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## CALCULUS II

by: Kennith Herman

28

0

11

# CALCULUS II MA 114

Kennith Herman
UK
GPA 3.54

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
11
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 11 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 114 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/228154/ma-114-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15
Lecture 8 L Hopital s rule 0 Recognize indeterminate forms 0 Compute limits using L7Hopital7s rule Some wellknown limits Recall some familiar limits 2 lim m z 2 40 1 1 lim nlt 95 1 zaO The rst two may be viewed as difference quotients and this allows us to know the limit For example7 sin2 i sin2 7 0 z 7 z 7 0 and thus with f sin27 Indeterminate forms Each of the limits above can be thought of as lim f zaa g where gr f96 1331 996 0 This is an indeterminate form of type As the above examples show7 these limits can have many values depending on the functions f and g This is why they are called indeterminate forms Ererez39se Given a value 17 can you choose f and 9 so that the indeterminate form has the value a If in the above limit we have 00 and 00 then this is an indeterminate form of type We say the same if one or both of the limits is foo Eat ample Indeterminate forms of type 000 If i131 f 0 and mlgnggltgt 00 then i133 fltzgtgltzgt is an indeterminate form This can be rewritten as 00 by considering fzgz 1 Emmise Can you also rewrite this indeterminate form as ooz39nfty Indeterminate forms 0f1 By rewriting fgw 69wh1f1 such an indeterminate form can be evaluated by if we understand the indeterminate forms discussed above Eat ample Compute lin1J1 x1 Solution We rewrite this as 61n1z maO since we evaluated the limit of the exponent above and the exponential function is continuous7 we have lin131 x m e The rule nally Theorem 1 If limmna M is an indeterminate form of type 00 of of type oooo 9w and we have 7 lim f z L zaa g then liming 572 exists and equals L We will not give the proof here If a is nite then the rule can be proven by a generalization of the mean value theorem Example Try to apply L7Hopital7s rule to What do we obtain What is the value of the limit Solution lHopital gives 1 and the correct value is 0 I Examples Example Use L7Hopital to compute e lim lnl xx7 lim 7 lim xe 7 lim 1 7 cos xxz an mace mace 40 Example Show that lim emx 00 race forn123 777 Lecture 10 Powers of sin and cos 0 Integrating non negative powers of sin and cos The goal In this section7 we learn how to evaluate integrals of the form sin z cos x dx The procedure will depend on several familiar trigonometric identities and the double angle formula for cos x Case 1 One of m or n is odd Let us suppose that m 2k l is odd In this case7 we rely on the Pythagorean identity7 sin2 z coszx 1 which allows to rewrite the integral sin2k1 z cos x dz as sinl 7 cos2 xkcos zdz This is sort of a mess7 but you should be able to see that the substitution u cos x will reduce this to evaluating the integral of a polynomial Erercz39se How does the procedure differ in n is odd and m is even This will be clear if we try an example Example Find the inde nite integral sin3 m cos3 x dx Solution Since both exponents are odd7 there are at least two ways to evaluate the integral We choose to rewrite coszx as 1 7 sin2 and then substitute u sinx7 du cos x dz This gives sin3cos3xdz sin317sin2xcoszdz u317 uz du u3 7u5 du 4 6 du 4 6 sin4 sin6 7 Cd 4 6 1 u Case 2 Both exponents are even For this case7 we rely on the double angle formula for cos H V cos2x 2 cos2 x 71 172sin2z D ln fact7 we will take these two forms and solve them for cos2 x and sin2 s respectively This gives us the formulae 1 2 cos2 x W 3 1 7 2 sin2 z 4 Many of you will recognize these as the half angle formula To make use of these formulae7 we will start with an integral of the form sin2k z cost x dx If we substitute the formulae 3 47 we end up with 1 7 2 kfrac1 cos 2x27 dx If we multiply out the powers7 we obtain an expression involving sin 2x and cos 2x which is of LOWER DEGREE If we repeatedly use this trick7 we end up with terms that we can treat by the method of case 1 Again7 an example Example Evaluate 27r sinzmdz 0 Solution We rst nd an anti derivative Using the double angle formula 4 we have 1 2 7 cos x sin2dz Td This integral is comparatively easy to evaluate 17 cos 2x z sin 2x d i O 2 95 2 4 Some of you may wish to use the substitution u 2x Now we make use of the anti derivative 2 sin2xdz g 7 m 0 2 4 to evaluate the de nite integral 27r i z s1n2x sinzxdz 7 2739 2 4 m0 71quot Emrcz39se Compute the integrals sin2cos2xdz sin5dz tansec2zdz tanzdz We close with an example which illustrates where such integrals arise in nature Example The voltage of an alternating electric current might have the form Vt Asinwt Because the voltage changes over time7 it is not clear how to assign a single number that represents the voltage of this current For example7 the 110 volt current in our houses is actually an alternating current where the voltage may reach as high as 155 volts A standard way of assigning a single voltage to V is to take the root mean square voltage This means7 square V7 take mean or average over one period and then take the square root This gives w 27rw 2 12 RMS Vt dt a Compute RMS b Show that if A is 1557 then EMS is about 110 Solution A short calculation shows RMS A An even shorter calculation shows 155 is about 1096 Problem If the nth partial sum of a nd an and 221 an Proof For n 1 we get 31 0 so a1 an MA114 Calculus 11 Problem 49 page 757 March 1 2007 oo senes 271 an 25 n71 n1 5n 7 0 For n gt 1 we know that 3 a1 a2 t t t an Therefore 8n7a1a2asman71 7 377371 7 n71 n7171 n17n711 7 nn71 n72n1 7 nn1 7 nn1 n27n7n27n72 n2n 2 n2n We now know what an is To nd 221 an we simply need to take the limit of the sequence of partial sums Therefore we get Thus n71 hmsn 1m naoo namn1 7 lim n n 1 namz 1 oo 00 Z Z 2 1 an 7 2 n1 n1n n Lecture 8 L Hopital s rule 0 Recognize indeterminate forms Especially7 007 0000 Be able to reduce limits of the form 0 007 10 and 1 to evaluating a limit of a quotient 0 State L7Hopital7s rule for quotients 0 Compute limits using L7Hopital7s rule Some wellknown limits Recall some familiar limits lim 2 zaO l 1 lim nlt 95 7 1 zaO licosx 1 133 2 7 5 The rst two may be viewed as difference quotients and this allows us to know the limit For example7 sin2 7 sin2 7 0 z i z 7 0 and thus with f sin27 Indeterminate forms Each of the limits above can be thought of as lim f zaa g where gr f96 1331 996 0 This is an indeterminate form of type As the above examples show7 these limits can have many values depending on the functions f and g This is why the expression is called an indeterminate form Ererez39se Given a value 17 can you choose f and 9 so that the indeterminate form has the value a If in the above limit we have 00 and 00 then this is an indeterminate form of type We say the same if one or both of the limits is foo Eat ample Indeterminate forms of type 000 If f 0 and 9a 00 then 93 fltzgtgltzgt is an indeterminate form This can be rewritten as 00 by considering fzgz 1 Emerez39se Can you also rewrite this indeterminate form as ooz39nfty Indeterminate forms 0f1 We rewrite aw 59w1nfw39 If f approaches 17 then lnf approaches 0 and we assume that g has a limit of 00 Thus on the right hand side7 the exponent is the indeterminate form 00 0 Example Compute lin1J1 x1 Solution We rewrite this as 61n1z maO since we evaluated the limit of the exponent above and the exponential function is continuous7 we have lin131 x m e The rule nally Theorem 1 If limmna g2 is an indeterminate form of type 00 or of type 0000 9w and we have 7 lim f z L zaa g then liming 572 exists and equals L We will not give the proof here If a is nite then the rule can be proven by a generalization of the mean value theorem Example Try to apply L7Hopital7s rule to What do we obtain What is the value of the limit Solution Misapplying L7Hopital7s rule gives 1 L7Hopital7s rule is not applcable because xew is not an indeterminate form at 0 The correct value of the limit is 0 I Examples Example Use L7Hopital to compute em 7 7 l 7 cos x lim lnl xx7 lim 7 lim xe 7 lim an mace mace an 2 Example Show that lim emx 00 race forn0123 7777 Solution The base case n 0 is obvious If we know that limmH00 3 007 then by L7Hopital7 61 2120 7m 113 n W ex If the limit is in nity for n7 it will also be 00 for n 1 Now the principle of mathematical induction tells us that the limit is 00 for all whole numbers n

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