The answers below are not comprehensive and are meant to indicate the correct way to solve the problem. sin

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1 Math : Practice Final Answer Key Name: The answers below are not comprehensive and are meant to indicate the correct way to solve the problem. Problem : Consider the definite integral I = 5 sin ( ) d. () Using 4 subintervals, find the left hand approimation L 4 to I. Answer: sin() + sin(/2) + sin(/3) + sin(/4) (2) Using 4 subintervals, find the right hand approimation R 4 to I. Answer: sin(/2) + sin(/3) + sin(/4) + sin(/5) (3) Using 4 subintervals, find the midpoint approimation M 4 to I. Answer: sin(2/3) + sin(2/5) + sin(2/7) + sin(2/9) (4) Using 4 subintervals, find the trapezoidal approimation T 4 to I. Answer: (/2) sin() + sin(/2) + sin(/3) + sin(/4) + (/2) sin(/5) (5) Using the error bounds discussed in class, estimate I R 4. Answer: f () = ( / 2 ) cos(/) / 2 Since [, 5] and since / 2 is decreasing f () for all [, 5]. Choose K =. Then: I R 4 2. (6) Using the error bounds discussed in class, estimate I M 4. Answer: f () = (/2 )( / 2 ) sin(/) + (2/ 3 ) cos(/) ( / 4 ) + 2/ 3 3

2 2 when [, 5]. Choose K 2 = 3. Then: I M /(24 6) Problem 2: Consider the initial value problem y = y t y() = Using 4 time steps of size t = 4, estimate y(). Answer: y() = y(/4) y() + f (, y()) (/4) = y(/2) y(/4) + f (/4, y(/4)) (/4) = (/6) y(3/4) y(/2) + f (/2, y(/2)) (/4) = (/6) (5/64) y() y(3/4) + f (3/4, y(3/4)) (/4) = [ (/6) (5/64)] + [( /6) (5/64) (3/4)] Problem 3: Find the arc length of the following curves: () f () = ln() from = / 3 to = 3 Answer: So the arc length is + f () 2 = / d Perform trig substitution with = tan θ to get:

3 3 Rewrite as: Which equals π/3 π/6 π/3 π/6 sec 3 θ tan θ dθ (tan 2 θ + ) sec θ tan θ dθ π/3 π/6 Finding antiderivatives we get: Plug in and obtain: tan θ sec θ + csc θ dθ sec θ + ln csc θ cot θ π/3 π/6 (2) f () = 2 from = to = Answer: (2/ 3) + ln(2 3) (2 + ln(/ 3)) + (f ()) 2 = The arc length is then d. Use trig substitution with = (/2) tan θ to obtain: π/4 (/2) + tan 2 θ sec 2 θ dθ which equals (/2) π/4 sec 3 θ dθ The formula for this would be given to you: sec 3 θ dθ = (/2) sec θ tan θ + (/2) ln sec θ + tan θ + C Use this formula to find the final answer.

4 4 (3) f () = (2/3) 3/2 from = to =. Answer: + (f ()) 2 = + Hence, the arc length is + d = ( + ) 3/2 = 3 4 Problem 4: Find the following volumes () Very Loud (2) Very soft (3) The object obtained by placing the base of equilateral triangles on a circle of radius. Answer: It is easier to consider half of this object (draw a picture!). If an equilateral triangle has base of length s it has area ( 3/4)s 2. The upper half of the unit circle is given by y = 2. So at point [, ] the triangle at that point has area A() = 3( 2 ) 4. Integrate this to find half the volume: ( 2 ) d = 4 2. The total volume is twice this amount. (4) The object obtained by rotating the graph of f () = e for [, 2] around the ais. Answer: The volume is given by the integral: Use integration by parts twice. 2 π 2 e 2 d (5) The object obtained by rotating the graph of f () = e for [, 2] around the y ais.

5 5 Answer: Solve the integral: 2 2π 2 e d (6) The object obtained by rotating the region between f () = and g() = around the ais. Answer: Solve the integral: You will need to epand the integrand. ( ) 2 d Problem 5: Solve the following work problems: () Suppose that a bucket which weighs 3 kg is attached to a 3 meter long rope which weighs a total of 3 kg. If the rope is hanging off a 3 meter tall building, how much work is required to haul the rope (with bucket attached) to the top of the building? Answer: The work required to move the bucket is J. We now consider just the rope. A section of rope at height y i of length y is weighs (3kg) y/3m = y kg. The force required to move this is 98 y N. It moves a distance of 3 y i. The work to move this piece of rope is 98(3 y i ) y. Adding up all the pieces of rope we find that the work is approimately: n 98(3 y i ) y i= To find the actual work take the limit as n and change the result to an integral: (3 y) dy = 49(3 y) 2 = 49(3) So the total amount of work for rope and bucket is 2352 J. (2) The lower half of a sphere of radius 4 m is buried in the ground so that its top is level with the ground. The tank has water in it, so that the water is 3 m deep at

6 6 its deepest point. How much work is required to pump the water out the top of the tank? (Water has a density of g/cm 3 ). Answer: Say that the ground is at height zero and that the center of the sphere is at the origin. A slab of water at height y i of thickness y has volume π(6 y 2 i ) y. Its mass is π(6 y 2 i ) y. It moves a distance of y i and so the total work to move the slab is: 98π(6 y 2 i )( y i ) y Add up, take the limit, and change to an integral as before: Finally, solve the integral. 4 98π(6 y 2 )(4 y) dy Problem 6: Find all solutions to the following differential equations: () y = 3y Answer: y = Ae 3t (2) y = yt Answer: y = Ae t2 /2 (3) y = (y 2 + )(t 2 + ) ( ) Answer: y = tan (/3)t 3 + t + C (4) y = (/y) ln t Answer: (/2)y 2 = t(ln t ) + C Problem 7: Perform the following integrations: () (2) cos() d 4 3 sin3 () d

7 7 (3) (4) (5) 3 ln() d arctan() d sin()e d Answer: For the second integral use a trig identity. For all the others use integration by parts. Problem 8: Find the following antiderivatives. () (2+)( 2 ++) d Answer: Use the method of partial fractions to find (2 + )( ) = 4/3 ( /3) (2/3) We will integrate each piece separately: 2 (4/3) 2 + d = (4/3)(/2) ln + + C For the other fraction use completing the square to write = ( + (/2)) 2 + (3/4) Substitute u = + (/2) to get the integral: Rewrite as: ( /3)(u (/2)) (2/3) u 2 du + (3/4) ( /3) u u 2 + (3/4) du u 2 + (3/4) du To solve the first integral, substitute v = u 2 + (3/4) to get the integral ( /3)(/2) v dv = ( 5/3) ln C To solve the second integral, factor a (3/4) out to get: (4/3) ( (4/3)u) 2 + du Substitute v = (4/3)u to obtain:

8 8 Hence the answer to the problem is: v 2 dv = arctan( + (/2)) + C + (7/3) ln 2 + (5/3) ln arctan( + (/2)) + C (2) (+3) 2 (4+) d Answer: The method of partial fractions allows us to rewrite the integral as: ( + 3) 2 (4) 4 + d This equals: 3 ln ln C ( + 3) Problem 9: Find the following antiderivatives. () d Answer: Trig substitution gives us: Substitute u = sin θ to get: Plugging back in gives: cos 3 θ ( sin 2 sin θ dθ = θ) cos θ dθ sin θ ( u 2 ) du = ln u u 2 u2 + C ln sin θ 2 sin2 θ + C We know = sec θ and so using a triangle and the Pythagorean theorem we get sin θ = and so the answer is 2 ln 2 ( 2 ) C

9 9 (2) (3) d Answer: Let u = The integral becomes du 2 u Which is d u + C = C. Answer: Complete the square and rewrite the integral as: Let u = 2: Let u = 2 sin θ 2 4 ( 2) 2 d (u + 2) 2 4 u 2 du (2 + 2 sin θ) 2 cos θ 4 4 sin 2 θ dθ which equals which equals: [ 2 θ + (2 + 2 sin θ) 2 cos θ 2 cos θ dθ = 2 + sin 2 θ dθ ] (/2)( cos(2θ)) dθ = 3θ 2 sin(2θ) + C This equals: Which equals ( ) 2 3 arcsin 2 sin θ cos θ + C 2 ( ) arcsin ( 2) + C 2 2

10 Problem : Find the following antiderivatives, without using an antiderivative table. () 3 e d Answer: Let u = 3. Then du = (/3) 2/3 d = (/3)u 2 d. The integral is then: 3u 2 e u du and you can solve this using integration by parts twice. (2) arcsin() d Answer: Use integration by parts with u = arcsin() and dv = d. (3) sin(4) cos(3) d Answer: Use the trig identities: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) and sin(a b) = sin(a) cos(b) sin(b) cos(a) to find that: which means: sin(a) cos(b) = (/2)[sin(a + b) + sin(a b)] (4) which you know how to integrate. ln( 2 ) d sin(4) cos(3) = (/2)[sin(7) + sin()] Answer: Write 2 = ( + )( ) so that the integral becomes: ln( + ) + ln( ) d Then use the formula for the antiderivative of ln or solve it using integration by parts. Problem : Write down the formulae for the Taylor polynomials p n () of e, sin, and cos centered at =. Problem 2:

11 () Find the Taylor polynomial p 5 () for f () = 3 centered at =. Answer: + 2 ( ) ( ) ! ( ) ! ( ) ! ( )5 (2) Find a bound on the error f (3/2) p 5 (3/2) using Taylor s theorem. Answer: We have f (6) () = /3 6 for. Choose, therefore, K 6 = 6. We then have: f (3/2) p 5 (3/2) 6(/2)6 6! Problem 3: Find the Taylor polynomial p 3 () for f () = arcsin() centered at =. Answer: p 3 () = + 3 /6 Problem 4: Determine whether or not the following improper integrals converge. If so, find what they converge to. () e d (2) (3) (4) Answer: Converges to. d Answer: Diverges + 2 d Answer: Converges to π. /2 d Answer: Converges to 2.

12 2 (5) (+) d Answer: Use the method of partial fractions to find: [ ] ( + ) d = ln ln + Use properties of natural log to rewrite it as: We then have: [ ln + ] = ln + + ln + lim ln s s = ln lim = s + s + s + s + Hence, the integral does not converge. Notice however that (+) d does converge. Problem 5: Determine if the following integrals converge or diverge. () (2) 3 + d Answer: Write the integral as 3 + d + d. The first integral is 3 + just a regular definite integral. The second integral converges by comparing the integrand to. Hence, the integral converges. 3 ln() 2 + d (3) Answer: Compare the integrand to ln() 2 can find. ( ) sin / d which has an antiderivative that you Answer: Perform the substitution u = / to rewrite the integral as: u 2 sin(u) du This is a definite integral of a continuous function and so the original integral converges.

13 3 (4) (5) (6) (cos()+5) 5 ) 3 d Answer: Notice that (cos() + 5) and so: (cos() + 5) 5 d 3 You can find an antiderivative for the second integrand and discover that by the comparison test the original integral converges. (cos()+5) 5 ) d Answer: Notice that (cos() + 5) By the comparison test (cos() + 5) 5 d and the last integral diverges, so the first one does as well. 2 2 d d d Answer: This integral is improper because the integrand is discontinuous at =. Consider therefore the integral: 2 d When <, we have that d 2. (Check!) Hence: d. The last integral diverges and so our integral does as well. Problem 6: Determine if the volume of the object obtained by rotating the graph of y = /2 cos() for [, ) around the ais is finite. Answer: Use the disc method. Consider the integral: π cos2 () This integral can be rewritten using the identity: d cos 2 () = ( + cos(2)) d 2

14 4 Hence: cos 2 () d = 2 d + sin() d 2 the first integral on the right hand side diverges. The second converges (but that s harder to see). So, the volume is not finite. Problem 7: Determine the limits of the following sequences. You may assume that the limit eists. () { f k+ f k } where f k is the kth Fibonacci number. Answer: Let L be the limit. Then: f k + f k L = lim k f k f k = + lim k f k = + L Hence, L 2 L = and so L = (using the fact that L. (2) { 2k k! } Answer: One method is to notice that by the ratio test k= converges. However an infinite series converges only if the sequence of its terms goes to zero. Hence lim k 2 k k! =. 2 k k! (3) {a k } where a = 3 and a k = 3 + a k. Answer: Let L be the limit. Then L = 3 + L and so L 2 L 3 =. Hence, L = (4) {a k } where a = 2 and a k = 2 + a k Answer: Let L be the limit. Notice that L = 2 + L. Hence L2 2L = and so L = = + 2. Problem 8: What is k= (2/3)k?

15 5 Answer: The answer is 3 by the formula for geometric series. Problem 9: Determine whether the following series converge or diverge. Be sure to eplain your reasoning. () k= k (2) (3) (4) Answer: Diverges by the integral test k= 3 k 2 Answer: Diverges by the integral test. k= k/2k Answer: Converges by the ratio test. k= k k 3 + Answer: Converges by comparison test with k= k 2. Problem 2: Determine whether the following series converge or diverge. () k= f k where f k is the kth Fibonacci number. (2) (3) Answer: Converges by the ratio test (see class notes). k= 3k k! Answer: Converges by the ratio test. k= k! Answer: Apply the ratio test and consider: lim (k)! k (k+)! = k! limk (k+)! = lim k k+ = (4) k= k k

16 6 Answer: Apply the ratio test and consider: lim k k k (k+) k+ = lim k ( ) k k k+ k+ = ( ( ) )( ) k lim k k k+ lim k k+ = e = Problem 2: Eplain why the following series converge. () k= ( )k k Answer: Use the alternating series test. (2) + /4 /9 /6 + /25 + /36 /49 / Answer: Write the series as k= a k and notice that a k = ±. Since k 2 k= a k = k= converges by the integral test, the absolute convergence k 2 theorem tells us that our series also converges. (3) k= ( 3/4)k Answer: This converges by either the alternating series test or by the fact that it is a geometric series with r <. Problem 22: Determine the radius of convergence for the following power series. Ecept for (4), also find the interval of convergence. () k= 2k k Answer: Write as a geometric series: k= (2)k and recall that it will converge when 2 <. I.e. < (/2). The radius of convergence is, therefore, (/2).

17 7 (2) The interval of convergence is ( 2 endpoints are not included. k= kk to 2 ). Be sure you understand why the Answer: By the ratio test, this converges when <. The radius of convergence is, therefore,. The interval of convergence is (, ) since the series k= ( )k k and k= k don t converge. (3) k= k!k Answer: By the ratio test this converges when (k + )! lim = lim (k + ) < k k! k However, when the above limit is and so the radius of convergence is and the interval of convergence is {}. (4) k= f k k where f k is the kth Fibonacci number Answer: Use the ratio test: (5) lim k f k+ f k f Now, as shown in class lim k+ k f k = φ, the golden ratio. Hence (/φ) is the radius of convergence. k= k k(k+) Answer: By the ratio test (and some work) this has radius of convergence equal to. When = we have the series k= k(k+). By the comparison test we have: k= k(k + ) k 2 < So the series converges when =. The absolute convergence test shows that it also converges when =. The interval of convergence is, therefore [, ]. k= Problem 23: Find power series representations for the following functions at the point indicated. State the radius of convergence of the power series.

18 8 () f () = e = Answer: k= k k! (2) f () = 2+ = The radius of convergence is. Answer: It s probably easiest to use Taylor polynomials. Some work will show that the nth Taylor polynomial has the form: The power series is thus: p n () = n k= 2 + = k= ( ) k 2 k 3 k+ ( ) k ( ) k 2 k 3 k+ ( ) k Use the ratio test to find that the series converges when: lim k 2 < 3 That is, when < (3/2). So the radius of convergence is (3/2). (3) f () = ln = Answer: Consider the power series for : k= Replace with to obtain the power series k ( ) k ( ) k k= for (/). Integrate to find the power series k= ( ) k ( )k+ k + for f () = ln. The radius of convergence is.

19 9 (4) f () = arctan() = Answer: Use the power series = k and substitute in 2 to get the power series: + 2 = ( ) k 2k Now integrate to get: arctan() = The radius of convergence is. k= k= ( ) k 2k + 2k+ (5) f () = e 2 = Answer: Make the substitution 2 into the power series for e. The radius of convergence is. (6) f () = sin( 3 ) = Answer: Make the substitution 3 into the power series for sin(). The radius of convergence is. (7) f () = + 3 = Answer: Make the substitution 3 into the power series for. The radius of convergence is. Problem 24: Find power series representatives based at the point indicated for an antiderivative of the function f (). () f () = 2 e 2 at = Answer: In the previous problem you should have found that: e 2 = 2k /k! k=

20 2 Multiply by 2 to find: Now integrate: (2) f () = sin( 3 ) at = Answer: 2 e 2 = 2 e 2 = k= k= 2k+2 /k! k= 2k+3 k!(2k + 3) + C ( ) k 6k+4 (2k + )!(6k + 4) (3) f () = + 3 at = Answer: k= ( ) k 3k+ 3k + Problem 25: The following are a list of statements that you may be asked to prove. The numbers involved may change. () If f () for [a, b] then the nth left hand approimation L n to I = b a f () d satisfies I L n (b a)2 2n (2) If f (3) () 8 then the 2nd MacLaurin approimation p 2 () satisifies f (5) p(5) ! (3) Suppose that r >, prove that k= rk converges if and only if r < and that when it converges it equals /( r).

21 2 Answer: It is a fact that + r r n = rn+ r Hence the nth partial sum of the series is: S n = n k= r k = rn+ r If r > we have lim n r n+ =. If r < we have lim n r n+ =. Hence: { r > lim S n = n r r < If r =, the series is k= =. Hence, the geometric series converges if and only if r <. (4) Prove that the sequence {a k } where a = 2, a k = a k is a bounded, decreasing sequence. (5) Prove that the sequence {a k } where a = 2 and a k = 2 + a k is a bounded, increasing sequence. (6) Prove that if f is continuous on [, ) then if f () d converges so does f () d (7) Use a geometric series to eplain why = (8) Let {a k } be a sequence of numbers all of which are whole numbers between and 9 (including and 9). Use the comparison test and a geometric series to eplain why k= a k/ k converges. This partially eplains why there are numbers with infinite, non-repeating decimal epansions. Answer: Notice that a k 9 for all k. Hence, by the comparison test: a k / k k= 9/ k = 9 (/) k =. k= k=