Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe get some Rcacellatio ad are left with ta d. Usig ta = sec, we have sec d d = ta + C. Substitutig back for x's p we get x 9 arcta p x 9 + C. Z cos (x) si (x) We keep oe copy of si x with, ad covert the remaiig si x to cos x. We get R cos x( cos x) si x. Multiplyig out ad separatig wehave R cos x si x R cos x si x. I each of these itegrals use the substitutio u = cos x, ad we get cos x + cos x + C. Z x cos(x) This seems a atural to do with itegratio by parts. I will start with u = x ad dv = Rcos(x), so du = ad v = si(x). The the itegral ca be writte as x si(x) si(x) = x si(x)+ cos(x)+c. Problem ( poits) Evaluate the itegrals Z (x ) This is a improper itegral The deomiator goes to zero at x = which is withi the rage of itegratio. We rst split it ito two itegrals that have the problem" at their edpoits, Z (x ) + Z (x ) Now we use a limit o each to hold back from the trouble spots Z b lim b! (x ) Z + lim a! + (x ) Now each of these itegral is oe to which the Fudametal R Theorem applies, so we ca evaluate it by dig atiderivatives usig the power rule. =(x ) so the limits become (x ) h i h lim (b ) ( ) + lim ( ) (a ) b! a! + The parts with (b ) ad (a ) go to zero as a ad b go to two. We are left with p + p = p +. i
Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew variable. Whe x =,u = si =, ad whe x = ß, u = si ß =. So ow wehave Z du +u You may recogize that as givig a arctaget. If you did ot recogize it, you could evaluate this with a triagle ad a trig substitutio ad get the same aswer. We get arcta() arcta() = ß = ß. Problem ( poits) For each of these sequeces (i) Tell whether it coverges or ot. (ii) If it coverges, tell what its limit is. You should be sure to give reasos, but you do ot eed the precisio of ffls ad the formal deitio of the limit of a sequece. a = si ß + Sice si(x) is a cotiuous fuctio, we ca d the limit of ß + ad take the si of that result. I this limit ß does ot chage. The limit of is The umerator is xed while the deomiator is growig without boud. So the limit of ß + is ß. The we just take the si ad get si ß = p as the aswer. ß ß a = si + Cosider what happes for differet values of If is odd, e.g. =,we have si( ß ). (For ay other odd, the agle" will differ from this oe by a multiple of ß which will give the same value for the si.) You ca work out what that is, but it is eough to ote that the agle is i the secod quadrat where the si is positive. But for ay eve value of, the agle will be exactly ß away from that oe, i the fourth quadrat where the si is egative. So this sequece oscillates betwee two values, oe positive ad oe egative, ad has o limit. a = Usig a little algebra, a =( ) which is the same as (+ x ) if x =. Hece the sequece coverges to e = e. Problem ( poits) For each of the followig series (i) Tell whether it coverges. (ii) If it coverges, tell what it coverges to. Be sure to give reasos for your aswers! This ca be treated as two separate series, each a coverget geometric series, where oe is subtracted from the other. = The rst is geometric with a = ad r = ad so coverges to ( ) = 6. The secod is geometric with a =ad r = ad so coverges to =. (We kow they are coverget because i each case jrj <.) Hece the sum of the origial series is 6 =.
Agai we look at this series as a combiatio of two series, but this time the rst oe is geometric with r = ad so diverges. The secod oe coverges ad so the combiatio diverges. You ca also ote that the terms of this series get larger as icreases ad do ot go to zero, so by the th term test it diverges. There are other tests that would also show itdiverges. = The terms of this series get larger without boud as!, rather tha goig to zero. So the th term test agai tells us it diverges. Problem ( poits) For each of the followig series (i) Tell whether it coverges absolutely, coverges coditioally, or diverges. Be sure to give reasos for your aswers! = cos() The absolute value of a for this series is j cos()j which is». A series with the latter terms would be a p-series with p = > ad so would coverge. Sice the terms of our series, after takig absolute values, are smaller tha those of a coverget series, our series coverges absolutely. = ( ) l( ) This time if we take the absolute value of the term a we get = =. Sice l( ) l() l() l() <, l() >. Sice the series P, the harmoic series, diverges, this series of absolute values diverges by compariso. So our series does ot coverge absolutely. But it might coverge if we do ot take absolute values. We check ad it satises the three coditios of Leibiz' theorem The terms are alteratig i sig, they are decreasig i magitude, ad their limit is zero. Hece by Leibiz' theorem the series as origially give coverges, but it does ot coverge absolutely, soitcoverges coditioally. = ( ) + The terms of this series approach as!, so they do ot have limit zero. Hece by the th term test this series caot coverge, i.e. it diverges. Problem 6 ( poits) For the power series (x +) Fid the radius of covergece, ad also d the iterval of covergece. reasoig. We apply the ratio test to the absolute values of the terms of the series. Be sure to show your ρ = lim! j a + a j = lim! jx+j + + jx+j = lim! jx +j = jx +j By the ratio test the series will coverge absolutely whe ρ <, i.e. <. Rearragig algebraically, < x+ < or < x + <. Subtractig we get 8 < x <. But the ratio test jx+j
does ot tell us what happes if ρ = so we eed to check the edpoits I this case at both x = 8 ad x = the terms of the series (alterately ± atx = 8 ad cosistetly at x = ) do ot go to zero so the series diverges. Hece the iterval of covergece is ( 8; ). This tells us that the series coverges for x values up to uits away from the ceter of the iterval (x = ) so the radius of covergece is. Problem 7 ( poits) Fid the Maclauri series for f(x) =l( x). Oe way to d this series is to ote that the derivative ofl( x) is ad that looks like x our formula for the sum of a geometric series. From that oe ca write out a series for x ad itegrate it term-by-term to get a series for l( x). Here, however, is the more direct way from the formulas we have for coefciets i a Maclauri series. I start with a table of the various derivatives ad their values at x = f () (x) f () () l( x) ( x) ( x) ( x) 6 ( x) 6 We ca see that each time we take the derivative from here o the chai rule with x i the deomiator will require us to multiply by but also the egative power correspodig to ( x) beig i the deomiator will require a sig, so the sigs will all stay. The power will go oe more step egative each time, so the umbers,, 6 will cotiue to get multiplied each time by oe higher umber ad produce factorials. Puttig i x = will always give for the deomiator of the fractio. Hece the th derivative, at zero, will always be egative ad will step through the factorials but oe step delayed, givig ( )!. Hece whe we compute the coefciets a for the Maclauri series we get a = ( )! =, for each.! Sice we had a = which does ot t that patter it is easiest to start the sum at rather tha. Thus the series is = = x x x x x x For what values of x does the series coverge absolutely? The ratio of successive terms gives us x + + x = jxj + ad the limit as! is ρ = jxj. Hece the series coverges absolutely whe jxj < or <x<. (At x = ± the series of absolute values is the harmoic series which diverges.) Problem 8 ( poits) Suppose we use the terms of the Maclauri series for f(x) =e x through the x term to compute a approximatio to e. Give a boud for how far the approximatio might differ from the true value. Use a calculus result to produce your boud No credit will be give for usig a calculator to give the
differece betwee these umbers. You may wish to use the fact that e<. Be careful i choosig a method ot to apply a alteratigseries result uless the series is truly alteratig! The Maclauri series for e x is +x + (x) + (x) 6 + (x) + To use this to compute e we would set x = so that x =. Hece we wat to estimate the remaider term R (; ) for f(x) =e x. I geeral R (x; ) = f x. For f(x) =e x, the fourth! derivative is 8e x. Hece the remaider term is 8ec () where c is some umber betwee ad. The largest this ca be is 8e (). Sice we do't kow e we use e <e< ad have the remaider term is less tha 8 () =