Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this problem, a = ad b =, so x = b a = With all of this ow i had, we have x x + dx ad j= j= j= x j = a + j x = + j f(x j ) x j= j= j= ( ) ( + ) j ( + ) j + ( + j + ) j ( + j + j + j 8 j + j= j j 8 j + ( + ) j + ) + ( 8 ) j + j= j ( ) 8 ( + ) + ( ) ( + )( + ) 6 ( + ) ( + )( + ) + = + = To check all of this work, we ow employ the Evaluatio Theorem: x x + dx = x x + x = 7 8 + + = = As we get the same aswer, we re happy ad appropriately cofidet that all of the work we ve doe is correct We re off to a good start! P i= =, P i = i= ( + ), P i = i= ( + )( + ) 6
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) The fuctio l x is sometimes defied for x > by itegratio: l x = x t dt Assume a > ad x > Derive the idetity l(ax) = l a + l x by carryig out ad justifyig each of the followig steps: (a) Show that d ax dx t dt = Hit: Use the Chai Rule x Solutio: First, by the Fudametal Theorem of Calculus, Part I, we kow d x dx t dt = x for the FTC states that for every cotiuous fuctio f, we have d x f(t) dt = f(x) dx a Now let u = ax, so that we ca compute d ax dx t dt = d u ( d u ) ( ) ( ) dx t dt du d = du t dt dx = u dx ax = a = ax x by the Chai Rule (ie, df dx = df du du dx ) (b) Explai why the result i part (a) implies that l(ax) = l x + C for some costat C x ax Solutio: By part (a), both l x = dt ad l(ax) = dt are atiderivatives t t of the same fuctio, amely /x Hece, by the Mea Value Theorem, they differ by a costat, C, so as claimed l(ax) l x = C which implies l(ax) = l x + C (c) By choosig a appropriate value of x, show that the value of the costat C i part (b) is l a Solutio: Settig x =, we fid that l a = l(a) = l + C = + C = C, so C = la ad l(ax) = l x + l a, as desired ( poits) Cosider the fuctio f(x) = + x o the iterval, (a) Evaluate the defiite itegral + x dx Solutio: Usig u-substitutio, with u = + x (so du = dx), we fid the idefiite itegral of + x to be + x dx = u / du = u/ / + C = u/ + C = ( + x)/ + C
Math, Calculus II Witer 7 Fial Exam Solutios I particular, ( + x)/ is a atiderivative of + x, so we may evaluate the defiite itegral usig the Evaluatio Theorem (ie, FTC, Part II) as + x dx = ( + x)/ = ( + )/ ( + )/ = 8 = (b) Use Simpso s Rule with = 6 to approximate + x dx Solutio: I this problem, we have a = ad b =, so x = b a x j = a + j x = j Now Simpso s Rule with = 6 gives = 6 = ad S 6 = / f() + f(5) + f() + f(5) + f() + f(5) + f() = + 5 + + 5 + + 5 + 6 = 799978 6 = 666565 so the Simpso s Rule approximatio is + x dx S6 = 666565 (c) Fid the maximum error, E S6, of your approximatio i part (b) How does this compare to the actual error? Solutio: We begi by fidig M, usig the formula give, which requires us to fid the fourth derivative of f(x) = + x = ( + x) / So we compute f (x) = ( + x) / f (x) = ( + x) / f (x) = ( + x) 5/ 8 f () (x) = 5 ( + x) 7/ 6 Thus, f () (x) = 5 6( + x) 7/ is largest whe ( + x)7/ is smallest, which is whe x = ad its value is, so M = 5 will work 6 With M ow i had, we ca compute the maximum possible error: E S6 M(b a)5 8 = 5 ( )5 6 86 = 785 8 = 976565 which is t all that large ad eve is pretty small Fially, we compare this with the actual error, E S6 = + x dx S6 = 666565 = 66667, which is certaily less tha the maximum possible error foud above (it is roughly oeith the potetial error) Recall that E S M(b a)5 8 where f () (x) M o a, b
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Cosider the regio R bouded by the curves y = x ad y = x (a) Sketch the curves ad shade i the regio R below: y = x y = x (b) Fid the area of the regio R Solutio: The area of R is equal to the sum of the areas of the two yellow regios, oe i the first quadrat ad the other i the third It is clear that i the third quadrat, y = x is o top while y = x is below, while i the first quadrat y = x is above ad y = x is the lower boud Fially, we determie our limits of itegratio by solvig x = x, so x =,, are the solutios Thus the area of R is A = = = x (x ) (x) dx + (x) (x ) dx x () ( x + ) + x ( ) () = Notice that we could have exploited the obvious symmetry of the regio R to compute A = (x) (x ) dx = / = /, as above (c) Set up, but do ot evaluate!, itegrals that compute i the volume of the solid obtaied by revolvig R about the y-axis; Solutio: Notice that y = x is always farther from the y-axis that the lie y = x So the outer radius is determied by the curve y = x, so it is R out = x = y / Likewise, the ier radius is r i = x = y Thus the volume is equal to the itegral V = y= π (R out ) π (r i ) dy = y= ( π y / y ) dy ii the volume of the solid obtaied by revolvig R about the x-axis Solutio: I cotrast to part (i), we see that the lie y = x is always farther from the x-axis tha the curve y = x, so the volume is give by V = x= π (x) π ( x ) dx = x= π ( x x 6) dx 5 ( poits) Cosider the sie curve, y = si x, o the iterval,π (a) Sketch y = si x, for x π Note: For your coveiece, there are s placed alog the x-axis at x =, π 6, π, π, π, 5π 6 ad π
Math, Calculus II Witer 7 Fial Exam Solutios y = si x (b) Write the legth of the sie curve from x = to x = π as a itegral Do ot evaluate! Solutio: Recall that wheever y = f(x) gives y as a fuctio of x, the arc legth of the curve from x = a to x = b is give by the itegral b a + f (x) dx So, i our curret situatio with f(x) = si x, we have f (x) = cos x, which meas the arc legth of the sie curve from x = to x = π is L = π x= + cos x dx = π x= + cos x dx (c) Use the Trapezoidal Rule with = 6 to estimate the itegral i part (b) Solutio: I this problem, we have a = ad b = π, so x = b a = π 6 ad x j = a + j x = π 6 j Now the Trapezoidal Rule, with = 6 ad f(x) = + cos x (ie, the itegrad), gives T 6 = π/6 f() + f(π/6) + f(π/6) + f(π/6) + f(π/6) + f(5π/6) + f(6π/6) = π + 75 + 5 + + 5 + 75 + = π 59657 = 89867 Hece, accordig to the Trapezoidal Rule, the legth of the sie curve from x = to x = π is approximately 89867 6 ( poits) Is the sum of differeces equal to the differece of the sums? (a) Show that ( + ) = + Solutio: Startig with the right-had side, let s get a commo deomiator ad fid + = + ( + ) which is what we wated to show (b) Show that ( + ) coverges = ( + ) () = = ( + ) ( + ) ( + ), Solutio: We see that the terms of this series look similar to those of the p-series, so let s use the Limit Compariso Test by evaluatig lim ( + ) ( + ) + =
Math, Calculus II Witer 7 Fial Exam Solutios Thus, as this limit exists ad is larger tha zero, the Limit Compariso Test says that both series do the same thig Now the p-series,, coverges sice p = > Hece the give series,, also coverges ( + ) (c) Show that both = ad + diverge = Solutio: The first of these series,, is the Harmoic Series, which we all already kow diverges The secod series,, looks a great deal like the Harmoic Series, + so we ll agai employ the Limit Compariso Test by evaluatig lim + + = As this limit exists ad is greater tha, the Limit Compariso Test tells us that both series do the same thig Sice we have already metioed that the Harmoic Series diverges, we coclude that this other series,, likewise diverges (d) Why ca t we write = = + = + =? + Solutio: We ca t write that the ifiite sum of differeces, as we are give o the left-had side of the equatio above, is equal to the differece of the correspodig ifiite sums sice we foud by parts (a) ad (b) that the left-had side coverges while either of the series o the right-had side coverge 7 (5 poits) Cosider the series (a) Why does m= m= m coverge? m Solutio: The series m= is a p-series with p = >, so it coverges m (b) It is kow that m= m = π 9 Approximate this value usig the partial sum s Solutio: The fourth partial sum of this series is s = m= so this says π /9 787599 m = + + + = 787599 (c) Estimate the error π 9 s usig the error boud R f(x) dx
Math, Calculus II Witer 7 Fial Exam Solutios Solutio: The error, usig the iequality give, is bouded by dx x b b x x dx b b b b = 9 = 58 Notice that the actual error is π 9 s = (8) (787599) = 577, which is smaller tha the maximum possible error i our estimate above (d) How large must we take to isure the error R = π 9 s is less tha? Solutio: To fid the value of sufficietly large to make sure that R <, we istead fid so that the itegral is less tha, sice we kow R dx, so x that this will esure the correspodig R is small eough Now dx x b b x x dx b b b b = Thus we are lookig for the value of so that < = = < = > = 9858 Hece, we must take 5, so = 5 will be large eough to isure that the error R 5 is less tha 8 ( poits) Cosider the series = ( ) (x + ) (a) Fid the series radius of covergece Solutio: To fid the radius of covergece, we use the Ratio Test: lim ( ) + (x + ) + ( + ) + ( ) (x + ) ( )(x + ) ( + ) = (x + ) = x + So, accordig to the Ratio Test, this series coverges (absolutely) wheever x+ <, so x + < or < x + <, ad it diverges wheever x + > We see that the width of the iterval of covergece will be 6, so the radius of covergece must be half of this, which is R = (b) Fid the iterval of covergece of the series Solutio: To fid the iterval of covergece, we first ote that we already kow the series coverges absolutely wheever < x+ <, so subtractig from all quatities we fid 5 < x < esures absolute covergece Now we oly have to check the two edpoits, as we kow, also by the Ratio Test, that the series diverges whe x < 5 ad whe x >
Math, Calculus II Witer 7 Fial Exam Solutios At x = 5, the series is = ( ) ( 5 + ) = = ( ) ( ) = = ( )( ) = which diverges sice it is the Harmoic Series Thus x = 5 is ot i the iterval of covergece At x =, the series is = ( ) ( + ) = = ( ) = = ( ), which coverges sice it is the Alteratig Harmoic Series Thus x = is i the iterval of covergece Therefore, the iterval of covergece is = ( 5, = {x : 5 < x } (c) Where does the series coverge absolutely? Where does it coverge coditioally? Solutio: By the Ratio Test, the series coverges absolutely for 5 < x < ad diverges whe x < 5 or x > By part (b), we also kow that the series diverges at x = 5, so it remais to determie whether we have absolute or coditioal covergece at x = To do this, we ca t use the Ratio Test sice it was already exhausted i part (a), so we have to resort to the defiitio by cosiderig ( ) ( + ) = ( ) =, which diverges sice it is the Harmoic Series Therefore, the series is ot absolutely coverget whe x =, so it is oly coditioally coverget there Thus, the series is absolutely coverget for 5 < x < ad it is coditioally coverget at x = = = =, 9 ( poits) Use series to carry out the itegratio Do t forget the +C! (a) x dx Solutio: This is x dx = (x ) dx = C + x x + dx = C + + = = = (b) si x x dx This is ( ) si x x dx = x ( ) x + ( ) x dx = dx ( + )! ( + )! = = ( ) = C + x ( ) x + dx = C + ( + )! ( + )! + = =
Math, Calculus II Witer 7 Fial Exam Solutios (c) (d) si(x ) dx Solutio: This is si(x ( ) x + ( ) x + ) dx = dx = dx ( + )! ( + )! = = ( ) = C + x + ( ) x + dx = C + ( + )! ( + )! + e x dx = Solutio: This is ( x ) ( ) x e x dx = dx = dx!! = = ( ) = C + x ( ) x + dx = C +!! + = = = (5 poits) Fid the 5-th degree Taylor polyomial (ie, oly up to the (x a) 5 -th term) geerated by f(x) = ( + x) / cetered at a = Solutio: To compute this, we eed to kow the values of the first 5 derivatives of f(x) = ( + x) / ad their values at a = : f(x) = ( + x) / = f() = f (x) = ( + x) / = f () = f (x) = ( + x) / = f () = f (x) = 8 ( + x) 5/ = f () = 8 f () (x) = 5 6 ( + x) 7/ = f () () = 5 6 f (5) (x) = 5 ( + x) 9/ = f (5) () = 5 Thus the 5-th degree Taylor polyomial geerated by f(x) = ( + x) / at a = is f() + f ()! = + / (x ) + f () (x ) + f ()!! x + / x + /8 = + x 8 x + 6 x 5 8 x + 7 56 x5 (x ) + f() ()! 6 x + 5/6 x + 5/ x 5 (x ) + f(5) () (x ) 5 5!