MATH 104 FINAL EXAM SOLUTIONS. x dy dx + y = 2, x > 1, y(e) = 3. Answer: First, re-write in standard form: dy dx + 1

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1 MATH 4 FINAL EXAM SOLUTIONS CLAY SHONKWILER () Solve the initial value problem x dy dx + y =, x >, y(e) =. Answer: First, re-write in standard form: dy dx + x y = x. Then P (x) = x and Q(x) = x. Hence, dx P (x)dx = x. Letting u =, du = xdx, so this integral becomes du = ln u = ln(), u where we can discard the absolute values since x >. Thus, v(x) = er P (x)dx = e ln() =. y = x dx. Letting u =, du = xdx, so this becomes udu = ] [ u + C Now, using the initial condition, = [ () + C ] = + C. = y(e) = ln e + C ln e = + C, so C =. y = +. () Consider the sequence {a n } where If it con- a n = en cos n n. Determine whether the sequence converges or diverges. verges, find the it of the sequence.

2 CLAY SHONKWILER Answer: Since cos n, e n cos n n en n = ( e ) n. Since e <, ( e n ) as n. by the Sandwich Theorem, a n. () Find the interval of convergence of the power series n= x n n. n= Answer: Using the nth root test, n x n = n x The series converges absolutely by the nth root test when x <, which is to say that x < or < x <. Now we need only check the endpoints. When x =, the series becomes ( ) n n = ( ) n = ( ) n, n= n= n=. which diverges. When x =, the series becomes n n =, n= which also diverges. the interval of convergence of the power series is < x <. (4) Evaluate the integral x csc xdx. n= Answer: We integrate by parts, letting u = x and dv = csc xdx. Then du = dx and v = cot xdx. Hence, the integral is equal to cos x x cot x + cot xdx = x cot x + sin x. Now, letting u = sin x, du = cos xdx, so the integral on the right becomes du = ln u + C = ln sin x + C. u x csc xdx = x cot x + ln sin x + C.

3 MATH 4 FINAL EXAM SOLUTIONS (5) Find the length of the curve y = x/ between x = and x =. Answer: Recall the length-of-curve formula ( ) dy L = + dx. dx Now, dy dx = ( x, so dy dx) = x. Hence, ] L = + xdx = ( + x)/ = 4 [8 ] =. (6) Evaluate the integral dx x x + 4. Answer: We complete the square in the denominator first: x x + 4 = (x x + 6) = (x 6) + 5. Letting u = x 6, du = dx, so the above integral becomes du u + 5 = tan u + C = ( ) x 6 tan + C (7) Evaluate the integral x 4 x 6x + 9 dx. Answer: Note that x 6x + 9 = (x ). we solve x 4 by partial fractions by letting x 6x+9 = A x + B and solving (x ) for A and B: x 4 = A(x ) + B = Ax + B A. Hence, A =. Since 4 = B A = B () = B 9, we see that B = 5. [ ] x 4 x 6x + 9 dx = x + 5 (x ) dx (8) Does the series n= = ln x 5 x + C. n + 8n 6n + converge or diverge? Answer: As n gets large, we expect that the constant terms and lower-order terms should be relatively insignificant, so we expect

4 4 CLAY SHONKWILER this series to act like n = 8n comparison between these two series: n+ 8n 6n+ 4n 4n. 8n + n = 8n 6n +. we do a it Using three applications of L Hôpital s Rule, this it is equal to 4n + 4n 4n 6 = 48n n 48 = 48 =. the two series either both converge or both diverge. Since = 4n 4 is a p-series with p = >, it converges so we n conclude that n + 8n 6n + converges as well. (9) Does the series n= n= n! 4n converge or diverge? Answer: Using the Ratio Test: (n + )! n 4(n+) n + 4 =. n! = we see that, by the Ratio Test, the series diverges. () Determine whether the series n= ( ) n n π n converges or diverges. If it converges, find the sum of the series. Answer: Re-write this series as ( ) n ( ) n n =. πn π n= Hence, this is a geometric series with a = and r = π <. Thus, the series converges to π = π+ π = π π +. () Consider a bacteria culture that starts with a single, isolated bacterium. Supposing that the rate of change in the population of the culture is proportional to its size and that there are bacteria in the culture after hour, how many bacteria should we expect in the culture after hours? (Hint: the number you get should be quite simple)

5 MATH 4 FINAL EXAM SOLUTIONS 5 Answer: Since the culture starts with a single bacterium, the population is modeled by Now, A(t) = e kt = e kt. = A() = e k() = e k, so k = ln. after hours, there should be A() = e k() = e ln = e ln = = bacteria in the culture. () Consider the region enclosed by y =, the x-axis and x = e. Find the volume of the solid obtained by rotating this region about the x-axis (Hint: there are two different methods for computing the volume of a solid of revolution, so if the first you try doesn t work, try the other). Answer: It turns out that the disc method is no good, so let s try the shell method instead. Recall that, using the shell method, we need to re-express this function in terms of y, since we re rotating about the x-axis. Now, the inverse of is e x, so y = and x = e y describe the same curve. Now, ln e = and the x-axis describes y =, so V = In this case, r = y and h = e y, so V = πrhdy. πye y dy = π ye y dy. Let u = y and dv = e y dy; then du = dy and v = e y, so this integral becomes [ ] V = π ye y e y dy = π [ye y e y ] = π [(e e) ( )] = π. () Evaluate the integral x dx. x 4 Answer: Use the trigonometric substitution x = sin. Then dx = cos d, so the integral becomes sin cos sin 4 cos d = cos sin 4 cos d = sin 4 d = cot csc d. Let u = cot. Then du = csc, so this integral becomes u du = u + C = cot + C.

6 6 CLAY SHONKWILER x Now, since = sin x, cot = x, and so we conclude that x x 4 dx = ( x ) / x + C. (4) Does the series tan n + n n= converge or diverge? Answer: Compute the integral: tan x dx = + x b b tan x + x dx. Now, if we let u = tan x, then du =, so +x tan ( x + x dx = udu = u tan + C = x ) + C. b [( tan x tan x ) ] b dx = b + x b [( tan b ) ( tan ) ] = b = π 8 π = π. (5) Consider the sequence {a n } where a n = ln n n. Does the sequence converge or diverge? If it converges, what is the it of the sequence? Answer: Using L Hôpital s Rule, ln n n = n n n the sequence converges to. DRL EA, University of Pennsylvania address: shonkwil@math.upenn.edu dx n = n = =. n