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foxtrotgeomseries
Course: M 166, Fall 2009School: Iowa State
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166 MATH IN FOXTROT 5/22/05 Paige says: "RELAX. IT WAS How much of a discount did she receive? We can use Calculus II to find out! 1 3k k =1 OFF." continued on next page... MATH 166 IN FOXTROT 1 3k k =1 is a geometric series. But it is not in the standard form of ar n =0 n (common mistake). To put it in this standard form, we must subtract 1 from the lower bound (k = 1). This...
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166 MATH IN FOXTROT
5/22/05
Paige says: "RELAX. IT WAS How much of a discount did she receive? We can use Calculus II to find out!
1 3k k =1
OFF."
continued on next page...
MATH 166 IN FOXTROT
1 3k k =1
is a geometric series. But it is not in the standard form of
ar
n =0
n
(common
mistake). To put it in this standard form, we must subtract 1 from the lower bound (k = 1). This means we have to add 1 to the `k' in the argument (we do the opposite of what we do to the bound). gives This us:
1 k +1 k =1-13
3
k =0
1
k +1
This can be rewritten as:
1 3 k =0
k +1
Using the properties of exponents:
1 1 3 3 k =0
Simplifying:
k
1
1 1 3 3 k =0
k
The `common ratio' in this geometric series is 1 , which is less than 1. Therefore this is 3 a ...
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Iowa State - M - 166
ANCTIL, MICHAEL MATH 166 FINAL REVIEW WORKSHEET Topics to Know: I. Applications of the Integral a) Area of a plane region b) Volume of a solid of revolution c) Length of a plane curve d) Area of a surface of revolution e) Work done by a variable forc
Iowa State - M - 166
Objectives for Math 1661Applications of the IntegralSet up and evaluate integrals to calculate 1. Area of a plane region 2. Volume of a solid of revolution 3. Length of a plane curve 4. Area of a surface of revolution 5. Work done by a variable
Iowa State - M - 166
MIKE ANCTIL TH MATH 166 REVIEW, VARBERG 9 ED., CHAPTERS 5, 75.1: The Area of a Plane Region Find the area bounded by the function y = x and y = x - 2 .2Step 1: GraphStep 2: Find the intercepts This is done by setting the equations equal to eac
Iowa State - M - 166
Correction to the solution to #10: The last two rows of the posted solution should be as follows: 27 27 36 3 9 3 27 3 = + = 9 2 6 2 2 21
Iowa State - M - 166
MIKE ANCTIL TH MATH 166 REVIEW, VARBERG 9 ED., CHAPTERS 5, 75.1: The Area of a Plane Region Find the area bounded by the function y = x and y = x - 2 .25.2: Volumes of Solids: Slabs, Disks, and Washers Find the volume of the solid whose base is
Iowa State - M - 166
MIKE ANCTIL 12.4 PRACTICE PROBLEMSChoose the Best Graph:(x + 3)2 + ( y + 2)24 16A) B)=1A)9 x 2 - 16 y 2 + 54 x + 64 y - 127 = 0B)C)D)C)D)MIKE ANCTIL 12.4 PRACTICE PROBLEMSDetermine the Correct Equation: Horizontal Ellipse Cent
Iowa State - M - 166
12.3: #13 Find points of tangency. Ellipse9x 2 + 4 y 2 = 1y-intercept: (0, 6)x y2 + =1 4 92(A)Equation of a tangent line for an ellipse (given in the book) at (x0, y0):xx0 yy 0 + = 1 (B) 4 9Plugging the y-intercept (0, 6) into equation
Iowa State - M - 166
MIKE ANCTIL MATH 166 PRACTICE PROBLEMThis problem is similar to problem #22 in the Integration by Substitution Section of the Techniques of Integration chapter of the book.x4x dx + 16Initially this problem appears to not match any direct fo
Iowa State - M - 166
Iowa State University Math 166 Instructor:Midterm Exam October 7, 2004 TA:Name: Section:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the reaso
Iowa State - M - 166
MIKE ANCTIL APRIL 28 FINAL REVIEW PROBLEMS1) Find the Sum of the Geometric Series:1 1 1 1 1 + 2 + 3 + 4 + . + n 7 7 7 7 7You should also be able to identify this series if given in this form:1 1 1 1 + + + + . 7 49 343 2401From this we can ea
Iowa State - M - 166
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1 Exercises 6.2 1. R is the region bounded by the graphs of y = x2+1, y = 0, x = 0, x = 2. Let S be the solid of revolution generated by revolving R about the x-axis. Find the volume of S. y R 2 y = x2 + 1 or x = y 1
Iowa State - MATH - 166
Math 166 Final Exam Spring 2006Name & Section: Instructor:Answer each question completely. Show all work and all steps, and state reasons for conclusions. No credit is allowed for mere answers with no work shown. Give numerical answers exactly, n
Iowa State - MATH - 166
1 Here let f(x) be a real-valued function with continuous derivatives of all orders on an open interval I of the x-axis. Let x0 be a real number in I. The Taylor series for f(x) about x0 is the power series (1)k =0f ( k ) ( x0 ) k ( x - x0 )
Iowa State - MATH - 166
1 Return now to the above power series (1)an =0n ( x x0 ) n .Suppose now that one of the following two cases occurs. (i) The power series (1) converges for all real numbers x. In this case, let R = . (ii) There is a positive real number R
Iowa State - MATH - 166
1Some ObservationsNow let be the curve which is the graph of the quadratic function y = f(x) = ax2 + bx + c where a0. Note as before that 2 b ax2 + bx + c = a x + x + c a 2 2 b b b a x2 + x + + c a = a 2a 2a
Iowa State - MATH - 166
Exercises 6.1 26. ( (1,-1) y (8,4) R (3,-1)xIntersection of x = y 2 - 2 y and x = y + 4 : y 2 - 2 y = y + 4. y 2 - 3 y - 4 = 0. ( y + 1) ( y - 4) = 0. y = -1 y = 4 or x=3 x = 8. Area of R = 4 -1( ( y
Iowa State - MATH - 166
Exercises 8.3 3.tdt 3t + 4Let u = 3t + 4. u 2 = 3t + 4. 2udu = 3dt dt = t= 1 2 ( u - 4). 3 tdt 2 udu. 31 2 ( u - 4) 2 udu 3 3 3t + 4 = u 3 2 ( u - 4u ) = du 9 u 2 = ( u 2 - 4 )du 9 2 1 = u 3 - 4u + C 9 3 3 1 2 8 = ( 3t + 4 ) 2 -
Iowa State - MATH - 166
1 Departmental Final Solution Spring 2006 1. ( x + 1)213dx= lim bb 21 dx ( x + 1) 3b 1 -2 = lim - ( x + 1) b 2 2 1 -2 = ( 2 + 1) 2 1 = . 18 2. k + 5 - k + 6 k =1n kk +1 n n n +1 1 2 2 3 3 4 n - 2 n -1
Iowa State - MATH - 166
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Iowa State - MATH - 166
1 Exercises 10.1 11. e 2n an = 2 . n + 3n - 1 e2x Let f(x) = 2 . x + 3x - 1 d 2x e dx lim f ( x) = lim x x d x 2 + 3x - 1 dx 2x 2e = lim x 2 x + 3 d 2e 2 x = lim dx x d ( 2 x + 3) dx 4e 2 x = lim x 2 = . lim a n = .( )()()x 18.1
Iowa State - MATH - 166
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Iowa State - MATH - 166
Exercise 10.36.k =100 ( k + 2)b32Let f(x) =3 . ( x + 2) 2 dx = 3 2 1 x + 2 100 b ( x + 2)10031 1 1 + 3 as b . b+2 100 + 2 34 Converges. = 3 1000k 2 10. 3 k =1 1 + k 1000x 2 Let f(x) = . 1+ x3 b 1000 3 1 f ( x)dx = 3 l
Iowa State - MATH - 166
1 Exercises 11.2 11.1.51cos( x) dx.Let f(x) = cos( x) . Let n be a positive integer. 1.5-1 .5 Let h = = . n n Let x 0 = 1, x1 = 1 + 1 h, x 2 = 1 + 2 h, x3 = 1 + 3 h, x 4 = 1 + 4 h, , x n -1 = 1 + (n - 1) h, x n = 1 + n h = 1.5. Let Tn
Iowa State - MATH - 166
( 2n) 6 2 4 3 3 3 3 dx x 13 dx - 1 1 - x + x - x + + ( - 1) n x cos 0 0 2! 4! 6! ( 2n)! 1 ( 2n ) 6 2 4 3 13 x 3 x 3 x 3 n x -dx = cos x - 1 - + - + + ( - 1) 0 2! 4! 6! (2n)! 1 (2n) 6 2 4 3 3 3 3 dx x 13 - 1
Iowa State - MATH - 166
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Iowa State - MATH - 166
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Iowa State - MATH - 166
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Iowa State - MATH - 166
Exercises 10.2 5. k + 2.k =1k -51 k -5 k -5 k = k+2 k+2 1 k 5 1- k 1 0 as k . = 2 1+ k Diverges. 7. k - k - 1 . k =2 n1 11 1 k - k - 1 . k =21 1 1 1 1 1 1 1 1 1 = - + - + - + - + - 2 1 3 2 4 3 n
Iowa State - MATH - 166
Exercises 6.4 33. : x = a ( t sin(t ) ) , y = a (1 cos(t ) ) for 0 t 2. is one arch of a cycloid. (See Problem 18 and Figure 20.) Revolve about the x - axis. Area of the surface of revolution generated = 2 2 0 2y(t) [ x (t)] 2 + [ y (
Iowa State - MATH - 166
Mathematics 166 Practice Test 3 1. Determine whether or not each of the following infinite series converges. Give a reason for your conclusion in each case. 1 (a) 4 n =1 n 3 1 (b) 4 n =1 n 3 + 1 n (c) 4 n =1 n 3 + 1(d) (e) (f)e (n =1
Iowa State - MATH - 166
1 Solution Final Exam Spring 2006 1.0e -4 x dx= lim e -4 x dxb 0b 1 = lim - e - 4 x b 4 0 1 1 = lim - e - 4b + . b 4 4 1 = . 4 1 1 1 2 3 = 2 1 = 2 2 = 3. 2. n =0 1- 3 3 3. (a) nbnn =11321 x3 21
Iowa State - MATH - 166
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