Prelim 2 Math Please show your reasoning and all your work. This is a 90 minute exam. Calculators are not needed or permitted. Good luck!

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1 April 4, Prelim Math Please show your reasoning and all your work. This is a 9 minute exam. Calculators are not needed or permitted. Good luck! Trigonometric Formulas sin x sin x cos x cos (u + v) cos u cos v sin u sin v cos u cos v (cos (u v) + cos (u + v)) sin sin (u + v) sin u cos v + cos u sin v sin u cos v (sin (u + v) + sin (u v)) cos x cos x sin x cos x sin x cos (u v) cos u cos v + sin u sin v u sin v (cos (u v) cos (u + v)) sin (u v) sin u cos v cos u sin v where h b a n Numerical Integration Trapezoidal Rule T h (y + y + y y n + y n ) Error M(b a) n and M is an upper bound of f (x) on [a, b]. Simpson s Rule S h (y + 4y + y + 4y + y y n + 4y n + y n ) Error M(b a)5 8n 4 where h b a n and M is an upper bound of f (iv) (x) on [a, b]. Problem ) (5 Points) Evaluate the following: (And please don t forget the +C.)

2 a) (5 Points) x sin (5x) dx Solution: Using IBP, u x, du dx, v 5 cos 5x, dv sin 5xdx x sin (5x) dx udv uv vdu x cos 5x cos 5x dx 5 5 x cos 5x + sin 5x x cos 5x + sin 5x + C 5 b) (5 Points) sin 5 x dx Solution: c) (5 Points) sin 5 x dx sin 4 x sin xdx ( cos x) sin xdx (Substitution u cos x) ( u ) ( )du ( u 4 + u )dx u5 5 + u u + C cos5 x 5 π + cos x cos (4x) cos (7x) dx. Solution: Using the trig formula π cos (4x) cos (7x) dx sin (x) π + + d) (5 Points) cos x + C π sin (x) π (cos (x) + cos (x))dx x x dx Solution: Using the trig substitution x sin θ for π θ π.

3 Then dx cos θdθ, x x sin cos θ sin θ cos θ as cos θ is positive on the domain of θ. x x dx sin θ dθ csc θdθ cot θ + C x x + C e) (7 Points) x sec 4 (x ) dx Let u x. Then du x dx. x sec 4 (x ) dx sec 4 (u) du sec (u)( + tan (u)) du Let v tan(u). Then dv sec (u)du. sec (u)( + tan (u)) du ( + v ) dv (v + v ) + c (tan(u) + tan (u)) + c (tan(x ) + tan (x )) + c

4 f) (8 Points) e x dx Let u e x. Then du e x dx. e x dx u(u ) du Next we use partial fractions. Finding a common denominator, u(u ) A u + B u A(u ) + Bu Setting u, we see that A. Setting u, we see that B. u(u ) du 6 u + u du 6 ln u u + C 6 ln e x e x + C Problem ) ( Points) Find the general solution of the differential equation x dy dx xy x e x + x. Dividing by x to get into standard form, dy dx y x xe x +. Thus P (x) x so P (x) dx ln(x), for x >. Thus v(x) e ln(x) x. 4

5 Multiplying by v(x), dy x dx y x e x + x d ( y ) e x + dx x x y x e x + x dx y x e x + ln( x ) + c y x e x + x ln( x ) + cx (In fact it is clear from the above that the above integrating factor of x also works for x < leading to the above general solution being equally valid there.) Problem ) (8 Points) It is desired to approximate 8 dx using the 4x trapezoidal rule. Find a reasonable number of subdivisions to use if we want to guarantee an error of at most 4. (An arithmetic expression will be fine - we realize you don t have a calculator.) Solution: On page of the exam one finds the error formula Error M(b a) n where h b a and M is an upper bound of f (x) on [a, b]. n Here that becomes Error M(7 ) n where M is an upper bound for f (x) 4 ( 4x). on the interval [, 8]. The denominator is smallest in absolute value when x, so M

6 Thus 4 4(7 ) n n 7 4 ( ) 7 n. (The above evaluates to a bit less than.) Problem 4 ( Points) Determine and explain briefly which of the following sequences converge. 4a) {a n } n + n 4 n where a n n4 + n n 4 n. Solution: n + n 4 n4 + n n 4 n (n + n 4)( n 4 + n + n 4 n ) n ( + n 4 ) n ( n 4 + n + n 4 n ) () As n tends to infinity, the first factor of () tends to while the second factor tends to infinity. Thus {a n } n diverges. 4b) {a n } n where a n 6n 5 n + 7. n Solution: 6 n 5 n + 7 ( n 5 ) n ( n 6) ( ) n 5 Note that tends to as 5 ( ) n 7 <, while tends to infinity as >. So the denominator of the RHS tends to infinity and a n tends to. We have that {a n } n converges. Problem 5) ( Points) 5a) What specifically is wrong with the following argument for computing an indefinite integral to aid in evaluating 8 9 x dx? x 4 Let x sec θ for θ appropriately chosen between π and π. Then dx 6

7 sec θ tan θ dθ and x x 4 dx (Then re-express in terms of x....) 4 sec θ sec θ tan θ dθ 4 sec θ 4 sec θ tan θ dθ 4 sec θ tan θ 4 sec θ dθ cos θdθ 4 4 sin θ + C Solution: The problem is in the simplification of 4 sec θ 4 to tan θ which appears on the second line. Since θ is between π and π, tan θ <. So 4 sec θ 4 tan θ. 5b) Give a clearly described example of the following: Two series Σ na n and Σ nb n which are both divergent but for which Σ n(a n + b n ) converges. Solution: One easy example is letting a n ( ) n and b n ( ) n+. Neither series converges because both have even partial sums of while the odd partial sums are of absolute value. But a n + b n always, so the series of sums converges to. Problem 6) ( Points) 6a) The general form of the partial fraction decomposition of A x 7 + B x + 7. Use the Heaviside method to determine the coefficients A and B. Solution: Multiplying both sides of the equation x + (x 7)(x + 7) A x B x + 7 x + x 7 is

8 by (x 7) and taking lim x 7 gives A. Similarly multiplying both sides of the equation by (x + 7) and taking lim x 7 gives 7 + B. 7 6b) Write down the general form of the partial fraction decomposition of x + 7x (x )(x )(x + 4) Please do not attempt to write down the system of linear equations or to solve these equations. Solution: The denominator factors as (x )(x )(x + 4) (x ) (x + )(x + 4) so the general form of the expansion is A x + B (x ) + C x + + Dx + E x F x + G (x + 4). Throughout problems 7 and 8 you may use without proof well known facts about how the rates of growth of logs, polynomials, and exponentials compare. Problem 7) (7 Points) Explain why the improper integral (ln x) dx x 4 converges. Your answer should make reference to techniques we have studied to show convergence of improper integrals. HINT: How would compare to the function you are integrating? x Solution: For x sufficiently large, ln x < x. Therefore, for x >, < ln x < x. x 4 x 4 x 8

9 But dx lim x b ( b dx) lim x b ( ). Thus, by b the direct comparison test, the original improper integral converges. Problem 8) ( Points) Let I n (ln x) n dx for n > a whole number. This improper integral converges for all such n. Show that I n ni n.. (Use the convergence of the improper integral without proving it, though you will still need to consider the definition of this improper integral.) Solution: I n lim a + lim a + (ln x) n dx lim lim (ln x) n dx a ( [x(ln x) n ] a a + a a + ni n a a xd(ln x) n ) ( x n(ln x) n ) dx x n(ln x) n dx n(ln x) n dx Note: Let x y n. As x +, y +. We have lim x + x(ln x)n lim y + ( ln y n ) n y n lim y + ( n)n ( ) n ln y y 9