1. (25 points) Consider the region bounded by the curves x 2 = y 3 and y = 1. (a) Sketch both curves and shade in the region. x 2 = y 3.

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1 Test Solutions. (5 points) Consider the region bounded by the curves x = y 3 and y =. (a) Sketch both curves and shade in the region. x = y 3 y = (b) Find the area of the region above. Solution: Observing the symmetry of the region with respect to the y-axis, we see that it will be easier if we integrate, with respect to y, the area in the first quadrant and then double it. So we ll be using horizontal rectangles of width x and height dy, so we solve the equation x = y 3 for x and find that x = y 3/, so the area is A = x dy = [y 3/ ] dy = [ y 5/ 5/ ] [ = 5/ ] [ ] = = 5/ 5 5. (c) Find the perimeter of the region above. Solution: The perimeter of the region is the length of the line segment along y = from the point (, ) to the point (, ) (these are the points of intersection of the two curves), which is clearly, plus the length of the curve x = y 3 between these same points. Again, due to the symmetry of x = y 3 with respect to the y-axis, we see that the arc length of this curve from (, ) to (, ) will be twice its arc length from (, ) to (, ). Hence the perimeter is P = + = + 3/ u= ( ) dx + dy = + dy [ ] [ u / 9 du = + 8 u 3/ 9 3/ ] 3/ u= + ( ) 3 y/ dy = + [ (3 ) 3/ ] = using the substitution u = + 9 y, so du = 9 dy and dy = 9 du. + 9 y dy = (5 points) A probability density function is a function f(x) with the following properties I. f(x) for A x B. II. B A f(x) dx =. (a) Show that the function f(x) = e x is a probability density function on the interval from A = to B =. Solution: First, we show that f(x) = e x is greater than or equal to zero for x <. Now > and e x > for every real number x, so f(x) = e x > for all real numbers. In particular, f(x) for x. Hence f satisfies Property I above.

2 Test Solutions Next, we show that f(x) dx = to demonstrate Property II. So let s compute e x dx = lim b e x [ dx = lim e x ] [ b = lim e b + e ()] = + e = as needed. Hence f(x) = e x is a probability density function on the interval [, ). (b) The median of a probability density function f is the number m such that B m f(x) dx = Find the median of the probability density function f(x) = e x from part (a). Solution: We want to find the median m for f(x) = e x, so we solve the equation = m f(x) dx = lim b m e x [ dx = lim e x ] [ b m = lim e b + e m] = e m for m. Hence we have = e m, so ln(/) = ln(e m ) = m and m = ln(/) = ln( ) = ln = ln = is the median of the probability density function f(x) = e x. 3. (5 points) Consider the series ( ) n n= (a) Show that the series converges. Solution: This is an alternating series, so let s use the Alternating Series Test. This first requires that we write the series in the form ( ) n b n, which means that b n =. Now we must show i. b n+ b n : Now b n+ = (n + )! = (n + ), while b n =. Clearly the denominator of b n+ is larger than the denominator of b n, so b n+ b n is true for all n. ii. lim b n = : Here we have lim b n = lim =, since the denominators as n. So, as both conditions of the Alternating Series Test are met, we may conclude that the series ( )n is convergent. (b) Is the series absolutely convergent or merely conditionally convergent? Solution: The series ( )n will be absolutely convergent if and only if its related series of absolute values, ( ) n =, converges. To test this series, let s apply

3 Test Solutions the Ratio Test by considering lim /(n + )! / = lim (n + )! = lim (n + ) = lim n + =. This limit is definitely less than, so by the Ratio Test, we conclude that the series is absolutely convergent. Hence, by the Absolute Convergence Theorem, converges, so our original series ( )n is also absolutely convergent. (c) Find the sum of the series correct to three decimal places. Solution: Before we simply start adding terms at random, let s determine how many we must add by invoking the Alternating Series Estimation Theorem, which states that S s n b n+. So, to find the sum of the series ( )n correct to three decimal places, we must have an error of no more than.5, so we want to find n so that b n+ <.5. Consider the following table: s 6 = ( )! n b n+ = (n + )! so we stop at n = 6 since b 7 <.5. Thus + ( )! + ( ) 3! + ( )3! + ( ) + ( )5 5! 6! = =.639 will be accurate to within ±b 7 = ±.98. However, our error bound, which is at most b 7 =.98, still allows for either.63 or.63 as the correct first three digits. To be safe, let s look at the next few partial sums and see what happens: s 7 = , s 8 = , s 9 =.63887, s = ,... From this check, it is clear that the sum of the series ( )n, correct to three decimal places, is.63.. (5 points) The accompanying figure shows the first five of a sequence of squares. The outermost square, which is the first, has an area of m. Each square (after the first) is obtained by joining the midpoints of the sides of the square before it. We could have used the Ratio Test in Part (a) and taken care of both parts (a) and (b) all at once, but where s the fun in that?

4 Test Solutions (a) Find the sum of the areas of the squares. Solution: In order to compute this sum, we first find a recursive relationship between the area of the n-th square and that of the (n + )-st square. Let A n denote the area of the n-th square. We claim that the area of the (n + )-st square is equal to one-half the area of the n-th square, i.e., that A n+ = A n for n. Suppose the n-th square has sides of length s n, so A n = s n is its area. Then the length of a side in the (n + )-st square, s n+, is the hypotenuse of a right triangle whose legs each have length s n / because the sides of the (n + )-st square are obtained by joining the midpoints of the sides of the n-th square. Therefore the area of the (n + )-st square is ( A n+ = s n+ = sn ) ( sn ) ( ) s + = n = s n = s n = A n by the Pythagorean Theorem, as we claimed. Thus the sum of the areas of the squares is a geometric series with initial term a = A = and ratio r =, so we have A + A + A A n + = A n = n= ( ) n = n= = / = 8 (b) Find the sum of the perimeters of the squares. Solution: In the solution to Part (a) above, we have shown that the lengths of the sides of the (n + )-st square are related to the lengths of the sides of the n-th square by the equation s n+ = A n+ = A n = s n for n. Hence s n+ = s n = s n for n. As the perimeter of the n-th square is P n = s n, we have P n+ = s n+ = s n = P n Hence the sum of the perimeters of the squares is also a geometric series. This time, the ratio is r =, but we haven t yet determined the initial term, which is the perimeter of the first square. For the first square to have area A =, we find s using the equation A = s, in which case s = A = =. Thus P = s = () = 8 is the perimeter of the first square and is the initial term of our geometric series. Therefore the sum of the perimeters of the squares is P n = n= n= ( ) n 8 = 8 =

5 Test Solutions 5. (5 points) A sequence {a n } is defined by a = and for n. a n+ = + a n (a) List the first five terms of the sequence. Solution: The first five terms of the sequence are a = =.35637, a = + = , a 3 = + + = , a = = , a 5 = = (b) Use induction to show that this sequence is bounded above by 3. Solution: When n =, we have a = =.35637, which is certainly less than 3. So the inequality a n 3 is true for n =. Now suppose that a n 3 is true and consider a n+ = + a n + 3 = 5 = Hence, if a n 3, then a n+ 3 as well. So, by induction, a n 3 for all n. (c) Assuming that {a n } is convergent (trust me, it is), find L = lim a n. Solution: Let L = lim a n. As a n+ = + a n and L = lim a n = lim a n+, we have L = lim a n+ = lim + an = + lim a n = + L Thus L = + L, so L L =. Yet L L = (L )(L + ), so either L = or L =. Since all of the terms of the sequence a n are obviously positive, L = doesn t make sense. Therefore, we have lim a n = L = 6. (5 points) Consider the series n= nx n n (n + ). (a) Find the radius of convergence of the series.

6 Test Solutions Solution: To find the radius of convergence of the power series, we ll use the Ratio Test, so consider (n + )x n+ lim n+ [(n + ) + ] nx n = lim (n + )x n+ n (n + ) n+ [n + n + ] nx n n (n + ) = lim n + x n+ n n + n x n n+ n + n + = lim n + ( x ) n + n n + n + = x = x By the Ratio Test, the power series converges (absolutely) when this limit is < and diverges whenever it is >, so we solve the inequality x < = x < = x < = < x < Thus the series converges when < x < and diverges when x < or x >. Hence the radius of convergence is equal to one-half the width of this interval, so the radius of convergence is R =. (b) Find the interval of convergence of the series. Solution: So far we have found that the series converges (absolutely, in fact, thanks to the Ratio Test) for < x < and diverges for x < and x >. So it only remains to check the endpoints, x = ±. When x =, the series is n()n n (n + ) = n n. While this isn t a p-series, it is + similar to one, namely the series n n =, which diverges (p = ). To show n that n n also diverges, we use the Limit Comparison Test and consider + n/(n + ) n lim = lim /n n + = >. Since the limit above is >, both series do the same thing, so both must diverge. Therefore, x = is NOT in the interval of convergence. When x =, the series is n( )n n (n + ) = ( )n n n, which is an alternating series, + so we ll use the Alternating Series Test. i. b n+ b n : In this case, b n = n n + and b n + n+ = n + n +. To show b n+ b n, it is equivalent to show that (n + )(n + ) = n 3 + n + n + is less than or equal to n(n + n + ) = n 3 + n + n, which will be true so long as n + n. ii. lim b n = : Clearly lim n n + =. Hence, by AST, the power series converges when x =. Therefore, the interval of convergence is [, ) = {x : x < }. (c) Where does the series converge absolutely? Where does it converge conditionally? Solution: By the Ratio Test, the series is absolutely convergent for < x <. Since the series converges when x = but not when x =, the series is only conditionally convergent at x =.