cos 2x ¡ ¦ § ¤ £ Using the identity cos2 2x 1 2 1 1 cos 2x 2 1 1 4 cos 4x 1 cos 4x dx 4 ¢ ¥ £ ¦ ¥ ¤ £ sin2 x cos 2x 1 1 cos 2x cos 2x 2 cos 4x gives ¦ ¥ ¤ £ Solution. Using the identity sin2 x 1 2 1 cos 2x we get 1 cos 2x 2 1 cos2 2x 2 ¢ ¡ sin2 x cos 2x dx MATH 31B SECTION 2 SECOND PRACTICE MIDTERM 2 Problem 2. Find the limit Solution. We have since the function Now, ex is continuous.  ©  © x 0 x 0 Applying L’Hopital’s rule gives © x Thus  © x 0 ¢ £ £ lim xx e0  © ¦ ¤¥ 0 x 0 1 ¢ £ ¥ £ lim 1x 1 x2  lim   ©  © x 0 x 0 x ¢ £ lim x ln x lim  © x 0 ln x 1x 0  0   £ £ £ lim xx lim eln x x lim ex ln x  © x 0 ¢ lim xx elimx x ln x MATH 31B SECTION 2 SECOND PRACTICE MIDTERM 3 Problem 3. Use a trigonometric substitution to evaluate the definite integral: 1 0 x2 ¥   1 ¡ x2 0 ¢  £ ¡    ¥ £ ¥ 0 1 0 0 π 4 1 sin 2θ 4 π2 π4 ¤ sin2 θ ¦ ¥ £ sin2 θd θ ¡ π2 sin2 θ cos θd θ  π2 π2 1 1 2 ¢ £  £  £ Solution.We let x ¡ sin θ, so that 0 θ π 2 and dx ¢ dx cos θd θ The integral then becomes cos 2θ d θ MATH 31B SECTION 2 SECOND PRACTICE MIDTERM 4 Problem 4. Evaluate the definite integral 1 0 r2 §  0 0 0 0 1  r2 as 0 π4 π4 π4 0 To evaluate π4 0 0 0 0 0 Hence the final answer is ¦ $¦ § ¤ ¥ 2 ln   ¤ 1 2 2 1 ¢¦ ¨#¦ § ¤ § ¤ £ sec3 θd θ 2 ln   ¡ ¡ " Solving for the integral π4 3 0 sec θd θ π4 gives 1 2 2 1 ¢ ¥ !¦ § ¤ § 2 ln 2 1 π4 ¡ § ¥  ¥ 2  sec3 θd θ π4 ¦ ¥ ¤ ¡ ¥ 2  π4   0 0 sec2 θ ¥  ¡  £ sec3 θd θ £ sec θ tan θ 0 1 sec θd θ sec θd θ sec3 θd θ ¡ ¡ π4 π4 π4 tan2 θ sec θd θ ¢ £ £ £ £ £ £ we use integration by parts. Let u Hence ¡ sec θ and dv sec2 θd θ. Then du  sec3 θd θ ¢ ¦ § ¤ ¥ sec3 θd θ ln 2 1 sec θ tan θd θ and v    §   ¥ sec3 θd θ ln sec θ tan θ 0 ¡ ¦ ¥ sec3 θ ¤ sec θ d θ π4 ¥ ¡ £ ¡ ¡ ¡ ¡ ¡ π4 £ £ £ £ tan2 θ 2 sec θd θ sec θ π4 sin2 θ dθ cos3 θ π4 ¢ £   ¡ £ Solution. We let r tan θ, so that 0 θ π 4 and d θ ¢ dr sec 2 θd θ Hence the integral is the same 1 cos2 θ dθ cos3 θ tan θ MATH 31B SECTION 2 SECOND PRACTICE MIDTERM 5 Problem 5. Sketch the graph of the function f x ln sin2 x . Indicate the limits at infinity, vertical asymptotes, maxima, minima and inflection points. Please use the coordinate axes drawn below. ' & £ %¦ ¤ -10 -5 5 10 -2 -4 -6 -8 -10 2 sin x cos x Solution. Let f x ln sin2 x . Then f x 2 cot x and f x csc x. It follows sin2 x that f x 0 for all x and hence the function is concave down and there are no inflection points. 0 at x π nπ, n 0 1 2 , where the function takes on the value of ln 1 Also, f x 2 2 0. Since sin x 1, it follows that f x 0 for all x and hence these points are the maxima. Lastly, since f x ∞ when x nπ, n 0 1 2 and f x ∞ when x nπ, n 012 , f x has vertical asymptotes at these points. £ £ @ ¥£ 20¦ ¤ 1) ) ¥ @ 4¦ ¤ ¢$$¢  ¢ £ 6 7 6 7 ¢¢ $$¢  £ 0¦ ¤ ) £  8¦ ¤ 6 7 6 7 @ £ ¦ ¤ § £ (¦ ¤ £ ¦¤ ¢¢ #$¢  @ A9¦ ¤ )   £ 5¦ ¤ ) 3 4¦ ¤ 1) ) 6 7 6 7 MATH 31B SECTION 2 SECOND PRACTICE MIDTERM 6 Problem 6. Use integration by parts to evaluate the definite integral 1 0 r3 § £ ¥  § 4 r2 0 0 5 4 0 ¦  ¤ ¢ §    ¡  § ¥ ¥   £ £ £ ¥ 5 udu    § £ Substituting u 4 r2 gives 2 325 u 3 4 2 2 55 5 3 3 7 5 16 3 3 5 4 3 ¢ § ¥ £ ¥   § r2 4 r2 2r 4 r2 dr 5 2r    1 § 4 r2 1 1 4 £ §  £  £ Solution. We let u r2 , dv . Hence v  rdr 4 ¢ dr r2 and du 2rdr. The integral is then r2 dr ...
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This note was uploaded on 02/22/2010 for the course MATH Math 31B taught by Professor Houdayer during the Spring '09 term at UCLA.

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