different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

Transcription

1 Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different methods (left endpoint, right endpoint, midpoint, trpezoid, Simpson s). 2. Find x 4x + 5dx. 3. Find ln xdx. 4. Find e x cos xdx. 5. Find x 3 e x2 dx. 6. Wht is the smllest vlue of n needed to ensure tht our numericl pproximtion method for 3 1 dx/x is within.1 = 1 4 using the different methods? 7. Find dx x. 8. Let popultion stisfy the eqution dn =.53N. Find the doubling time. 9. A logistic eqution is given by dp on the initil vlue. = r(1 P/K)P. Describe how the solutions depend 1. Describe the dynmics if there is constnt hrvesting occurring nd how it depends on h. 11. Solve dy = t/y. 12. When cn we compre n integrl with 1 how? dx x p to show convergence? Divergence? And True/Flse 13. True Flse One needs to lern the method of mthemticl induction in order to find pproximtions of res under functions using Riemnn sums for specific n (sy n = 5). 14. True Flse Antiderivtives re useful in t lest three plces: solving simple DE s, finding speeds nd distnces trvelled during free-flls, nd voiding using Riemnn sums when finding res fter we lern bout the Fundmentl Theorem of Clculus.

2 Mth 1A with Professor Stnkov Wednesdy, 12/6/ True Flse Despite the fct tht (ln x ) = 1/x for ny x, the integrl 1/xdx for x is strictly speking, not equl to ln x +C becuse the function 1/x is discontinuous, cusing us to use two different constnts for the negtive nd positive rel numbers. 16. True Flse The function e x2 hs no ntiderivtive in the form of n elementry function becuse no one hs been ble to find it. 17. True Flse Any continuous function on (, b) is integrble (i.e., it hs n ntiderivtive), but the converse is not true becuse some continuous functions do not hve derivtives. 18. True Flse To show tht the rule Integrl of product is the product of integrls is flwed it suffices to produce one counterexmple where it does not work. 19. True Flse The formul for the re of trpezoid (the product of the verge of the bses nd the height) cn be shown by dding up the res of the two tringles into which digonl divides the trpezoid. 2. True Flse Riemnn sums re somewht cumbersome tools for finding pproximtions of res, yet they re bsolutely necessry to link ntiderivtives to res. 21. True Flse To clculte the definite integrl x2 dx, we must find n ntiderivtive of 25 x 2 nd use the FTC I to evlute it t the ends of the intervl [ 5, 5]. 22. True Flse There re t lest three wys to compute π π sin(x)dx. 23. True Flse Splitting n integrl long its intervl s in b c f(x)dx mkes sense only when c is between nd b. b c 24. True Flse When we do not know n ntiderivtive of function nd we cnnot find the limit of the corresponding Riemnn sums on [, b], serching for geometric interprettion of the desired definite integrl is lso pointless. 25. True Flse Bounding f(x) on [, b] from bove nd below by some constnts M nd m produces only n estimte of b f(x)dx. 26. True Flse Geometric res cn be, fter ll, negtive if they pper underneth the x-xis nd bove some function f(x). 27. True Flse We cn turn limits of Riemnn sums into definite integrls nd vice vers. 28. True Flse The formul for the re of right trpezoid ppers in the geometric interprettion of the definite integrl 7 12 xdx.

3 Mth 1A with Professor Stnkov Wednesdy, 12/6/ True Flse When proving the bby definite ILL± nd IL*c, we need to descend ll the wy down to limits of Riemnn sums, nd pply long the wy the corresponding LL± nd LL*c. 3. True Flse FTCI sys tht if you strt with function F(x), then differentite it, nd then integrte it (ssuming ll these opertions re OK), you get bck the originl function F(x). 31. True Flse FTCII sys tht if you strt with function f(x), integrte f(u) from to x, nd then differentite with respect to x (ssuming ll these opertions re OK), you will get bck the originl function f(x). 32. True Flse The re-so-fr function F (x) = x f(u) du is > when f(x) increses, nd it is < when f(x) decreses. 33. True Flse A vlid PST sys tht if two functions re equl, then their derivtives re equl nd lso their integrls re equl (ssuming tht ech exists), so one my ttempt to tke derivtives (or integrls) on both sides of the equlity. 34. True Flse The Substitution Rule is relly ILL becuse it undoes for ntiderivtives wht CR does for derivtives. 35. True Flse (ln x ) = 1/ x for ll x. 36. True Flse Some ides for substitution tht work well in number of exmples re substituting wht is under rdicl, the denomintor of frction, nd n expression whose derivtive ppers in the numertor. 37. True Flse Checking your nswers fter hving done substitution is wste of time. 38. True Flse The re-so-fr function F (x) = x f(u) du is concve up where f(x) is incresing, nd it is concve down where f(x) is decresing. 39. True Flse There re two different wys to clculte definite integrls by SR (substitution rule): forget temporrily bout the bounds of integrtion, find n ntiderivtive, nd use FTCI; or go directly forwrd with SR while not forgetting to chnge the bounds of integrtion. 4. True Flse After hving done IP (integrtion by prts), checking your nswers by differentition is wste of effort since you hve lredy used vlid method(s) to clculte these integrls. 41. True Flse To justify IP on indefinite integrls, we pplied PR (product rule) nd FTCI. 42. True Flse When deciding which of two functions in n integrl h 1 (x)h 2 (x) dx will ply the role of u = f(x) nd which of v = g (x) in IP, we follow our intuition becuse integrtion is complicted process nd there re no guidelines to follow when doing IP.

4 Mth 1A with Professor Stnkov Wednesdy, 12/6/ True Flse If in the integrl h 1 (x)h 2 (x) dx we see tht h 1 (x) hs simpler (or comprble in difficulty) ntiderivtive while h 2 (x) s derivtive is simpler thn h 2 (x), we go for IP with u = f(x) = h 2 (x) nd v = g (x) = h 1 (x). 44. True Flse If in the integrl h 1 (x)h 2 (x) dx ech of the functions h 1 (x) nd h 2 (x) hs eqully complicted derivtives nd integrls s itself, then there is no point in pplying IP, since it will turn the integrl into n eqully hrd integrl. 45. True Flse When one of the functions h 1 (x) nd h 2 (x) in the integrl h1 (x)h 2 (x)dx is x 2 nd we wnt to solve the problem vi IP, then we must set u = f(x) = x 2, becuse if we do v = g (x) = x 2 this will mke g(x) = x 3 /3, which is more complicted thn x 2 nd hence it will complicte our problem. 46. True Flse The formuls for the error bounds for the vrious pproximtion rules for f(x)dx cn be used to find the exct errors for these pproximtions. b 47. True Flse The Trpezoidl Rule sum is the verge of the Right Endpoint nd Left Endpoint sums for b f(x)dx. 48. True Flse To estimte n pproximtion mens to find out t most how fr it cn be from the exct vlue, nd hence this is not useful since we either don t know the exct vlue, or if we knew it, we wouldn t be even pproximting, much more so estimting n pproximtion of it. 49. True Flse Simpson s Rule uses degree 4 polynomils to better pproximte the shpe of the grph of f(x), s evidenced by the fourth derivtive nd the 4th power n 4 in the formul for the error of the Simpson s pproximtion: E S K 4(b ) 5. 18n 4 5. True Flse The lrger the difference between the mximum nd the minimum of f (x) on [, b], the bigger the estimtes of the errors for E L nd E R will turn out to be. 51. True Flse To find out how fr we hve to go with the number of subintervls n of [, b] in order to ensure tht our pproximtion is close enough to the true vlue of b f(x)dx, we need to set up n inequlity using formul for the error bounds nd solve it for n. 52. True Flse For the sme number n of subintervls, the Midpoint Rule tends to be more precise thn the Trpezoidl Rule, but Simpson s Rule is lwys more precise thn the Trpezoidl Rule. 53. True Flse If function is concve down, we cn obtin n overestimte by pplying either the Right Endpoint Rule or the Left Endpoint Rule. 54. True Flse Infinitely mny continuous functions do NOT hve ntiderivtives in elementry functions, but they still do hve ntiderivtives, s shown by using the re-so-fr function nd pplying FTC II to it.

5 Mth 1A with Professor Stnkov Wednesdy, 12/6/ True Flse We cn solve the exponentil growth model DE y (t) = ky(t) only by guessing tht y(t) is n exponentil function. 56. True Flse The logistic model DE is modifiction of the exponentil growth model, tking into ccount tht environmentl resources my be limited to llow unrestricted exponentil growth forever. 57. True Flse To solve the logistic model DE P (t) = kp (1 P ), we need to integrte K both sides nd pply integrtion by prts on the RHS. 58. True Flse The logistic model DE cn be modified to ccount for constnt hrvesting rte h by subtrcting h from the RHS of the DE. 59. True Flse The growth rte of P (t) in the logistic model is the logrithmic derivtive of P (t). 6. True Flse The reltive growth rte in the exponentil decy model remins constnt for ll t. 61. True Flse The method of seprble DE cn be pplied only when the RHS of DE dy = f(y, t) cn be somehow written s product of function in y lone nd function in t lone. 62. True Flse The hlf-life of rdioctive element during exponentil decy depends on the initil mount of this element. 63. True Flse We cn get pretty good ide of the solutions to the hrvesting modifiction of the logistic model DE P (t) = kp ( ) 1 P K h by fctoring the qudrtic polynomil in P on the RHS nd studying where it is positive, negtive, or. 64. True Flse In the modified the logistic model dp ( = kp (t) 1 P (t) K ) h, there is vlue of the hrvesting rte h for which P (t) hs unique equilibrium, bove which ll solutions re decresing to this equilibrium nd below which the popultion becomes extinct. 65. True Flse We cn show tht 1 dx converges in t lest three wys: by 5 x 1.1 brute force clcultion using the definition of n improper integrl, by representing 1 dx s prt of 1 dx nd then using formul 5 x x 1.1 from clss for the vlue of the ltter integrl, or by compring it with the more fmilir to us integrl 1 dx. 5 x True Flse If we cnnot compute the exct vlue of n improper integrl b f(x)dx, we could try to compre it with nother integrl b g(x)dx, but tht, if successful, would only tell us if the originl integrl converges or diverges. 67. True Flse The vlue of sin x dx depends on where we stop the vrible t when clculting the limit of the proper integrls t sin x dx.

6 Mth 1A with Professor Stnkov Wednesdy, 12/6/ True Flse To show tht e x2 dx converges, it is enough to compre it with e x dx. 69. True Flse If g(x) f(x) on [, ) nd converges. f(x)dx converges, then g(x)dx lso 7. True Flse Improper integrls in Sttistics re used, for exmple, to compute the res under probbility distribuition functions. 71. True Flse The qudrtic ( formul ) is useful when fctoring the RHS of the DE dp = kp (t) 1 P (t) h. K 72. True Flse A semistble equilibrium is obtined for the modified logistic model when the hrvesting constnt h is such tht the qudrtic eqution in P, kp ( ) 1 P K h, hs unique root for P. 73. True Flse The vlue of convergent two-sided improper integrl f(x) dx for continuous function f(x) my depend on where we split the integrl s sum of two one-sided improper integrls f(x) dx + f(x) dx; however, the divergence of such n integrl does NOT depend on the prticulr we choose. 74. True Flse If we lredy know tht f(x) dx converges, then we cn compute t it by choosing symmetric bus stops ; i.e., s lim t f(x) dx; yet, t until we know tht the integrl converges we cnnot do this nd we must compute insted lim t t t t f(x) dx. 75. True Flse The improper integrl e x2 dx converges becuse the integrnd function is even nd the integrl on the right hlf on the number line e x2 dx is lredy shown to converge. 76. True Flse Integrls cn be improper in more thn two plces, but in this clss we will concentrte mostly on improper integrls of functions without infinite discontinuities becuse PDFs will be generlly continuous or piecewise continuous. 77. True Flse 3 dx x 1 = ln 2.