KOÇ UNIVERSITY MATH 106 FINAL EXAM JANUARY 6, 2013

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1 KOÇ UNIVERSITY MATH 6 FINAL EXAM JANUARY 6, 23 Durtion of Exm: 2 minutes INSTRUCTIONS: No clcultors nd no cell phones my be used on the test. No questions, nd tlking llowed. You must lwys explin your nswers nd show your work to receive full credit. Use the bck of these pges if necessry. Print (use CAPITAL LETTERS) nd sign your nme. GOOD LUCK! Solutions by Ali Alp Uzmn (Check One): Section (Burk Özbğcı- MWF: 2:3-3:2): Section 2 (Burk Özbğcı- MWF: :3-5:2): Section 3 (Ali Ülger - MWF: :3-:2) : Section (Vrg Klntrov - MWF: :3-2:2) : Section 5 (Ali Göktürk - MWF: 9:3-:2): PROBLEM POINTS SCORE TOTAL

2 Problem (5 pts) Find the lim x Solution x 2 cos( x2) if it exists. Note tht cos( x 2 ), for ny rel number x. Multiplying ech side with x 2, we get x 2 x 2 cos( x 2 ) x 2, for ny rel number x. Since ±x 2 s x, by the Squeeze Theorem we hve lim x x 2 cos( x 2 ). e sin x Problem b (5 pts) Find the lim x x Solution b if it exists. Note tht this limit is of the / indeterminte form. So we pply L Hospitl s Rule to get e sin x lim x x lim e sin x cos x x Problem 2 ( pts) Use implicit differentition to find n eqution of the tngent line to the curve x 2 + xy + y 2 3 t the point (x, y) (2, ). Solution 2 x 2 + xy + y 2 3 d dx (x2 + xy + y 2 ) d dx (3) 2x + (y + xy ) + (2yy ) (x + 2y) + (2x + y)y y x + 2y 2x + y y (x,y)(2,) 5 The slope of the tngent line to y f(x) is /5 t (x, y) (2, ) nd hence the eqution of the tngent line is y 5 x

3 Problem 3 ( pts) Find the derivtive of the function Solution 3 g(x) +2x x t sin t dt Let f(x) : x sin x. Then g(x) +2x x f(t)dt. Let F (x) : x f(t)dt, tht is, F is n ntiderivtive of f, by the Fundmentl Theorem of Clculus. Since g(x) +2x x f(t)dt +2x f(t)dt + x f(t)dt F ( + 2x) F ( 2x), we conclude tht g (x) d F ( + 2x) F ( 2x) dx 2F ( + 2x) + 2F ( 2x) 2f( + 2x) + 2f( 2x) 2( + 2x) sin( + 2x) + ( 2x) sin( 2x) Problem ( pts) If f is continuous on,, use the substitution u x to show tht Solution xf(sin x)dx 2 f(sin x)dx Let u x. Then du dx, x u. We proceed s follows: xf(sin x)dx ( u)f(sin( u))(du) ( u)f(sin( u))du ( u)f(sin u)du (since sin( t) sin t for ll t ) ( x)f(sin x)dx (renming the dummy vrible) f(sin x)dx xf(sin x)dx 2 xf(sin x) xf(sin x) 2 ()f(sin x)dx f(sin x)dx

4 Problem b (5 pts) Evlute the integrl x sin x + cos 2 x dx Solution b sin x Let u x nd dv dx. Then du dx nd v rctn(cos x). By + cos 2 x the integrtion by prts formul we get x sin x + cos 2 x dx x rctn(cos x) + rctn(cos x) dx rctn() 2 Here we use the fct tht rctn(cos x) dx. Problem 5 ( pts) Find the re of the region bounded by the curves y cos x nd y cos 2 x between x nd x. Solution 5 Note tht cos x cos 2 x for x /2 nd cos x cos 2 x for /2 x. Then the desired re should be computed s follows: A cos x cos 2 x dx /2 /2 2 (cos x cos 2 x)dx + (cos x cos 2 x)dx sin x x 2 sin 2x /2 /2 /2 ( cos x + cos 2 x)dx (cos x cos 2 x)dx sin x x 2 sin 2x /2

5 Problem 6 ( pts) Show tht the eqution 3x + 2 cos x + 5 hs exctly one rel root. Solution 6 Let f(x) : 3x + 2 cos x + 5. Note tht f() < nd f() >, which shows tht there exists c, such tht f(c), by the Intermedite Vlue Theorem. So the eqution 3x + 2 cos x + 5 hs t lest one solution. Suppose tht there re two distinct solutions c < c 2 of this eqution. Then by Rolle s Theorem there is number d c, c 2 such tht f (d). This contrdicts to the fct tht f (x) 3 sin x > for ny rel number x. Hence the eqution 3x+2 cos x+5 hs exctly one rel root. Problem 7 (5 pts) Show tht sin x x for every rel number x. Solution 7 Let f(x) : x sin x. Then f() nd f (x) cos x, which implies tht f x sin x for ll nonnegtive x. Problem 7b (5 pts) Show tht 2 sin x x dx Solution 7b From prt (), we know tht x sin x for ll x. Then sin x x nd thus we hve for ll x /2 / sin x x dx /2 / dx

6 Problem 8 (5 pts) Determine whether the integrl is convergent or divergent. Solution 8 Note tht + x dx + x x x 2 for x. Then convergent by the Comprison Theorem, since + x dx is is convergent by p-test. x 2 dx Problem 8b (5 pts) Determine whether the integrl is convergent or divergent. Solution 8b 3 x dx Notice tht x hs discontinuity t x, so we rewrite the given expression s 3 3 x dx x dx + x dx We sy tht the given integrl is convergent if both of these summnds re convergent, otherwise it is divergent by definition. Since the first summnd x dx lim x dx lim x 3 3 is divergent, we conclude tht the integrl ( ) 3 lim 3 () x dx is divergent.

7 Problem 9 (5 pts) If f() g() f() nd f nd g re continuous, show tht Solution 9 f(x)g (x)dx f ()g() + f (x)g(x)dx We will strt from the left side of the eqution nd use the method of integrtion by prts twice to obtin the result. Let u f, dv g dx du f dx, v g. Then fg dx u dv (u v ) (u ()v () u ()v ()) (f()g () f()g ()) v du v du (since f() f() ) g f dx Now let u 2 f, dv 2 g dx du 2 f dx, v 2 g. Then v du v du fg dx g f dx u 2 dv 2 (u 2 ()v 2 () u 2 ()v 2 ()) (f ()g() f ()g()) (f ()g()) gf dx f ()g() + gf dx. (u 2 v 2 ) gf dx gf dx (since g() ) v 2 du 2