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Trigonometnc Functions of Important Angles 0 cos6 sin6 tan6 0 1 0 0 w6 T3 w4 g g 1 7T3 7T2 0 1 unde ned 27r3 37r4 1 5W6 7T 1 0 0 W6 57T4 1 47r3 37T 2 0 1 unde ned 57T3 77r4 1 ms e a 27F 1 0 0 Calculus I Spring 2003 MA1133910 12 Russell Brown Week 6 and 7 o Homework C will be due on Monday 24 February 2003 For this assignment it will be useful to know the quadratic formula The quadratic formula gives the solutions of a quadratic equation azZ bx c 0 as 7bi Vb 7 4ac z 2a The expression b2 7 4ac inside the radical is known as the discriminant The discriminant tells us how many roots the equation has Sign of b2 7 4ac Number of roots b274acgt0 2 b274ac0 1 b274aclt0 0 The quiz on Tuesday 25 February 2003 will cover Appendix D and section 24 The error of the week Not being able to work homework problems on the exam Two of the problems that l graded were not done very well nding the average velocity and nding a tangent line The problem on average velocity was similar to 11 5a 7a The problem of nding tangent lines is discussed in 22 37 39 45 47 49 Both of these were part of the assigned homework To do well in this course one must work lots and lots and lots of problems To study for the exam do not read your solutions Work the problems again or work the even problems next to homework problems Please remember that free tutoring is available in MathSkeller I am there from 3 to 4 on Monday No one has come to ask a question except possibly the week that I missed Homework D 25 60 66 Due 28 February 2002 We will omit section 210 Newton7s method This will allow us to catch up before the second exam 0 A puzzle 7 What happens when you di erentiate fg twice three times four times Do you see a pattern 1 The fundamental theorems of calculus o The fundamental theorems of calculus 0 Evaluating de nite integrals o The inde nite integral a new name for anti derivative o Differentiating integrals Theorem 1 Suppose f is a continuous function on cub FTC 1 we if N dt then 9 f FTC H IfF is an anti derivative of f then b ft dt Fb 7 Fa d w 1 g A Ecit 3 x3 dz 0 Proof An idea of the proofs FTC I Write h 1 M 9 96 i 9 96 i 7 m f 7 hm ftdt Example Compute Compute We will show 1 zh E ft dt we The reader should write out a similar argument for the limit from the below If f is continuous7 then f has maximum and minimum values Mh and mh on the interval pp h Using the order property of the integral7 lim ha0 1 zh mh E ftdt Mh As It tends to 07 we have limhn0 Mh limhn0 mh u since f is continuous It follows that We dt we lim ha0 FTC ll We know from FTCl that f has one anti derivative ft dt We let em 7 flttgtdt7Fltzgt where F is some anti derivative as in FTC ll The derivative of G G z f 7 f 0 for all z in ab This uses FTC l and the hypothesis that F is an anti derivative of f Since the derivative of G is identically zero we can conclude that G is a constant If we set x a in the de nition of G we nd Ga 7Fa so that we can conclude the constant is 7Fa If we set x b in the de nition of G then we conclud b 7Fa m dt 7 Fb Adding Fb to both sides give the conclusion of FTC ll I 11 Inde nite integrals We use the symbol fz dx to denote the inde nite integral or anti derivative of f The inde nite integral is a function The de nite integral is a number According FTC ll we can nd the numerical value of a de nite integral by evaluating the inde nite integral at the endpoints of the integral Since this procedure happens so often we have a special notation for this evaluation FWLQ FUD Fa Eat ample Find malim and mallq Solution ba 7 a2 zy 7 2 According to FTC l anti derivatives exist provided f is continuous The box on page 351 should be memorized In fact you should already have memorized this information when we studied derivatives in Chapter 3 and when we studied anti derivatives in Chapter 4 Example Verify 1 x cosx2 dx i sin2 Solution According to the de nition of anti derivative we need to see if d 1 1 2 2 fisins xcosx d 2 lt gt lt gt This holds by the chain rule I 12 Computing integrals The main use of FTC ll is to simplify the evaluation of integrals We give a few examples Eat ample a Compute 07r sinx dx b Compute 4 222 1 Ch 1 39 Solution a Since 7 cosx sinz we have 7cosx is an anti derivative of sinx Using the second part of the fundamental theorem of calculus gives sinx dx 7 cosl0 2 0 b We rst nd an anti derivative As the inde nite integral is linear we write 222 1 With this anti derivative we may then use FTC ll to nd 2 42 1dx 2521L2212 1 5 4 dz2z32x 12d 2232dzz 12dx5252212C 4 m 4 4 345224127 g2 1285205745105 1345 Here is a more involved example that illustrates the progress we have made Eat ample Find 1 V L lim 7 sin k n we l Solution We recognize that 1 71 7 Z sinkn n k1 is a Riemann sum for an integral The points xk k 07 7n divide the interval 01 into 71 equal sub intervals of length 171 Thus7 we may write the limit as an integral n 1 lim l Zsinkn sinz dz Hoe n k1 0 To evaluate the resulting integral7 we use FTCll An anti derivative of sinz is 7 cosx7 thus 01 sinx dx 7 cosli0 17 cos1 13 Differentiating integrals FTC 1 plays an important role in the proof of FTC ll It is also used to nd the derivatives of integrals Eat ample Find d w 2 d w 2 d 1 O s1ntdt A2s1ntdt Edt ls the function L 1m dt increasing or decreasing ls the graph of L concave up or concave down 14 The net change theorem Since F is always an anti derivative of F 7 one consequence of part ll of the funda mental theorem of calculus is that if F is continuous on the interval 1 b then bF t dt Fb 7 Fa This helps us to understand some common physical interpretations of the integral For example7 if pt denotes the position of an object More precisely7 if an object is moving along a line and p gives the number of meters the object lies to the right of a reference point7 then 19 1 is the velocity of the object The de nite integral pltbgt 7pltagt 7 vlttgt dt lt1 denotes the net change in position ofthe object during the interval 1 b Note that ifi is measured in meterssecond7 then units on vtdt would be meterssecond gtlt seconds so the equation 1 is a sophisticated version of the familiar fact that distance rate gtlt time To give a less familiar example7 suppose we have a rope whose thickness varies along its length Fix one end of the rope to measure from and let denote the mass in kilograms of the rope from 0 to z meters along the rope If we take the derivative7 ii limhno mx h 7 771xh7 then this represents an average mass of the rope near z whose units are kilogramsmeter If we integrate this linear density and observe that 7710 07 then we recover the mass 1 dm mx 0 de November 167 2006