Review Problems for the Fial Math - 3 7 These problems are provided to help you study The presece of a problem o this hadout does ot imply that there will be a similar problem o the test Ad the absece of a topic does ot imply that it wo t appear o the test Fid the area of the itersectio of the iteriors of the circles x + y ) = ad x 3) + y = 3 Does the series 3 + + 9 3 3 + 6 3 + + coverge absolutely, coverge coditioally, or diverge? 3 Fid the sum of the series 9 7 + 8 3 + I each case, determie whether the series coverges or diverges a) = l) /3 b) + 3 ) + + 3) c) = + ) d) 3 8 + 8 3 8 + 3 + + e) + 6 f) = = + cose ) Fid the values of x for which the series coverges absolutely 6 Compute the followig itegrals a) e x cosx b) x x =!) x ) )!
c) x 6x x ) x + ) d) si x) 3 cosx) e) si x) cosx) f) 3 x x ) 3/ 7 Let R be the regio bouded above by y = x +, bouded below by y = x, ad bouded o the sides by x = ad by the y-axis Fid the volume of the solid geerated by revolvig R about the lie x = 8 Compute x + 8x x) 9 Compute + 3x x) If x = t + e t ad y = t + t 3, fid dy ad d y at t = a) Fid the Taylor expasio at c = for e x b) Fid the Taylor expasio at c = for What is the iterval of covergece? 3 + x Fid the area of the regio which lies betwee the graphs of y = x ad y = x +, from x = to x = 3 3 Fid the area of the regio betwee y = x + 3 ad y = 7 x from x = to x = 3 The base of a solid is the regio i the x-y-plae bouded above by the curve y = e x, below by the x-axis, ad o the sides by the lies x = ad x = The cross-sectios i plaes perpedicular to the x-axis are squares with oe side i the x-y-plae Fid the volume of the solid Fid the iterval of covergece of the power series = x 3) ) 3 6 Fid the slope of the taget lie to the polar curve r = si θ at θ = π 6 7 A tak built i the shape of the bottom half of a sphere of radius feet is filled with water Fid the work doe i pumpig all the water out of the top of the tak 8 Let 3 x = t, y = t t3 Fid the legth of the arc of the curve from t = to t = 9 Fid the area of the surface geerated by revolvig y = 3 x3, x, about the x-axis a) Covert x 3) + y + ) = to polar ad simplify b) Covert r = cos θ 6 si θ to rectagular ad describe the graph Fid the area of the regio iside the cardioid r = + cosθ ad outside the circle r = 3 cosθ
Solutios to the Review Problems for the Fial Fid the area of the itersectio of the iteriors of the circles x + y ) = ad x 3) + y = 3 Covert the two equatios to polar: x + y y + =, x + y = y, r = r si θ, r = si θ x 3) + y = 3, x 3x + 3 + y = 3, x + y = 3x, r = 3r cos θ, r = 3 cos θ - 3 - - - Set the equatios equal to solve for the lie of itersectio: si θ = 3cos θ, taθ = 3, θ = π 3 The regio is orage-slice -shaped, with the bottom/right half bouded by r = si θ from θ = to θ = π 3 ad the top/left half bouded by r = 3 cos θ from θ = π 3 to θ = π Hece, the area is A = π/ si θ) dθ + 3 cos θ) dθ = cosθ) dθ + 3 π/ si θ) dθ + 6 + cosθ) dθ = [θ + ] si θ 6 π 3 889 π/ cosθ) dθ = + 3 [θ + ] π/ si θ = Does the series 3 + + 9 3 3 + 6 3 + + coverge absolutely, coverge coditioally, or diverge? 3 + + 9 3 3 + 6 3 + + = ) + 3 + 3 =
The absolute value series is Sice = 3 + 3 + for large values of, I ll compare the series to = 3 + 3 = 3 + = The it is fiite ) ad positive > ) The harmoic series = diverges By Limit Compariso, the series = 3 diverges Hece, the origial series does ot coverge absolutely + Returig to the origial series, ote that it alterates, ad Let f) = 3 + = 3 + The f ) = 3 ) + 3 ) < for > Therefore, the terms of the series decrease for, ad I ca apply the Alteratig Series Rule to coclude that the series coverges Sice it does t coverge absolutely, but it does coverge, it coverges coditioally 3 Fid the sum of the series 9 7 + 8 3 + = 9 9 7 + 8 3 + 3 + 9 ) 7 + = 9 3 ) = 9 3 = I each case, determie whether the series coverges or diverges a) = l) /3 Apply the Itegral Test The fuctio f) = [, + ) Sice is positive ad cotiuous o the iterval l) /3 f ) = 3 l) 7/3 l), /3 it follows that f ) < for Hece, f decreases o the iterval [, + ) The hypotheses of the Itegral Test are satisfied
Compute the itegral: [ 3 p l) /3 p d = d = l) /3 p l) /3 ] p ) = 3 p lp) /3 l) /3 To do the itegral, I substituted u = l, so du = d) Sice the itegral coverges, the series coverges, by the Itegral Test b) + 3 ) + + 3) Apply the Ratio Test The th term of the series is a = 3 ) 3), = 3 l ) /3 so the + )-st term is Hece, a + = 3 ) 3 + ) ) 3) + ) 3) c) a + 3 ) 3 + ) ) 3) 3 + ) ) = = a 3) + ) 3) 3 ) + ) 3) = 3 + + The itig ratio is 3 + + = 3 The it is less tha, so the series coverges, by the Ratio Test = + ) Apply the Root Test The it is a / = + ) + { = + ) } { = + ) } = e ) Sice e = e <, the series coverges, by the Root Test d) 3 8 + 8 3 8 + Sice + 3 3 + = 3, it follows that a is udefied the values oscillate, approachig ± 3 Sice, i particular, the it is ozero, the series diverges, by the Zero Limit Test
e) = 3 + + + 6 Apply Limit Compariso: 3 + + + 6 3 / + 3/ + / = = 3 + 6 The it is fiite ad positive The series = Therefore, the origial series diverges by Limit Compariso f) = + cose ) cose ) + cose ) 6 + cose ) 6 diverges, because it s times the harmoic series Therefore, = = Compariso diverges, because it s a p-series with p = < + cose ) diverges by Direct Fid the values of x for which the series coverges absolutely Apply the Ratio Test: =!) x ) )! a + a = + )!) x + + ))!!) x )! + )! =! ) )! x + + )! x = + ) x + ) + ) The itig ratio is + ) + ) + ) x = x The series coverges absolutely for x <, ie for < x < 9 The series diverges for x < ad for x > 9 You ll probably fid it difficult to determie what is happeig at the edpoits! However, if you experimet compute some terms of the series for x = 9, for istace you ll see that the idividual terms are growig larger, so the series at x = ad at x = 9 diverge, by the Zero Limit Test 6 Compute the followig itegrals 6
a) b) e x cosx x x d + e x cosx ց e x si x ց + e x cosx e x cosx = ex si x + ex cosx e x cosx = ex si x + ex cosx, e x cosx = ex si x + ex cosx + C e x cosx, x = x si θ) cos θ dθ = si θ) si θ) cos θ dθ = si θ) dθ = cosθ) [x = si θ, = cosθ dθ] cosθ) dθ = θ ) si θ + C = θ si θ cos θ) + C = si x x x + C x c) - x x 6x x ) x + ) x 6x x ) x + ) = a x + b x ) + c x +, x 6x = ax )x + ) + bx + ) + cx ) Settig x = gives 6 = 3b, so b = Settig x = gives 7 = 9c, so c = 3 Therefore, x 6x = ax )x + ) x + ) + 3x ) Settig x = gives = a + 3, so a = Thus, x 6x x ) x + ) = x x ) + 3 ) = l x + + 3 l x + + C x + x 7
d) si x) 3 cosx) si x) 3 cosx) = si x) cosx) si x ) = cosx) ) cosx) si x ) = u )u si x) [ ] du u = cosx, du = si x, = si x ) du = u u ) du = si x cosx) ) 3 cosx)3 + C u 3 u3 ) + C = e) si x) cosx) si x) cosx) = cos8x) + cos8x) = cos8x) ) = si 8x) = cos6x) = x ) si 6x + C 8 8 6 f) 3 x x ) 3/ I eed to complete the square Note that = ad ) = The 3 x x = x + x + 3) = x + x + ) = [ x + ) ] = x + ) So = 3 x x ) 3/ = x + ) ) 3/ cosθ dθ) = si θ) ) 3/ cosθ) 3cosθ dθ) = [x + = siθ, = cosθ dθ] x + cosθ) dθ = secθ) dθ = taθ + C = - x + ) x + 3 x x + C 7 Let R be the regio bouded above by y = x +, bouded below by y = x, ad bouded o the sides by x = ad by the y-axis Fid the volume of the solid geerated by revolvig R about the lie x = x = h = x + ) + x h -x r = - x 8
Most of the thigs i the picture are easy to uderstad but why is r = x? Notice that the distace from the y-axis to the side of the shell is x, ot x Reaso: x-values to the left of the y-axis are egative, but distaces are always positive Thus, I must use x to get a positive value for the distace As usual, r is the distace from the axis of revolutio x = to the side of the shell, which is + x) = x The left-had cross-sectio exteds from x = to x = You ca check that if you plug x s betwee ad ito r = x, you get the correct distace from the side of the shell to the axis x = The volume is V = π x)x + ) + x ) = π = π x x 3) [ = π x x ] x = π 6838 8 Compute x + 8x x) x + 8x x) = x + 8x x) 8x x + 8x + x = 8x x + 8x + x x x x + 8x + x x + 8x + x = = 8 + 8x + = x + 8x x x + 8x + x = 8 + = 9 Compute + 3x x) Let y = + x) 3x, so ly = l + x) 3x = 3x l + ) x The ly = 3x l + ) l + ) x = x 3x Therefore, = + x 3x + 3x = x) y = e l y) = e 6 x ) = 6 + x = 6 If x = t + e t ad y = t + t 3, fid dy ad d y at t = dy = dy dt dt = + 3t + e t 9
Whe t =, dy = + e d y = d ) dy = ) dt d dt )) dy = d dt ) dy dt = d + 3t dt + e t + e t = Whe t =, d y = 6 + e + e) 3 + e t )6t) + 3t )e t ) + e t ) + e t = + et )6t) + 3t )e t ) + e t ) 3 a) Fid the Taylor expasio at c = for e x e x = e x )+ = e e x ) = e + x ) + x ) b) Fid the Taylor expasio at c = for! + 3 x ) 3 3! What is the iterval of covergece? 3 + x ) + 3 + x = + x ) = + x = x ) = x + The series coverges for < x ) ) 3 x x + ) <, ie for 3 < x < Fid the area of the regio which lies betwee the graphs of y = x ad y = x +, from x = to x = 3 8 6 3 As the picture shows, the curves itersect Fid the itersectio poit: x = x +, x x =, x )x + ) =, x = or x =
O the iterval x 3, the curves cross at x = I ll use vertical rectagles From x = to x =, the top curve is y = x + ad the bottom curve is y = x From x = to x = 3, the top curve is y = x ad the bottom curve is y = x + The area is A = x + ) x ) 3 + x x + ) ) = 3 3 Fid the area of the regio betwee y = x + 3 ad y = 7 x from x = to x = 3 7 6 3-3 As the picture shows, the curves itersect Fid the itersectio poit: x + 3 = 7 x, x =, x = I ll use vertical rectagles From x = to x =, the top curve is y = 7 x ad the bottom curve is y = x + 3 From x = to x = 3, the top curve is y = x + 3 ad the bottom curve is y = 7 x The area is 7 x) x + 3)) + 3 x + 3) 7 x)) = [ x x ] + [ x x ] 3 = + = x) + 3 x ) = The base of a solid is the regio i the x-y-plae bouded above by the curve y = e x, below by the x-axis, ad o the sides by the lies x = ad x = The cross-sectios i plaes perpedicular to the x-axis are squares with oe side i the x-y-plae Fid the volume of the solid x,) y The volume is V = e x ) = e x = [ ] ex = e ) 393
Fid the iterval of covergece of the power series Apply the Ratio Test to the absolute value series: x 3 + + ) + ) 3 x 3 ) 3 = = x 3) ) 3 ) 3 x 3 + + + x 3 = The series coverges for x 3 <, ie for < x < 8 At x =, the series is It s harmoic, so it diverges At x =, the series is = = 8 ) 3 = 8) ) 3 = = = ) This is the alteratig harmoic series, so it coverges Therefore, the power series coverges for x <, ad diverges elsewhere + 8 x 3 = x 3 8 6 Fid the slope of the taget lie to the polar curve r = si θ at θ = π 6 Whe θ = π 6, r = si π 3 3 = dr Sice dθ = cosθ, whe θ = π 6, dr dθ = cos π 3 = The slope of the taget lie is ) ) ) 3 3 dr + ) dy r cos θ + si θ = dθ r si θ + cosθ dr = ) ) ) = 3 8867 3 3 3 dθ + ) 7 A tak built i the shape of the bottom half of a sphere of radius feet is filled with water Fid the work doe i pumpig all the water out of the top of the tak y x y r dy -
I ve draw the tak i cross-sectio as a semicircle of radius extedig from y = to y = Divide the volume of water up ito circular slices The radius of a slice is r = y, so the volume of a slice is dv = πr dy = π y ) dy The weight of a slice is 6π y ) dy, where I m usig 6 pouds per cubic foot as the desity of water To pump a slice out of the top of the tak, it must be raised a distace of y feet The - is ecessary to make y positive, sice y is goig from to ) The work doe is [ ] W = 6π y) y ) dy = 6π y 3 y) dy = 6π y y = 96π 783 foot pouds 8 Let 3 x = t, y = t t3 Fid the legth of the arc of the curve from t = to t = so ) + dt dt = 3t ad dy dt = 3 t, ) dy = 3t + 3 ) dt t = 3t + 3 t + 9 6 t = + 3 t + 9 6 t = + 3 ) t Therefore, The legth is ) + dt + 3 ) t dt = ) dy = + 3 dt t [ t + t3 ] = 8 9 Fid the area of the surface geerated by revolvig y = 3 x3, x, about the x-axis - - The derivative is dy = x, so dy ) + = x + 3
The curve is beig revolved about the x-axis, so the radius of revolutio is R = y = 3 x3 The area of the surface is S = ) x π 3 x3 + = π 3 7 ) du u / x 3 x 3 = π 6 7 u / du = π 6 [ u = x +, du = x 3, = du ] ; x =, u =, x =, u = 7 x3 π ) 7 3/ 79 9 [ ] 7 3 u3/ = a) Covert x 3) + y + ) = to polar ad simplify x 3) + y + ) =, x 6x + 9 + y + 8y + 6 =, x + y = 6x 8y, r = 6r cos θ 8r si θ, r = 6 cosθ 8 si θ b) Covert r = cos θ 6 si θ to rectagular ad describe the graph r = cosθ 6 si θ, r = r cos θ 6r si θ, x + y = x 6y, x x + y + 6y =, x x + + y + 6y + 9 = 3, x ) + y + 3) = 3-3 - - -3 - - -6 The graph is a circle of radius 3 cetered at, 3) Fid the area of the regio iside the cardioid r = + cosθ ad outside the circle r = 3 cosθ r= + cos 3 - - - r=3 cos
Fid the itersectio poits: 3 cosθ = + cosθ, cosθ =, cos θ =, θ = ±π 3 I ll fid the area of the shaded regio ad double it to get the total The shaded area is The cardioid area is π + cosθ) dθ = The circle area is π/ cardioid area from π ) 3 to π circle area from π 3 to π ) π + cosθ + cosθ) ) dθ = π [ θ + si θ + θ + )] π si θ = π 9 3 6 3 cosθ) dθ = 9 Thus, the shaded area is π/ The total area is π 8 = π 78 + cosθ + ) + cosθ) dθ = + cosθ) dθ = 9 [θ + ] π/ si θ = 3π 8 9 3 6 π 9 ) 3π 3 6 8 9 ) 3 = π 6 8 The best thig for beig sad is to lear somethig - Merly, i T H White s The Oce ad Future Kig c 8 by Bruce Ikeaga