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MA224_StudyGuide_Exam1

Course: MA 224, Spring 2012
School: Purdue
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Word Count: 1151

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I Exam Study Guide 1.Antiderivative If F (x) = f (x), in MA223, we call f (x) the derivative of F (x); in MA224, we also call F (x) an antiderivative of f (x). Just notice that all the antiderivatives of f (x) are in the form F (x) + C , where C is an arbitrary constant, if F (x) = f (x). 2.The Indenite Integral If F (x) = f (x) as above, we write f (x)dx = F (x) + C, and call both sides of the above equality the...

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I Exam Study Guide 1.Antiderivative If F (x) = f (x), in MA223, we call f (x) the derivative of F (x); in MA224, we also call F (x) an antiderivative of f (x). Just notice that all the antiderivatives of f (x) are in the form F (x) + C , where C is an arbitrary constant, if F (x) = f (x). 2.The Indenite Integral If F (x) = f (x) as above, we write f (x)dx = F (x) + C, and call both sides of the above equality the indenite integral of f (x). 3.Rules for Integrating Common Functions The constant rule: k dx = kx + C for a constant k xn dx = The power rule: The logarithmic rule: The exponential rule: xn+1 + C if n = 1 n+1 1 dx = ln |x| + C for all x = 0 x 1 ekx dx = ekx + C for constant k = 0 k 4.Algebraic Rules for Indenite Integration The constant multiple rule: The sum rule: k f (x)dx = k (f (x) + g (x))dx = The dierence rule: f (x)dx for a constant k f (x)dx + (f (x) g (x))dx = g (x)dx f (x)dx g (x)dx Exercise Page 381: 12, 21 5.Initial Value Problems If we are given f (x) and one point (a, f (a)), we get f (x) by forming the indenite integral of f (x) where C can be found using (a, f (a)). Exercise Page 382: 33, 34, 38, 48; Page 395: 52 6. Using substitution to integrate f (x)dx Most integrals are impossible to obtained directly. We can use substitution by letting u to be some part of the integrand f (x). The aim of this method is to convert the original 1 complicated integral to a much simpler one so that we can use the formulae above. Here are some guidelines: (a) If possible, try to choose u so that u (x) is part of the integrand f (x). (b) Consider choosing u as the part inside a radical, the denominator of a fraction, or the exponent of an exponential function. 1 dx, dont choose u = x ln x, let u = ln x instead. (c) Dont oversubstitue, e.g., in x ln x (d) If your choice of u doesnt work. Dont give up. Try another (and another) until you get the right one. Exercise Page 394-395: 3, 6, 12, 16, 17, 21, 25, 38, 31, 33 7. The Denite Integral Although closely related to each other, the denite integral b f (x)dx is fundamentally dierent than the indenite integral f (x)dx. The former is a a real number while the latter is an expression having x as a variable and an arbitrary constant C in the end. To evaluate an indenite integral, we need the Fundamental Theorem of Calculus: b b f (x)dx = F (x) a a where F (x) is an antiderivative of f (x), i.e., F (x) = f (x). Usually we let F (x) be the part having the constant C left out in the indenite integral f (x)dx = F (x) + C . Also notice b means F (b) F (a). that F (x) a Exercise Page 410: 9, 10, 12 8. Evaluating the Denite Integral Using Substitution There are two methods for doing denite integral using substitution. One changes the integral limits from a and b to u(a) and u(b) along with the change of variables from x to u. The other evaluates the associated indenite integral rst and then use the fundamental of theorem calculus to get the real number for the denite integral. Refer to the solution to Problem 1 of Quiz 2 posted online for the detailed explanation. Note that DONT confuse these two methods. Once you picked your favorite, stick to it! Exercise Page 411: 23, 24, 27 9. Algebraic Rules for Denite Integration Denite integrals share the algebraic rules of the indenite integrals: b The constant multiple rule: b The sum rule: b kf (x)dx = k a b (f (x) + g (x))dx = a f (x)dx for a constant k a b f (x)dx + a g (x)dx a 2 b b b f (x)dx (f (x) g (x))dx = The dierence rule: g (x)dx a a a Furthermore, denite integrals have another three rules which I never mentioned in class. So they wont appear in Exam I. But its good to know them: a f (x)dx = 0 for all a. a a b f (x)dx = a f (x)dx. b b c f (x)dx = c f (x)dx + a a f (x)dx: we can split an integral by splitting the interval b whereas the sum rule above split the integral by splitting the integrand. Exercise Page 411: 31, 38 10. Net Change Its easy to see that another way to write the fundamental theorem of calculus is b f (b) f (a) = f (x)dx. a This is all about the net change problems in real life. Exercise Page 412-413: 60, 64 11. Geometric Interpretation of the Denite Integral b f (x)dx mean geometrically? Well, if f (x) 0 on the interval a x b, What does b a f (x)dx is the area of the region on the x-y coordinate plane bounded above by the graph a of f (x), below by the x-axis (the graph of y = 0), left by the vertical line x = a and right by the other vertical x = b. In general, given two functions f (x) and g (x), the area of the region bounded by the graph of them is obtained by integrating the dierence between f (x) and g (x). If over the interval of intersection, f (x) g (x), we integrate f (x) g (x); if f (x) g (x), we integrate g (x) f (x). But be careful! In some cases, the region bounded have several components, among which the relationships between f (x) and g (x) are dierent. Exercise Page 411: 46; Page 427: 11, 12, 13 12. The Average Value of f (x) The average value V of f (x) over a x b is simply 1 V= ba 3 b f (x)dx a Exercise Page 428-430: 24, 25, 49, 51 13. Consumer Willingness to Spend Given the demand function D(q ), the consumer total willingness to spend on q0 units of products is q0 D(q )dq. A(q0 ) = 0 14. Surpluses If q0 units are sold and D(q ) is the consumers demand function for the commodity, then the Consumers Surplus is q0 D(q )dq D(q0 )q0 . CS = 0 If q0 units are sold and S (q ) is the producers supply function for the commodity, then the Producers Surplus is q0 P S = S (q0 )q0 S (q )dq. 0 The equilibrium price pe is the common value for D(q ) and S (q ). So to get pe , we equate D(q ) = S (q ) and then we get the solution for q . Plugging this q back in either one of D(q ) and S (q ), we get pe . Exercise Page 442: 1,2,8,11,15,16,18 4
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Purdue - MA - 224
Exam III Study Guide1.The Method of Least SquaresGiven two distinct points (x1 , y1 ) and (x2 , y2 ) on the x y plane, we can nd an equation ofa straight line going through them. But what if we are given n points (x1 , y1 ), , (xn , yn )?Well, in most
Purdue - MA - 224
MA 22400Midterm Exam 3Solution to Exam 31Spring 2012MA 22400Midterm Exam 3Spring 20121. (10 points) The equation of the least-squares line for the n points (x1 ,y1 ), (x2 ,y2 ), . ,(xn ,yn ) is y = mx + b, wheren xy x ym=n x2 ( x) 2x2nb=y
Purdue - MA - 224
Solution to Quiz 11Problem 1. (5 points) Find the indenite integral of f (x) = (x2 x)( x 1).x2Solution. The indenite integral of f (x) is found as follows:f (x)dx =====Problem 2. (5 points) Integrate11(x2 x)( 2 )dxxx1(x 1 1 + )dxx1(x
Purdue - MA - 224
Common Erros from Quiz 1ex+ C , but simply ex + C . Dont confuse it withx1.ex dx is not2.f (x)g (x)dx is NOTf (x)dxekx dx =ekx+ C.kg (x)dx. In general, we dont have a formula to dealwith the integral of the product of two functions. We only
Purdue - MA - 224
Solution to Quiz 2Problem 1. (5 points) Find the given denite integral10x2dx.(x3 + 1)2Solution. There are two ways to nd the denite integral if substitution is needed. We willfollow both methods.Method 1. This is the one getting the most votes in
Purdue - MA - 224
Common Erros from Quiz 21.11du is NOT ln |u2 | + C , it is + C . But we have2uu2.f (x)g (x)dx is NOTf (x)dx +1du = ln |u| + Cug (x)dx. In general, we dont have a formula todeal with the integral of the product of two functions. But we do ha
Purdue - MA - 224
Solution to Quiz 3Problem 1. (4 points) Circle the average value of f (x) = ex (1 ex ) over the interval1 x 1:e + e1A.2B. e e1 + 1e e1 2C.2e11D.+e .22Solution. The average value V of f (x) over the given interval is found as follows11f
Purdue - MA - 224
Common Error to Quiz 31dx = 1 + C1dx = x + C1
Purdue - MA - 224
Solution to Quiz 4xex dx.Problem 1. (5 points) FindSolution. A rule of thumb is that if you have ex in the integrand, let it be in dv instead ofletting u = ex . So we have the following as advertised in classu = x dv = ex dxdu = dx v =ex dx = ex (w
Purdue - MA - 224
Common Error to Quiz 4DONt confuse derivative with integral:ln xdx =1+CxIn fact,ln xdx = x ln x x + CThis is obtained by integration by parts.1
Purdue - MA - 224
Solution to Quiz 4Problem 1. (5 points) Find the improper integral+13x24dx.Solution. By the denition of improper integral,+N1=3 dx4N +x23x 2 dxlim4N2lim N +x422= lim ( + )N +4N=1=Problem 2. (5 points) Use the trapezoidal
Purdue - MA - 224
Common Error to Quiz 51. DONT confuse integral with derivative:325x 2 1x dx = 3+ C = x 2 + C.52 132Instead,3x 2 +12+ C = + C.x dx = 3x2 + 1322. In the trapezoidal rule, the coecient pattern is 1,2,2,2,.,2,2,1,not the others.1
Purdue - MA - 224
Solution to Quiz 62Problem 1. (5 points) Given f (x, y ) = ln yex ,(1) nd fx and fy .Solution. To nd fx , only x is treated as the variable while y is thought of as a constant.So2fx (x, y ) = ln y 2xex .Similarly, to nd fy , x is treated as a cons
Purdue - MA - 224
Common Error to Quiz 61. Given f (x, y ), to nd fx, we treat y as a constant. But this doesnt2mean that there is no y -part. For instance, given f (x, y ) = ln yex ,2fx = 2xex ,but2fx = ln y 2xex .2. Sometimes the usage of product rule is necessa
Purdue - MA - 224
dzProblem 1. (5 points) Given z = xy ; x = e2t , and y = et , use chain rule to ndwhendtt = 0.Solution.dxdydz= zx+ zydtdtdt= y 2e2t + x(et )Since t = 0, we have x = e20 = 1 and y = e0 = 1. Therefore,dz= 1(2 1) + 1(1)dt=1Problem 2. (5 p
Purdue - MA - 224
Solution to Quiz 8Problem 1. (5 points) Use the Lagrange multiplier method to nd the values of x, y wheref (x, y ) = x2 + y 2 gets the minimum under the constraint xy = 1.Solution. Letting g (x, y ) = xy and knowing f (x, y ) = x2 + y 2 , we nd the exp
Purdue - MA - 224
Solution to Quiz 9Problem 1. (5 points) Evaluate the given double integral2xx2 ydydx0Solution.2022x(x ydydx =02x00====1ydy )dx0x22=xy)dx2002xx2 ( 0)dx2023xdx022x424 02x2 (Problem 2. (5 points) Given the geom
Purdue - MA - 224
Solution to Quiz 10Problem 1. (5 points) Find the radius of convergence of the following power series:k=4xk2k1Solution.Lets nd the following L(x) which depends on x:xk+1 2k1|k 2(k+1)1 xkx= lim | |k 2x=| |2We know if L(x) &lt; 1, i.e., |x/2| &lt;
Purdue - MA - 224
Solution to Quiz 11Problem 1. (5 points) Write down the power series of1dx1xin summation notation.Solution.Let us rst write the integrand1as its power series and then integrate term by term:1x1dx =1xxn dxn=0xn dx=n=0=n=0xn+1n+1Probl
Purdue - MA - 224
Solution to Quiz 12Problem 1. (5 points) Solve the dierential equationdy= 2ydxSolution.Dividing both sides by y and multiplying by dx, we seperate the variables as follows:dy= 2dx.ydy= 2dx.yThen we get ln |y | = c + 2x, which yields |y | = ec
Purdue - MA - 224
Important limits(1) For k, p &gt; 0,Nplim kN = lim N pekN = 0N eN This is because in the long run (N ), ekN grows much fasterthan N p.Examplex2x1/3lim= 0; lim x/4 = 0.N e3xN e(2) For p &gt; 0,ln(N )= lim ln(N )N p = 0N N N plimThis is becau
Purdue - MA - 224
MA 22400Midterm Exam 1Solution to Exam 11Spring 2012MA 22400Midterm Exam 1Spring 20121. (14 points) Evaluate the following indenite integrals12 )dx2xxSolution Since none of the two multiplicative terms is part of the derivative of the other
Purdue - MA - 261
MA 261 - Spring 2011Study Guide # 3You also need Study Guides # 1 and # 2 for the Final Exam1. Line integral of a function f (x, y ) along C , parameterized by x = x(t), y = y (t) and a t b, isf (x, y ) ds =(bf (x(t), y (t)Cadxdt)2(+dydt)
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 1Spring 2011Problem 1. Find the angle between the vectors 1, 1, 1 and 1, 1, 2 . (Hint: Lookat the dot product of these two vectors.)Since1, 1, 1 1, 1, 2 = 1 + 1 2 = 0,we know these two vectors are perpendicular. Thus, the angle
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 2Spring 2011Problem 1. Given two planes x + y + z = 2 and x + y = 1,(a) Find the two normal vectors associated to each plane, respectively. (Notice thatx + y = 1 is the same thing as x + y + 0 z = 1)Solution. We read the normal v
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 3Spring 2011Problem. Consider the ellipsex2 y 2+= 1, z = 041in the three dimensional space. One system of parametric equations of it is r(t) =x(t), y (t), z (t) wherex(t) = 2 cos ty (t) = sin t , t [0, 2 ].z (t) = 0(a) Fi
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 3Spring 2011Problem. Consider the ellipsex2 y 2+= 1, z = 041in the three dimensional space. One system of parametric equations of it is r(t) =x(t), y (t), z (t) wherex(t) = 2 cos ty (t) = sin t , t [0, 2 ].z (t) = 0(a) Fi
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 4Spring 2011Problem 1. The position function of a particle is given byr(t) = 3 sin t, 3 cos t, 4t .(1) Find the velocity v(t).Solution. v(t) = r(t) = 3 cos t, 3 sin t, 4 .(2) Find the speed |(t)|.vSolution. |v(t)| = 32 + 42 =
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 5Spring 20112Problem 1 (Spring 2006). If u(x, y ) = yexy ,(a) nd ux .22Solution. ux = yy 2 exy = y 3 exy .(b) nd uxy .222Sollution. uxy = (ux )y = 3y 2 exy + y 3 2xyexy = y 2 exy (3 + 2xy 2 ).Problem 2 (Spring 2001). Give
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 6Spring 2011Problem 1. Suppose y is a function of x and they satisfy F (x, y ) = 0.Take the partial derivative with respect to x of both sides of the above equation (UseChain Rule ) to show thatFdyx= F .dxyProof. Given F (x
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 7Spring 2011Problem 1. Given f (x, y ) = x2 y 2 ,(a) Find the critical point of f (x, y ).Solution. Setting f (x, y ) to be zero, we have f (x, y ) = fx (x, y ), fy (x, y ) =2x, 2y = 0, 0 . So, (x, y ) = (0, 0) is the only critic
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 8Spring 2011Problem 1 (modied from Problem 2 in the second midterm of Spring22009). We will nd the maximum of f (x, y ) = xy on the ellipse x + y 2 = 1 using the4Lagrange multiplier method.2(a) Let g (x, y ) = x + y 2 . Write
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 9Spring 2011Problem 1 (Test II.8, Spring 2008). [Warning: No use of cell phonebrowsing the internet during the quiz!] Let D be the part of disk centered at 0with radius 2 that lies to the right of the line x = 1. Then which of the
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 10Spring 2011Problem 1. Find the length of a wire C in the shape of a helix described by theparametric equationC : x = cos t, y = sin t, z = t, 0 t 4Solution. To get the length of the wire, let us integrate the constant function
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 11Spring 2011Problem 1.(a) Given F(x, y ) = y exy , xexy , nd a function f (x, y ) such that f (x, y ) = F(x, y ).Solution. Since we are told f (x, y ) = F(x, y ), we have fx = yexy and fy = xexy .After integrating the rst equali
Purdue - MA - 261
MA261 0021&amp;0022 Quiz 12Spring 2011Problem 1.(a) Given a function f (x, y, z ) = xyz , nd its gradient f (x, y, z ) and then the curlof the gradient, namely curl( f (x, y, z ).Solution. Given f (x, y, z ) = xyz , we know f (x, y, z ) = fx , fy , fz =
Purdue - MA - 303
MA 303: Dierential and Partial DierentialEquations for Engineering and SciencesFall 2011, Final Examination(Instructor: Aaron N. K. Yip) This test booklet has TWENTY FIVE questions totaling 200 points for the wholetest. You have 120 minutes to do thi
Purdue - MA - 303
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 1MA 303 Fall 2011 (Aaron N. K. Yip)Friday, Sept. 2, in class1. (a) Do Textbook (Boyce-DiPrima, 9th-ed.) section 3.3, page 165, #34 (on Euler Equation)(b) Use the above result to nd the general solutions y (t) of the the following d
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 2MA 303 Fall 2011 (Aaron N. K. Yip)Friday, Sept. 16, in class1. Find the general solutionfollowing matrices:31B= 0 300ofdXdt= BX ,dXdt= CX anddXdt= DX where B, C, D are the03003121 , C = 0 3 1 , D = 0 3 3 300200
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 3MA 303 Fall 2011 (Aaron N. K. Yip)Friday, Sept. 30, in class1. For each of the following system, nd the general solution and plot the phase plot:(a)dX=dt1122(b)dX=dt112 2(c)dX=dt111 3(d)dX=dt3 11 1XXXX
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 4MA 303 Fall 2011 (Aaron N. K. Yip)Friday, Oct. 28, in class1. Consider the systems from the textbook (Boyce-DiPrima, 9th-ed.) section 9.4, page 530, #1,and #2.For each problem, do the following:(a) Find all the critical points;
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 5MA 303 Fall 2011 (Aaron N. K. Yip)Monday, Nov. 7, in class1. This question is to explore the Laplace Transform of periodic functions. A function f (t) iscall periodic with period T if for all t &gt; 0, f (t + T ) = f (t), i.e. the fu
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 6MA 303 Fall 2011 (Aaron N. K. Yip)Monday, Nov. 28, in class1. (a) Consider the following dierential equation:u (x) + u(x) = 0;0 &lt; x &lt; ;u(0) = 0,u ( ) = 0.(b) Consider the following dierential equation:u (x) + u(x) = 0;u (0)
Purdue - MA - 303
Purdue - MA - 303
Worked Out Homework 7MA 303 Fall 2011 (Aaron N. K. Yip)Friday, Dec. 9, in class, LAST DAY OF CLASS(Beware: I am using the following expression for Fourier series for any 2L-periodic function:an cosnxnx+ bn sinLLan cosf (x) = a0 +nxnx+ bn sin
Purdue - MA - 303
Purdue - MA - 303
Purdue - MA - 303
Properties and Formulas for Laplace Transformest g (t) dt.est f (t) dt and G(s) = Lcfw_g (t) =Let F (s) = Lcfw_f (t) =00In the following a and b are arbitrary constants and n is some positive integer.L cfw_af + bg = aF (s) + bG(s)(1)L cfw_f = s
Purdue - MA - 303
Purdue - MA - 303
Purdue - MA - 303
Purdue - MA - 303
Purdue - MA - 490
The Equations for Large Vibrations of StringsAuthor(s): Stuart S. AntmanReviewed work(s):Source: The American Mathematical Monthly, Vol. 87, No. 5 (May, 1980), pp. 359-370Published by: Mathematical Association of AmericaStable URL: http:/www.jstor.or
Purdue - MA - 490