Math Practice Exam 2 - solutions

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1 C Roettger, Fall 205 Math 66 - Practice Exam 2 - solutions State clearly what your result is. Show your work (in particular, integrand and limits of integrals, all substitutions, names of tests used, with details). You can name tests by their three-letter acronym. Calculators allowed, but no notes or books. Give exact results. Pick four questions out of this collection. For the real exam, time will be 50 minutes. Problem Find the partial fraction decomposition of x 3 + 3x 2 + 9x + 2 (x 2)(x + 3)(x 2 + 4) Solution. Step (-): This is already a proper rational function (degree of numerator is 3, lower than the degree of the denominator), so long division is not necessary. Step (0): The denominator is already factorized, again no work necessary here. Step (): Set up trial solution for linear factors (luckily, there are no repeated factors) A x 2 + B x + 3 Step (2): Set up trial solution for quadratic factors Cx + D x Step (3): Add all trial solutions, set equal to original function, multiply by denominator, and simplify x 3 +3x 2 +9x+2 = A(x+3)(x 2 +4)+B(x 2)(x 2 +4)+(Cx+D)(x 2)(x+3). Step (4): Solve for the unknowns A, B, C, D. We could compare coefficients of like powers, see below. But it s a little shorter to substitute values for x: substitute x = 2 to get 80 = 40A, so A = 2. Substitute x = 3 to get

2 3 = 3B, so B =. You can even substitute any other convenient value like x = 0 to get 2 = 2A 8B 6D = 32 6D so D = 5. Then do this one more time to find C = 0. Alternative The standard method of comparing coefficients of like powers works like this. The equations for the coefficients on both sides are = A + B + C () 3 = 3A 2B + C + D (2) 9 = 4A + 4B 6C + D (3) 2 = 2A 8B 6D (4) Then we could eg eliminate D by doing (2) (3), (4)+ six times equation (2), getting = A + B + C (5) 4 = A 6B + 7C (6) 80 = 30A 20B + 6C (7) Work with this and add (5) to (6), -30 times (5) to (7) to eliminate A (just one possibility there are many others) 5 = 5B + 8C (8) 50 = 50B 24C (9) Finally, add three times (8) to (9) to eliminate C and you got B =, then C = 0. The dominoes begin to fall! go back and substitute the values for B, C into (5), getting A = 2 and finally D = 5 from (2). Whichever way you do this, the final answer is x x x + 2 (x 2)(x + 3)(x 2 + 4) = 2 x 2 x x and this would eg be useful for computing x 3 + 3x 2 + 9x + 2 (x 2)(x + 3)(x 2 + 4) dx.

3 Problem 2 Find the following integrals (possibly ±, possibly not defined). Solution. a) b) 2 = 2 2 x (x 2 + 4) 8 s ln s 3/2 dx du = u3/2 ds = lim a 0 [ u /2 ln 2 a ] 8 = du =. u 2 4. It is also OK to say that b) is divergent. c) is divergent - the integrand approaches ± at v = π/2. Doing limits from left and right to π/2, as in lim b π/2 b 0 tan v dv = lim b π/2 [ln cos v ]b 0 = and π lim tan v dv = lim a π/2 + a a π/2 +[ln cos v ]π a = we find out that the contributions cancel each other out, but this integral is still not defined. Problem 3 Find the limit of the following sequences or explain why they diverge. a) 3 2n + 4n 3 lim n 2 + 4n 3n 3 b) lim 2 k ln k k Solution. a) converges to 4/3 (ratio of leading coefficients, divide numerator and denominator by highest power n 3 ). b) converges to 0. This can be seen by applying l Hopital s Rule to the function g(x) = 2 x ln x = ln x 2 x

4 with the numerator having derivative /x and the denominator ln(2)2 x, you get /x lim g(x) = lim x x ln(2)2 = 0. x Problem 4 Determine whether the following series are convergent or divergent. Name any test you used and give the details needed to reproduce your argument. a) b) c) d) e) n=0 k= n= m= 3 2n + 4n n 3n 3 2 k + k 4 k n n sin n m2 m ( cos(r)e r cos(r + )e r ) r=0 In part d), determine how many terms have to be computed to approximate the sum of the series so that the difference R N = m=n+ m2 m is less than OK, that gives away that d) indeed converges. But why? Solution. a) The terms of the series do not converge to zero (see the previous problem, part a)). So this series does not converge by the n-th Term Test. b) This can be split into two series, the first one is geometric with ratio

5 r = /2. integral For the second one, use the ICT. That means considering the x4 x dx which we can compare to the convergent integral ( ) x 4 3 x 4 x dx = dx. 3 using x 3 x, or alternatively to [ x4 x4 x x dx = ln 4 ] + ln 4 4 x dx. using x x. A third alternative would have been to use the LCT on that integral with eg 3 x giving a limit of 0 and still a converging integral. Therefore the answer for the entire series in the problem is also Yes, it converges. c) Again, we need the ICT, so compare to x x 2 + sin x dx. Then it s best to use the LCT with g(x) = /x. Clearly, lim x x x 2 +sin x g(x) x 2 = lim x x 2 + sin x =. So the given integral diverges, just like /x dx. Therefore, the original series diverges as well. d) Compare to x2 x dx = ] [ x2 x ln 2 + ln 2 2 x dx = [ x2 x ln 2 2 x ln 2 (2) which converges. So the series converges by the Integral Comparison Test. For estimating R N, you use the same integral, as in ] m2 m [ x2 x ln 2 2 x ln 2 (2) m=n+ = N2 N ln 2 N + 2 N ln N ln 2 2 ]. (0)

6 (see the argument below for why that is true for all N 2). expression is less than 0.00 for The last 0.5N ln 2 ln(0.00 ln 2 2). and that leads to N 22.05, so N = 23 terms are enough. To nail down why inequality (0) is true, consider Since the derivative f(x) = 2 0.5x (x ln 2 + ). f (x) = 0.5 ln(2)2 0.5x ln 2 0 for all x 2, f(x) is nondecreasing on [2, ). Since also f(2) 0, we know f(x) 0 for all x 2. Therefore also 2 x f(x) = 2 0.5x 2 x (x ln 2 + ) 0. and the desired inequality (0) follows. e) This is a telescoping series. The partial sums are S N = N ( cos(r)e r cos(r + )e r ) = cos(n + )e N r=0 and they clearly have limit, so that series converges. Problem 5 Express the number a = as a common fraction (use a geometric series). Solution. This can be rewritten as a = 0.23( ) = ( /000) = Problem 6 We can fit a circle of radius /2 inside a unit square. Inside that circle, we can fit a smaller square, into which we fit a smaller circle. Continue this process of inscribing smaller and smaller squares and circles indefinitely.

7 a) Find the radius of the second inscribed circle. b) What is the total perimeter of all circles in this figure? Solution. a) The first circle has radius /2, so circumference π. The second square has side length 2/2. The second circle has half of that for radius, ie 2/4, so circumference π 2/2. b) In every subsequent step, the side length of the square, the radius of the circle and the circumference of the circle shrink by that factor q = 2/2. If we start counting the steps at k = 0, then in step k, we have a circle with circumference πq k. The total length of all circles is πq k = π q = π 2 = π 2( 2 + ) = π(2 + 2) 2 k=0 (any form of the exact answer is OK). Not part of the problem - the wording of the problem does not determine the exact orientation of the successive squares, only their size and the position of the center. If you decide eg to rotate every square by a constant angle with respect to the preceding one, you get interesting figures like the one below. The answer to the problem is not affected. Problem 7 Approximate the integral S = 4 0 x + dx using Simpson s rule with n = 8 subintervals. Then use the formula E S M 4(b a) 5 80n 4 to find how many subintervals would be needed to guarantee an approximation error of less than Solution. Simpson s Rule with 8 subintervals gives the approximation ln(5) S = Then we compute the fourth derivative of /(x + ), which is f (4) (x) = 24 (x + ) 5.

8 Clearly, it has its maximum value on [0, 4] at x = 0 because it is a decreasing function. So M 4 = 24. Then M 4 (b a) 5 80n 4 = n can be solved for n to give n We need to round this up to the next higher even number, so n = 36 subintervals or more will guarantee the desired accuracy.

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