Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + = lim b du b x= u = lim b [l u ] b x= = lim b [l x 2 + ] b = lim b [(l b 2 + ) (l 2 + )] = lim b l(b 2 + ) = for as b we have b 2 + ad we kow l =. Thus (b) Fid the value of b lim b b x 2 +. x 2 + diverges. Solutio: I cotrast to our result above, ad as a explaatio of why we do t defie = lim b b b, we ow compute lim b b b x 2 + = lim b [l x 2 + ] b b = lim b [(l b 2 + ) (l ( b) 2 + )] = lim b [l b 2 + l b 2 + ] = lim b [] =. Had we defied = lim b b b, the rather tha cocludig that our itegral diverges as we did i part (a), this aswer would have suggested its value should be. (c) Fid the value of b lim b b x 2 +. b Solutio: Now cosider lim b b x 2 + = lim b [l x 2 + ] 2b b = lim b [(l (2b) 2 + ) (l ( b) 2 + )] = lim b [l 4b 2 + l b 2 + ] = lim b l 4b2 + = l 4 sice 4b lim 2 + b b 2 + = lim b 8b 2b = lim b 8 2 = 4 usig l Hôpital s Rule ad the fuctio l x is cotiuous at x = 4. This ow, is why we must defie the covergece or divergece of improper itegrals of the type f(x) dx by splittig them ito two parts - f(x) dx = a f(x) dx + a f(x) dx for ay value of a R - ad requirig both of these two parts of our improper itegral to coverge before we may claim that f(x) dx coverges ad compute its value. Otherwise, we could make its value to be ay umber we choose i the example above, as l 4 yet both appeared reasoable eough before we computed their values. b 2 + 2. (25 poits) Cosider the regio R bouded o the left by the parabola x = y 2 + ad o the right by the lie x = 5. Set up, but do ot evaluate!, itegrals that compute (a) the area of R; Solutio: The area of R is the area betwee the two curves x = 5 ad x = y 2 +. To determie the area of R, we must kow where the curves itersect, so we solve 5 = y 2 +: y 2 = 5 = 4 whe y = ±2. Therefore, our limits of itegratio will be y = to y = 2, ad i this iterval we see that x = 5 x = y 2 + sice this is true whe y = betwee y = ad 2. Hece, A = [(5) (y 2 + )] dy.
Math 3, Sectio 2 (b) the volume of the solid obtaied by revolvig R about the y-axis; Solutio: The volume obtaied by revolvig R about the y-axis is the volume of the solid the solid formed by revolvig the outer curve, x = 5, about the y-axis mius the volume of the solid formed by revolvig the ier curve, x = y 2 +, about the y-axis. Therefore, V = [ π[x = 5] 2 dy] [ π[x = y 2 + ] 2 dy] = π[(5) 2 (y 2 + ) 2 ] dy, which we could further simplify as V = π[25 (y 4 + 2y 2 + )] dy = π[24 2y 2 y 4 ] dy. (c) the surface area of the solid obtaied by revolvig R about the y-axis. Solutio: The surface area of the solid is made up of two parts, the surface area of the exterior ad that of the iterior. So the total surface area is the sum of these: so that SA = [ 2π[x = 5] ds] + [ 2π[x = y 2 + ] ds] SA = [ 2π[5] () 2 + dy] + [ 2π[y 2 + ] (2y) 2 + dy]. This first term i the itegral above is 2π[5] dy = π dy = π[y]2 = π[(2) ()] = 4π, which is the result we kow it should be usig the formula for the lateral surface area of a cylider: SA cylider = 2πrh for our r = 5 ad h = 4. You eed t have actually computed this, but it is ice to see it cofirms the formula we already kow to be true, so I icluded it here. 3. (25 poits) Cosider the fuctio f(x) = x 3 o the iterval [, 4], ad the regio R below the graph of f o this iterval. Fid: (a) The average value f ave o [, 4] ad the umber x ave i [, 4] at which it occurs (i.e., f(x ave ) = f ave ). Solutio: The average value of f o [, 4] is f ave = 4 4 f(x) dx = 4 4 x3 dx = 4 [x4 /4] 4 = 4 [(44 /4) ( 4 /4)] = 4 [64 ] = 6. Thus f ave = 6, ad we ow seek the value of x ave i the iterval [, 4] such that f(x ave ) = f ave = 6. As f(x) = x 3, we wat to solve x 3 ave = 6, which is x ave = 3 6 = 2.59842. (b) The value of x split i [, 4] for which the vertical lie x = x split divides R ito two regios of equal area. Solutio: The value of x split is that umber i the iterval [, 4] such that x split f(x) dx = 4 2 f(x) dx = 2 [x4 /4] 4 = 2 [64] = 32. The left had side, x split x 3 dx = [x 4 /4] x split = x 4 split /4. Hece all that is left is to solve the equatio x4 split /4 = 32, i.e., x4 split = 28 so x split = 4 28 = 3.36358566. Observe that the value of x ave ad that of x split differ from oe aother. 4. (25 poits) The accompayig figure shows the first three rows ad part of the fourth row of a sequece of rows of semicircles.
Math 3, Sectio 2 /8 /4 /2 There are 2 semicircles i the th row, each of radius /2. Fid the sum of the areas of all the semicircles. Solutio: The area of oe of the semicircles i the th row, which thus has radius r = /2, is half that of a full circle with the same radius. Therefore, this area is 2 [π(/2 ) 2 ] = π 2 (/22 ). As there are 2 such semicircles i the th row, the sum of the area of all of the semicircles i the th row is 2 [ π 2 (/22 )] = π 2 (2 /2 2 ) = π 2 (/2 ) = π 2 (/2). With this ow foud, the total of the areas of all of the semicircles is = π 2 (/2) = π 2 (/2) (/2) = π/4 /2 = π 2, for this sum is a geometric series with iitial term a = π 2 (/2) = π 4 the series occurs whe = ) ad ratio r = /2. (sice the first term of 5. (25 poits) Cosider the series (3x 2). = (a) Fid the series radius of covergece. (3x) + + Solutio: We fid the radius of covergece usig the Ratio Test: lim (3x) lim + (3x 2) = 3x 2. Recall that, by the Ratio Test, that this series coverges absolutely whe the limit is < ad diverges whe the limit is >. So the series coverges whe 3x = 3(x/3) = 3 x/3 <, which is whe x (2/3) < /3. Hece R = /3 is the radius of covergece of the power series. (b) Fid the iterval of covergece of the series. Solutio: Startig with the iequality x (2/3) < /3, which is equivalet to /3 < x (2/3) < /3 ad hece to /3 < x <, this must be icluded i our iterval of covergece ad o more save, perhaps, the edpoits. Let us, therefore, test these: x = /3 ad x =. Whe x = /3, our power series takes the form [3(/3)] = = ( ) =, which is the Alteratig Harmoic Series. If we recogize it already, we may coclude from past experiece that this series does coverge, so the edpoit x = /3 must be icluded i the iterval of covergece. If we fail to recogize this series by ame, we should at least otice that it is alteratig, so we employ the Alteratig Series Test - the b = ( ) / = / are all positive ad we see: i. b + = + b = is satisfied, ad =
Math 3, Sectio 2 ii. lim b = lim = is true as well. We may the coclude the series coverges. Whe x =, the power series is [3()] = = = = =. This is the Harmoic Series (a.k.a., the p-series with p = ), so it diverges. Thus the iterval of covergece for this power series is [/3, ) = {/3 x < }. (c) Where does the series coverge absolutely? Where does it coverge coditioally? Solutio: By the ratio test is part (a), the series coverges absolutely o the iterval (/3, ) = {/3 < x < }. The oly other value of x for which the series coverges is at x = /3, but here it does ot coverge absolutely (because ( ) = is the Harmoic Series which we have already metioed diverges). Thus the series coverges coditioally at x = /3. 6. (25 poits) The series coverges to si x for all x. si x = x x3 + x5 x7 7! + x9 9! x! + (a) Fid the first six terms of a series for cos x. For what values of x should the series coverge? Solutio: Sice cos x = d dx [si x], we obtai the series for cos x by differetiatig the oe for si x term-by-term: cos x = d x3 [x dx + x5 x7 7! + x9 9! x 3x2 + ] = + 5x4 7x6 + 9x8 x +! 7! 9!! so that exploitig properties of factorials,! = [( )!] i particular, this is cos x = x2 6! + x8 8! x! +. Accordig to our theorem for term-by-term differetiatio, the series for the derivative of a fuctio has the same radius of covergece as that of the origial fuctio. Sice we are told the series for si x coverges everywhere, its radius of covergece was R =. Thus the radius of covergece for the series we just obtaied for cos x is also R =, which meas that this series also coverges (to cos x) for all values of x. (b) By replacig x by 2x i the series for si x, fid a series that coverges to si 2x for all values of x. Solutio: The series for si 2x is obtaied from the oe above for si x by substitutig 2x for x throughout: si 2x = (2x) (2x)3 ad rewritig (2x) = 2 x we get + (2x)5 (2x)7 7! + (2x)9 9! (2x)! + si 2x = 2x 8x3 + 32x5 28x7 7! + 52x9 9! 248x! +. Though I did t ask, let me metio that this series coverges for all values of x.
Math 3, Sectio 2 (c) Usig the result i (a) ad series multiplicatio, calculate the the first six terms of a series for 2 si x cos x. Compare your aswer with the aswer i (b). Solutio: We fid this series by multiplicatio of the series for si x by that for cos x, ad look for that part of degree 6: 2 si x cos x = 2(x x3 + x5 x2 )( 6! + ) With a bit of computatio, this product is 2[x( x2 )+ x5 x2 ( 6! 2[x ( 2 + )x3 + ( + + ) ] = 2[(x x3 2! + x5 6! x3 x2 + ) ( 6! + x5 + )+( )] =. We otice that this ) ( x3 x5 2! 2 + )x5 ] = 2x 8x3 + 32x5 is the same series as we obtaied i part (b), as it should be for from trigoometry we recall that si 2x = 2 si x cos x holds for all values of x, so that the two series should be equal to oe aother. Extra Credit ( poits) This page is the extra credit part of the test. Carefully remove this page from the rest of your exam, brig it home with you after you fiish the exam problems, ad compute these itegrals over the weeked. Retur this page to me whe you come to class o Moday. You may ot work together but you may use your book to do these problems.. Does the series l(!) coverge? Solutio: Observe that l(!) = l 2 + l 3 + + l. Now l 2+l 3+ l ( ) l l sice the fuctio l x is a icreasig fuctio. Therefore, l(!) l. Yet, l diverges by the Itegral Test, as 2 dx b x l x = lim dx b 2 x l x = lim [l l b x ]b 2 = lim [(l l b) (l l 2)] = b diverges (usig the u-substitutio u = l x ad du = x dx to compute the itegral above). Therefore, by the Direct Compariso Test for series, sice l(!) l for every 2 ad the series l diverges as we have just show usig the Itegral Test, our give series must also diverge. l(!) 2. For what values of p does the series l( p ) coverge? l( p ) = (p l ) = p l. Solutio: Recall the property of l x that l( p ) = p l. Hece Thus, l( p ) = p l, which diverges as we have just show above (the p is etirely irrelevat!). So there is o value of p for which the series l( p ) coverges.