托马斯微积分-Thomas` CALCULUS 课后习题答案cap11b
托马斯微积分ThomasCALCULUS课后习题答案附录
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托马斯微积分第13版第七章答案
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CHAPTER 7 TRANSCENDENTAL FUNCTIONS
7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal line test. 2. Not one-to-one, the graph fails the horizontal line test. 3. Not one-to-one since (for example) the horizontal line y 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal line test. 5. Yes one-to-one, the graph passes the horizontal line test. 6. Yes one-to-one, the graph passes the horizontal line test. 7. Not one-to one since the horizontal line y 3 intersects the graph an infinite number of times. 8. Yes one-to-one, the graph passes the horizontal line test. 9. Yes one-to-one, the graph passes the horizontal line test. 10. Not one-to one since (for example) the horizontal line y 1 intersects the graph twice. 11. Domain: 0 x 1, Range: 0 y 12. Domain: x 1, Range: y 0
托马斯微积分
Figure 2.43: The balloon in Example 3.
Chapter 2 ET. Finney Weir Giordano, Thomas’ Calculus, Tenth Edition © 2001. Addison Wesley Longman All rights reserved. Chapter 2ET, Slide 7
Figure 2.31: sin (x°) oscillates only /180 times as often as sin x oscillates. Its maximum slope is /180. (Example 9)
Chapter 2 ET. Finney Weir Giordano, Thomas’ Calculus, Tenth Edition © 2001. Addison Wesley Longman All rights reserved. Chapter 2ET, Slide 15
Figure 2.51: The position of the curve y = (a h – 1) /h, a > 0, varies continuously with a.
Chapter 2 ET. Finney Weir Giordano, Thomas’ Calculus, Tenth Edition © 2001. Addison Wesley Longman All rights reserved. Chapter 2ET, Slide 16
微积分课后习题六答案
微积分课后习题六答案微积分课后习题六答案微积分是一门重要的数学学科，它研究的是函数的变化和极限。
在学习微积分的过程中，课后习题是巩固知识和提高能力的重要途径。
本文将为大家提供微积分课后习题六的答案，希望能帮助大家更好地理解和掌握微积分知识。
1. 求函数f(x) = x^2在区间[0,1]上的定积分。
解：根据定积分的定义，我们可以将区间[0,1]等分成n个小区间，每个小区间的长度为Δx = (1-0)/n = 1/n。
然后，我们在每个小区间中选择一个代表点xi，计算函数在该点的函数值f(xi)，并将其乘以小区间的长度Δx。
最后，将所有小区间的函数值乘以对应的长度Δx后相加，即可得到定积分的近似值。
当n趋向于无穷大时，这个近似值将趋向于定积分的真实值。
即：∫[0,1] x^2 dx = lim(n→∞) ∑[i=1,n] f(xi)Δx= lim(n→∞) ∑[i=1,n] (xi)^2 * (1/n)= lim(n→∞) (1/n) * ∑[i=1,n] (xi)^2由于区间[0,1]上的任意小区间长度都是相等的，所以我们可以将其简化为：∫[0,1] x^2 dx = lim(n→∞) (1/n) * ∑[i=1,n] (i/n)^2= lim(n→∞) (1/n) * ∑[i=1,n] i^2/n^2= lim(n→∞) (1/n^3) * ∑[i=1,n] i^2根据数学公式∑[i=1,n] i^2 = n(n+1)(2n+1)/6，代入上式，得到：∫[0,1] x^2 dx = lim(n→∞) (1/n^3) * [n(n+1)(2n+1)/6]= lim(n→∞) (2n^3 + 3n^2 + n)/(6n^3)= 1/3所以，函数f(x) = x^2在区间[0,1]上的定积分为1/3。
2. 求函数f(x) = e^x在区间[0,2]上的定积分。
解：与上题类似，我们可以将区间[0,2]等分成n个小区间，每个小区间的长度为Δx = (2-0)/n = 2/n。
托马斯微积分 几何与多元微积分B上 - 副本
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参考答案
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1).理解序列、子序列、有界序列的概念， 以及递归法定义序列。 2). 会判别序列的敛散性, 会求序列极限。
8.3 无穷级数 8.4 非负项级数 8.5交错级数、绝对收敛和条件收敛
1).理解常数项级数的概念、了解收敛级数的性质。 2). 对正项级数、交错级数会判断其敛散性。 3). 对任意项级数会判断绝对收敛与条件收敛。
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1. The point ()2,4P lies on the curve x y =. If Q is the point (x , use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the value .99.3=x2. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,3].3.Find the limit.()379lim 25++→x x x4.Find the limit.()51lim 20+-→x x x x5. If ,22)(12++≤≤x x x f for all x find the limit.)(lim 1x f x -→6. Find the limit. 22lim |2|x x x →--7. Evaluate the limit.()xx x 11022lim --→-+8. Use the definition of the derivative to find (2)f '-, where 3()2f x x x =-.9. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).10. Use a graph to find a number N such that 3.031235622<---+x x x whenever N x >.11. If ()g x ()g x '.12. A machinist is required to manufacture a circular metal disk with area 21000cm .a) What radius produces such a disk? b) If the machinist is allowed an error tolerance of 25cm ±in the area of the disk, how close to the idealradius in part (a) must the machinist control the radius?13. Use a graph to find a number δ such that6.0314<-+x whenever .2δ<-x14. For the limit, illustrate the definition by finding values of δ that correspond to .25.0=ε31lim(43)2x x x →+-=15. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩16. For x = 5, determine whether f is continuous from the right, from the left, or neither.17. If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour,then Torricelli's Law gives the volume of water remaining in the tank after t minutes as2651000,100)(⎪⎭⎫ ⎝⎛-=t t V , 600≤≤tFind the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t .18.Find the derivative of the function.25314)(x x x f +-=19. If 2313)(tt f += find )(t f '.20. At what point is the function ()|6|f x x =- not differentiable.ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form A1. 0.2501562. 3/m s3. 2634. -∞5. 16. Limit does not exist7. -1/48. 109. 1016y x =-10. 9≥N11.(),3/5-∞12. cm , 0.0445cm13. 81.0≤δ14. 030.0≤δ15. 03at and16. neither17. ⎪⎭⎫ ⎝⎛--=65165200000t y 18. 310-x19. )3(26t t +- 20. 61.The point P (4, 2) lies on the curve .x y = If Q is the point ()x x ,, use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the value of .01.4=x2.The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t is measured in seconds. Find the average velocity over the time period [1,1.5].3.Find the limit, if it exists.44lim |4|x x x →-- 4. Find )(a f '.252)(x x x f -+=5. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).6. Evaluate the function 222)(--=x x x f at the given numbers (correct to six decimal places). Use the results to guess the value of the limit ).(lim 2x f x →7. The graph of f is given. State the numbers at which f is not differentiable.⎪⎭⎫ ⎝⎛→x x x 3cos lim 909. If 66)(12++≤≤x x x f for all x find the limit.)(lim 1x f x -→10.Evaluate the limit.()867lim 25++→x x x11. Evaluate the limit.()x x x 11022lim --→-+12.If an arrow is shot upward on the moon, with a velocity of 70 m/s its height (in meters) after t seconds is given by .99.070)(2t t t H -= With what velocity will the arrow hit the moon?13. The cost (in dollars) of producing x units of a certain commodity is .08.013336,4)(2x x x C ++= Find the average rate of change with respect to x when the production level is changed from 101=x to .103=x14.Let ()20()30333if x f x xif x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate each limit, if it exists.0lim ()x f x +→ b.) 0lim ()x f x -→15.If f and g are continuous functions with 3)3(=f and []3)()(3lim 3=-→x g x f x , find ).3(g16.Evaluate the limit. 9lim 9+-→x x17.Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.1ε=.then Torricelli's Law gives the volume of water remaining in the tank after t minutes as2651000,100)(⎪⎭⎫ ⎝⎛-=t t V , 600≤≤tFind the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t .19. If ()g x ()g x '.20. For the function f whose graph is shown, state the following.)(lim 4x f x -→ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form B1. 0.2498442.2.625 m/s3. Limit does not exist4. a 101-5.1016y x =- 6.(1.6, 0.7465), (1.8, 0.7257), (1.9, 0.7161), (1.99, 0.7079), (1.999, 0.707195), (2.4, 0.674899), (2.2, 0.690261), (2.1, 0.698482), (2.01, 0.706225), (2.001, 0.707018), Limit = 0.707107 7. 1,0,3-8. 09. 110. 21311. -1/412. -7013. 29.3214. a.) 3 b.) 015. 616. 017. 0.025δ=18. ⎪⎭⎫ ⎝⎛--=65165200000t y 19. (),3/5-∞20. -∞1. A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate inbeats per minute. The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with t = 38 and t = 42.Select the correct answer.a. -89b. 180c. 90d. 100e. 89f. 952. If an arrow is shot upward on the moon with a velocity of 55 m/s, its height in meters after t seconds is given by .04.0552t t h -= Find the average velocity over the interval [1, 1.04].Select the correct answer.a. 54.9194b. 55.0284c. 54.8174d. 54.9184e. 54.90843. The displacement (in feet) of a certain particle moving in a straight line is given by 3/8s t = where t is measuredin seconds. Find the average velocity over the interval [1, 1.8].Select the correct answer.a. 0.865b. 0.654c. 0.765d. 0.756e. 0.745f. 0.7554. For the function f whose graph is shown, find the equations of the vertical asymptotes.Select all that apply.a. x = -7b. x = 9c. x = 5d. x = -3e. x = 10f. x = -25. Find the limit, if it exists55lim |5|x x x →--Select the correct answer.a. 5b. 1-c. 1-d. 0e. limit does not exist6. Find the limit.lim x →-∞Select the correct answer.a. -1/2b. 3c. 3-d. 0e. limit does not exist7. Evaluate the limit.()()62lim 231-+→x x xSelect the correct answer.a. 27b. -45c. -135d. 29e. -1258.If 88)(12++≤≤x x x f for all x , find )(lim 1x f x -→. Select the correct answer.a. 1b. 8c. -1/8d. -1/16e. The limit does not exist9. Evaluate the limit.⎪⎭⎫ ⎝⎛→x x x 5cos lim 90Select the correct answer.a. -5b. 1c. 0d. 5e. The limit does not exist10. Use a graph to find a number δ such that 2.021sin <-x whenever δπ<-6x .Round down the answer to the nearest thousandth.Select the correct answer.a. 218.0≤δb. 368.0≤δc. 401.0≤δd. 251.0≤δe. 425.0≤δ11. A machinist is required to manufacture a circular metal disk with area 1000 cm 2. If the machinist is allowed an error tolerance of ±10 cm 2 in the area of the disk, how close to the ideal radius must the machinist control the radius?Round down the answer to the nearest hundred thousandth.Select the correct answer.a. cm 08898.0≤δb. cm 08908.0≤δc. cm 08999.0≤δd. cm 08913.0≤δe. cm 09913.0≤δ12. Consider the function x e x f /121)(+=. Find the value of -→0)(lim x x f . Select the correct answer.a. 1.5b. -0.1c. 0.1d. 0.9e. 0.513.Choose an equation from the following that expresses the fact that a function f is continuous at the number 6.Select the correct answer.a. 6)(lim =∞→x x fb. )6()(lim 6f x f x =→c. )6()(lim f x f x =∞→d. 0)(lim 6=→x x fe. ∞=→6)(lim x x f14. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Select the correct answer.a. 03andb. 0onlyc. 3onlyd. 03and -e. 3only -15. Use the definition of the derivative for find (2)f '-, where 3()2f x x x =-.Select the correct answer.a. 4b. 10c. -4d. -10e. none of these16.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5)-∞d. (),3/5-∞e. ()0,∞17. Find an equation of the tangent line to the curve 353+-=x x y at the point (2, 1).Select the correct answer.a. 138+=x yb. 139--=x yc. 137-=x yd. 137+-=x ye. 157-=x yStewart - Calculus ET 6e Chapter 2 Form C18. The cost (in dollars) of producing x units of a certain commodity is 201.019571,4)(x x x C ++=. Find theinstantaneous rate of change with respect to x when x = 103. (This is called the marginal cost .)Select the correct answer.a. 26.06b. 20.06c. 21.06d. 18.06e. 31.0619. If the tangent line to )(x f y = at (8, 4) passes through the point (5, -32), find )8(f '.Select the correct answer.a. 24)8(='fb. 20)8(='fc. 12)8(-='fd. 12)8(='fe. 32)8(='f20. At what point is the function ()|6|f x x =- not differentiable.Select the correct answer.a. 6b. 6-c. 1d. 1-e. 0ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form C1. c2. d3. f4.a, b, c, f5. e6. a7. c8. a9. c10. a11. a12. e13. b14. a15. b16. d17. c18. c19. d20. a1. The position of a car is given by the values in the following table.Find the average velocity for the time period beginning when t = 2 and lasting 2 seconds.Select the correct answer.a. 35.5b. 47.5c. 39d. 37.5e. 33.52. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,3].Select the correct answer.a. 3/m sb. 3.5/m sc. 1/m sd. 1.5/m se. none of these3. Find the limit.()71lim 20++→x x x xSelect the correct answer.a. 0b. 71c. 71- d. -∞ e. ∞4. Find the limit.lim x →-∞Select the correct answer.a. -1b. 0c. 1/2d. -∞e. -1/25.The slope of the tangent line to the graph of the exponential function xy 8= at the point (0, 1) is x x x 18lim 0-→. Estimate the slope to three decimal places.Select the correct answer.a. 1.293b. 2c. 2.026d. 1.568e. 2.079f. 2.5566. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c. 16y x =-d. 1016y x =+e. none of these7.Find the limit.()10lim tan 1/x x +-→Select the correct answer.a. 0b. ∞c. /2πd. /3πe. π .8. Let |1|1)(2--=x x x F . Find the following limits.),(lim 1x F x +→ )(lim 1x F x -→Select the correct answer.a. both 2b. 2 and 1c. 2 and – 2d. 2 and – 1e. both 19. Use continuity to evaluate the limit.()x x x sin 4sin lim 13+→πSelect the correct answer.a. π13b. - 1c. 0d. ∞e. 110.Let ()20()30333if x f x xif x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate the limit, if it exists.0lim ()x f x -→Select the correct answer.e. 3-11. For what value of the constant c is the function f continuous on ()?,∞∞-⎩⎨⎧>-≤+=2527)(2x for cx x for cx x fSelect the correct answer.a. 1=cb. 2=cc. 6=cd. 2-=ce. 7=c12. Find a function g that agrees with f for 25≠x and is continuous on ℜ.xx x f --=255)(Select the correct answer.a. x x g -=51)(b. x x g +=251)(c. x x g +=51)(d. xx g -=51)( e. x x g -=55)(13.Which of the given functions is discontinuous?Select the correct answer.a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x fb. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous14.Find the limit. 13lim 232-++∞→t t t tSelect the correct answer.a. ∞b. 0c. 3-d. 3e. 215.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5)-∞d. (),3/5-∞e. ()0,∞16. The cost (in dollars) of producing x units of a certain commodity is .03.013280,4)(2x x x C ++= Find theaverage rate of change with respect to x when the production level is changed from x = 102 to x = 118.Select the correct answer.a. 29.6b. 19.6c. 18.6d. 26.6e. 24.617. Evaluate the limit.|2|lim 2+-→x xSelect the correct answer.a. 2b. 4c. - 2d. 0e. The limit does not exist18. If a ball is thrown into the air with a velocity of 58 ft/s, its height (in feet) after t seconds is given by .11582t t H -=Find the velocity when t = 4.Select the correct answer.a. 27ft/sb. 30ft/sc. 31ft/sd. 25ft/se. 37ft/s19. Is there a number a such that 626lim 223-++++-→x x a ax x x exists? If so, find the value of a and the value of the limit. Select the correct answer.a. a =14, limit equals 1.4b. a =17, limit equals 1.6c. a =28, limit equals 1.4d. a =28, limit equals 1.6e. There is no such number20.If ()g x ()g x '.Select the correct answer.a. ()1/25()352g x x -'=-- b. ()1/21()352g x x '=-- c. ()2()35g x x '=-- d. ()25()352g x x -'=-- e. none of theseANSWER KEYStewart - Calculus ET 6e Chapter 2 Form D1. d2. a3. e4. e5. e6. e7. c8. c9. c10. a11. c12. c13. b14. b15. d16. b17. d18. b19. d20. a1.A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. If P is the point (15, 263) on thegraph of V, fill the table with the slopes of the secant lines PQ where Q is the point on the graph with the corresponding t .Enter your answer to two decimal places.2. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].3. If an arrow is shot upward on the moon with a velocity of 57 m/s, its height in meters after t seconds is given by 282.057t t h -=. Find the instantaneous velocity after one second.Select the correct answer.a. 55.46b. 55.35c. 55.25d. 55.36e. 55.374. Given that, 3)(lim 7-=→x f x and 9)(lim 7=→xg x . Evaluate the limit.)()()(2lim 7x f x g x f x -→5. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c.16y x =- d. 1016y x =+ e. none of these6. Let ()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate the limit, if it exists.0lim ()x f x -→Select the correct answer.e. 3-7. For the function f whose graph is shown, find the following.)(lim 7x f x →8.For x = 5, determine whether f is continuous from the right, from the left, or neither.9. Evaluate the limit.()xx x 11077lim --→-+10. Let |1|1)(2--=x x x FFind the following limits.)(lim ),(lim 11x F x F x x -+→→11. Use a graph to find a number δsuch that 3|0.6< whenever |2|x δ-<.Round down the answer to the nearest hundredth.12. Is there a number a such that 6810lim 223-++++-→x x a ax x x exists? If so, find the value of a and the value of the limit.Select the correct answer.a. a =49, limit equals 1.6b. a =13, limit equals 2.2c. a =49, limit equals 2.2d. a =19, limit equals 1.6e. a =49, limit equals 2.713. How close to 2 do we have to take x so that 5x + 3 is within a distance of 0.025 from 13?14. Find a function g that agrees with f for 25≠x and is continuous on .ℜxx x f --=255)( 15. Use the given graph of x x f =)( to find a number δ such that 4.0|2|<-x whenever .|4|δ<-x16.If ()g x ()g x '.17.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5]-∞d. (),3/5-∞e. ()0,∞18. At what point is the function |6|)(x x f -= not differentiable.19. How close to - 9 do we have to take x so that ()?10000914>+x20.Find the derivative of the function.25314)(x x x f +-=ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form E1. 5, -42.3, 10, -43.6,20, -18.4, 25, -24.1, 30, -17.532. 2.625 m/s3.d 4.-1/2 5. e6. a7.-∞ 8.neither 9. -1/4910. 2, -211. 81.0≤δ12. c13. 005.0|2|<-x14. ()x g +=5115. 44.1≤δ 16. ()1/25()352g x x -'=-- 17. d18. 619. 1.0|9|<+x20. 310-x1. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].2. If a ball is thrown into the air with a velocity of 45 ft/s, its height in feet after t seconds is given by 21545t t y -=. Find the instantaneous velocity when 4=t .3. If 5.4)(lim 3=-→x f x , then if )(lim 3x f x → exists, to what value does it converge?Select the correct answer.a. 6.5b. 4.5c. 1d. 2e. 64. For the function f whose graph is shown, find the limit.)(lim 9x f x +-→5. The function has been evaluated at the given numbers (correct to six decimal places). Use the results to guess the value of the limit.112)(--=x x x f________)(lim 1=→x f xSelect the correct answer.a. 1.255039b. 1.911314c. 1.969944d. 1.473889e. 16.Evaluate the limit.()()104lim 251-+→x x x7. Find the limit.lim x →-∞8.Find the limit.()10lim tan 1/x x +-→9.Evaluate the limit and justify each step by indicating the appropriate properties of limits.393198lim 22-++-∞→x x x x x10. Find an equation of the tangent line to the curve 34x y =at the point ()256,4--.11. Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.1ε=.12. Use a graph to find a number δsuch that 1.021sin <-x whenever δπ<-6x .Round down the answer to the nearest thousandth.13. Use the definition of the limit to find values of δ that correspond to 75.0=ε.Round down the answer to the nearest thousandth.()234lim 31=-+→x x x14. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩15. If f and g are continuous functions with 2)2(=f and [],2)()(2lim 2=-→x g x f x find )2(g .16. Find the limit.)(lim 22bx x ax x x +-+∞→17.State the domain.()sin F x =18.Find the derivative of the function using the definition of derivative.22919)(x x x f +-=19. Find a function g that agrees with f for 4≠x and is continuous on ℜ.xx x f --=42)(20. If an arrow is shot upward on the moon, with a velocity of 70 m/s its height (in meters) after t seconds is given by.99.070)(2t t t H -= With what velocity will the arrow hit the moon?ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form F1. 2.625 m/s2. -753. b4. -∞5. e6. 28125-7. -1/28.2π 9. 38 10. 512192+=x y11. 0.025δ=12. 112.0≤δ13. 085.0≤δ14.03at and 15.2 16. 2b a - 17. ),6[∞18.94-x 19.x g +=2120.-701. If 5.4)(lim 2=-→x f x , then if )(lim 2x f x →exists, to what value does it converge?Select the correct answer.a. 2b. 1c. 5d. 4.5e. 1.52. Consider the following function.()111111)(2≥<≤--<⎪⎩⎪⎨⎧--=x x x x x x x fDetermine the values of a for which )(lim x f ax →exists.3. Evaluate the limit and justify each step by indicating the appropriate properties of limits.443398lim 22-++-∞→x x x x x4. Find )(a f '.233)(x x x f -+=5. Guess the value of the limit.3055tan 3lim xx x x -→Select the correct answer.a. 121b. 135c. 134d. 130e. 1256. Given that 8)(lim 7-=→x f x and 10)(lim 7=→x g x .Evaluate the limit.())()(lim 7x g x f x +→7.Evaluate the limit.()()101lim 231-+→x x x8. Evaluate the limit.⎪⎪⎭⎫⎝⎛--→45lim 233x x x9. Find the derivative of the function using the definition of the derivative.2610)(x x x f +-=10.Let |9|81)(2--=x x x FFind the following limits.)(lim ),(lim 99x F x F x x -+→→Select the correct answer.a. 18 and 9b. 18 and - 18c. both 18d. 18 and – 9e. 81 and 911.Use the given graph of x x f =)(to find a number δsuch that 4.0|2|<-x whenever .|4|δ<-x12. Use a graph to find a number δsuch that 5.0|314|<-+x whenever .|2|δ<-x13. For the limit, illustrate the definition by finding values of δthat correspond to .5.0=ε()234lim 31=-+→x x x14. Find the slope of the tangent line to the curve 35x y = at the point (-4, -320).15. At what point is the function |8|)(x x f -= not differentiable.16.Which of the given functions is discontinuous?a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x f b. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous17.Select the right number for the following limit and prove the statement using the ,δε definition of the limit. 3183lim 23--+→x x x xSelect the correct answer.a. 6b. 8c. 5d. 9e. 1818.Prove the statement using the ,δε definition of the limit.0|2|lim 2=-→x x19.Prove the statement using the ,δε definition of the limit.()241lim 25=--→x x20.Use continuity to evaluate the limit.()x x x sin 3sin lim 17+-→πSelect the correct answer. a. π17- b. ∞ c. -1 d. 0 e. 1ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form G1. d2. ()()()∞--∞-,11,11,3.38 4.a 61- 5.e 6. 27. -728.22/5 9. 112-x10. b11. 44.1≤δ12. 6875.0≤δ13. 056.0≤δ14. 24015. 816. b17. d18. Given 0>ε, we need 0>δsuch that if | x - 2 | δ< then | | x - 2 | - 0 | ε<. But | | x - 2 | | = | x - 2 |. So this is true ifwe pick .εδ=19. Given 0>ε, we need 0>δsuch that if | x - ( - 5 ) | δ< then | ( x 2 - 1 ) - 24 | ε< or upon simplifying we need | x2 – 25| ε<whenever | x + 5 | δ<. Notice that if | x + 5 | < 1 , then- 1 < x + 5 < 1 - 11 < x - 5 < - 9 | x - 5 | < 11. So take =δmin {ε / 11, 1}. Then | x - 5 | < 11 and | x + 5 | ε</ 11, so | ( x 2 - 1 ) - 24 | = | ( x + 5 ) ( x - 5 ) | = | x + 5 | | x - 5 | < (ε / 11 ) ( 11 ) =ε. Therefore, by the definition of a limit, ().241lim 25=--→x x 20.dStewart - Calculus ET 6e Chapter 2 Form H1. The point P (4, 2) lies on the curve x y =. If Qis the point (,x , use your calculator to find the slope of thesecant line PQ (correct to six decimal places) for the value of 99.3=x .Select the correct answer.a. m PQ = 0.250157b. m PQ = 0.250156c. m PQ = - 0.250154d. m PQ = - 0.250156e. m PQ = 0.2501542. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].3. The displacement (in feet) of a certain particle moving in a straight line is given by 83t s =where t is measured in seconds. Find the instantaneous velocity when t = 3.4. If ,5.7)(lim 2=+→x f x then if )(lim 2x f x →exists, to what value does it converge?Select the correct answer. a. 5 b. 8.5 c. 8 d. 11.5 e. 7.55.If f and g are continuous functions with 3)2(=f and [],5)()(3lim 2=-→x g x f x find ).2(g6. The slope of the tangent line to the graph of the exponential function xy 4=at the point (0, 1) is .14lim 0x x x -→ Estimate the slope to three decimal places. Select the correct answer.a. 1.045b. 1.136c. 0.786d. 1.126e. 1.3867. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c. 16y x =-d. 1016y x =+e. none of these8. Find the limit.lim x →-∞9. How close to 2 do we have to take x so that 5x + 3 is within a distance of 0.075 from 13?10. Evaluate the limit and justify each step by indicating the appropriate properties of limits.693958lim 2-++-∞→x x x x x11. Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.01ε=.12. Use the given graph of 2)(x x f =to find a number δsuch that 2112<-x whenever δ<-1x .Round down the answer to the nearest hundredth.13.If ()g x ()g x '.14.If ()g x ()g x '.15.Let ()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate each limit, if it exists.a.) 0lim ()x f x +→b.) 0lim ()x f x -→16.Which of the given functions is discontinuous?Select the correct answer.a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x f b. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous17.If a ball is thrown into the air with a velocity of 62 ft/s, its height (in feet) after t seconds is given by21662t t H -=.Find the velocity when t = 5.18.Use continuity to evaluate the limit.()x x x sin 6sin lim 8+→πSelect the correct answer.a. ∞b. - 1c. 1d. 0e. π819. Find a function g that agrees with f for 16≠x and is continuous on ℜ.xx x f --=164)( 20. Consider the function .11)(/1x e x f +=Find the value of )(lim 0x f x +→.Select the correct answer.a. -0.8b. -0.5c. 0.3d. 0e. 0.8ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form H1. b2. 2.625 m/s3. 3.3754. e5. 46. e7. e8. -1/29.015.0|2|<-x 10. 38 11. 0.0025δ=12. 22.0≤δ13. ()1/25()352g x x -'=-- 14. (),3/5-∞15. a.) 3 b.) 016. b17. -9818. d19. x g +=4120. d。
大学_微积分(刘书田著)课后习题答案下载
微积分（刘书田著）课后习题答案下载微积分（刘书田著）课后答案下载人类对自然的认识永远不会止步，微积分这门学科在现代也一直在发展着。
以下列举了几个例子，足以说明人类认识微积分的水平在不断深化。
在黎曼将柯西的积分含义扩展之后，勒贝格又引进了测度的概念，进一步将黎曼积分的含义扩展。
例如著名的狄利克雷函数在黎曼积分下不可积，而在勒贝格积分下便可积。
[6]前苏联前苏联著名数学大师舍盖索伯列夫为了确定偏微分方程解的存在性和唯一性，建立了广义函数和广义导数的概念。
这一概念的`引入不仅赋予微分方程的解以新的含义，更重要的是，它使得泛函分析等数学工具得以应用到微分方程理论中，从而开辟了微分方程理论的新天地。
美国美籍华裔数学大师陈省身所研究的微分几何领域，便是利用微积分的理论来研究几何，这门学科对人类认识时间和空间的性质发挥着巨大的作用，并且这门学科至今仍然很活跃。
前不久由俄罗斯数学家佩雷尔曼完成的庞加莱猜想便属于这一领域。
中国中国的数学爱好者发现了积乘和微商，使微积分的内容进一步拓展。
微积分（刘书田著）：基本内容数学分析研究函数，从量的方面研究事物运动变化是微积分的基本方法。
这种方法叫做数学分析。
从广义上说，数学分析包括微积分、函数论等许多分支学科，但是现在一般已习惯于把数学分析和微积分等同起来，数学分析成了微积分的同义词，一提数学分析就知道是指微积分。
微积分微积分的基本概念和内容包括微分学和积分学。
微分学的主要内容包括：极限理论、导数、微分等。
积分学的主要内容包括：定积分、不定积分等。
微积分（刘书田著）：现代发展点击此处下载微积分（刘书田著）课后答案。
微积分课后题答案
微 积 分 课 后 习 题 答 案习 题 一 （A ）1．解下列不等式，并用区间表示不等式的解集：（1）74<-x ； （2）321<-≤x ；（3）)0(><-εεa x ； （4）)0,(0><-δδa x ax ；（5）062>--x x ；（6）022≤-+x x ．解：1)由题意去掉绝对值符号可得：747<-<-x ，可解得j .113．x <<-即)11,3(-. 2）由题意去掉绝对值符号可得123-≤-<-x 或321<-≤x ，可解得11≤<-x ,53<≤x .即]5,3[)1，1(⋃-3）由题意去掉绝对值符号可得εε<-<-x ,解得εε+<<-a x a .即)a , (εε+-a ；4）由题意去掉绝对值符号可得δδ<-<-0x ax ,解得ax x ax δδ+<<-00，即ax a x δδ+-00 , () 5）由题意原不等式可化为0)2)(3(>+-x x ,3>x 或2-<x 即）（3， 2） ， (∞+⋃--∞. 6）由题意原不等式可化为0)1)(2(≤-+x x ,解得12≤≤-x .既1] , 2[-.２．判断下列各对函数是否相同，说明理由： （1）x y =与x y lg 10=； （2）xy 2cos 1+=与x cos 2；（3）)sin (arcsin x y =与x y =；（4）)arctan (tan x y =与x y =；（5）)1lg(2-=x y 与)1lg()1lg(-++=x x y ； （6）xxy +-=11lg 与)1lg()1lg(x x x +--=．解：1)不同，因前者的定义域为) , (∞+-∞，后者的定义域为) , 0(∞+； 2)不同,因为当))(2 , )212((ππ23k k x k ++∈+∞-∞- 时，02cos 1 >+x ,而0cos 2<x ；3）不同，因为只有在]2, 2[ππ-上成立； 4）相同；5)不同，因前者的定义域为) , (11) , (∞+⋃--∞)，后者的定义域为) , 1(∞+； 6）相同3．求下列函数的定义域（用区间表示）： （1）1)4lg(--=x x y ； （2）45lg 2x x y -=；（3）xx y +-=11； （4）)5lg(312x x x y -+-+-=； （5）342+-=x x y ；（6）xy xlg 1131--=；（7）xy x-+=1 lg arccos 21； （8）6712arccos2---=x x x y ．解：1)原函数若想有意义必须满足01>-x 和04>-x 可解得 ⎩⎨⎧<<-<41 1x x ,即)4 , 1()1 , (⋃--∞.2)原函数若想有意义必须满足0452>-x x ,可解得 50<<x ,即)5 , 0(.3)原函数若想有意义必须满足011≥+-xx,可解得 11≤<-x ,即)1 , 1(-. 4)原函数若想有意义必须满足⎪⎩⎪⎨⎧>-≠-≥-050302x x x ,可解得 ⎩⎨⎧<<<≤5332x x ,即) 5 , 3 (] 3 , 2 [⋃,3]. 5)原函数若想有意义必须满足⎪⎩⎪⎨⎧≥--≥+-0)1)(3(0342x x x x ,可解得 ⎩⎨⎧≥-≤31x x ,即(][) , 3 1 , ∞+⋃-∞.6)原函数若想有意义必须满足⎪⎩⎪⎨⎧≠-≠>0lg 100x x x ,可解得⎩⎨⎧><<10100x x ,即) , 10()10 , 0(∞+⋃. 7)原函数若想有意义必须满足01012≤≤-x 可解得21010--≤<x 即]101 , 0()0 , 101[22--⋃- 8)原函数若想有意义必须满足062>--x x ,1712≤-x 可解得) 4 , 3 (] 2 , 3 [⋃--.4．求下列分段函数的定义域及指定的函数值，并画出它们的图形： （1）⎪⎩⎪⎨⎧<≤-<-=43,13,922x x x x y ，求)3( , )0(y y ；（2）⎪⎪⎩⎪⎪⎨⎧∞<<+-≤≤-<=x x x x x x y 1, 1210,30,1，求)5( , )0( , )3(y y y -．解：1）原函数定义域为：)4 , 4(-3)0(==y 8)3(==y .图略2)原函数定义域为：) , (∞+-∞31)3(-=-y 3)0(-==y 9)5(-=y y(5)＝-9.图略5．利用x y sin =的图形，画出下列函数的图形：(1)1sin +=x y ； （2）x y sin 2=； （3）⎪⎭⎫⎝⎛+=6sin πx y ．解：x y sin =的图形如下(1)1sin +=x y 的图形是将x y sin =的图形沿沿y 轴向上平移1个单位(2)x y sin 2=是将x y sin =的值域扩大2倍。
微积分习题答案
习题1-1１．(1) ［－３，３］；（２）（－∞，０）∪（２，＋∞）；（３）（－２，１）；（４）（－１．０１，－１）∪（－１，０．９９）２．（１）［－１，０）∪（０，１）；（２）（１，２］；（３）［－６，１）．３．（１）（－∞，１）∪（１，２］，f(0)=0,f（２)＝１．当a＜０时，f(a)=1a,当0≤a≤1时，f(a)=2a,当１＜a≤2时，f(a)=1．(2) (-2，2)，f(0)=1,f((-a)２，当1＜a＜2时，f(a)=a２－１．４．1．５．（１）偶函数；(2) 非奇非偶函数；(3) 奇函数．８．(1) y=13arcsinx2；(2) y=log２x1-x(3) f－１（x）=１２（x＋１），－１≤x≤1，2-2-x，1＜x≤2．９．（１）y＝１０１＋x２（－∞，＋∞）；（２）y=sinxln２，（－∞，＋∞）；（３）y=arctana２＋x２（－∞，＋∞）．习题1-2１．(1) y=3u，u=arcsinv，v=ax；（２）y=u３，u=sinv,v=lnx;（３）y=au,u=tanv,v=x2;（４）y=lnu,u=v2,v=lnw,w=t3２．（１）［－１，１］，（２）［２kπ，（２k＋１）π］，k∈Z;(3) ［－a，１－a］；（４）（－∞，－１］．３．(1) φ（x)=6+x-x2;(2)g(x)=(1+x)2+(1+x)+1;(3)f(x)=x2-2．习题1-3１．R(x)=4x-12x2．２．R(x)≈１３０x，１１７x＋９１００，０≤x≤７００，７００＜x≤１０００．３．Ｌ＝Ｌ（Ｑ）＝－１５Ｑ２＋８Ｑ－５０，＝－Ｑ５＋８－５０Ｑ．习题2-1略．习题2-2２．f(x)=-1,１，x≤０x＞０，则limx→０f(x)=1，但limx→０－f(x)=-1，limx→０＋f(x)=1，故limx→０f(x)不存在．３．limx→０(x２＋a）＝a,limx→０-e１x＝０，a=0．２． ， ， ， ， ， ， ， ．３．(1)无穷大量．(2) x→０＋时为无穷大量，x→１时为无穷小量．x→＋∞时为无穷大量．（３）x→０＋时为无穷大量，x→０－时为无穷小量．(4) 无穷小量．（５）无穷小量．(6) 无穷小量．习题2-4５．（１）３/５；（２）０；（３）∞；(4) 1/3；(5) 4/3６．（１）16；(2) ∞；（３）３；（４）－２２；（５）３x２．（６）４３；（７）n（n＋１）２；（８）１；（９）１；（１０）－１；（１１）０．习题2-5１．53；2．２５；３．１；４．２２；５．２１２；６．e－１；７．e３；８．lna；９．２lna；１０．０；１１．e－１2；１２．１；１３．１；１４．１；１５．e１e；１６．e－１．习题2-6３．tanx－sinx=O（x３）４．(1) ab；(2) k２２；(3) 2；(4) 24；(5) 1；(6) 1；(7) 49；(8) 3．习题2-7４．(1) x=1(可去)，定义f(1)=2;x=2(第二类)；(2) x=0(可去)，定义f(0)=1;x=kπ，k≠0，为整数(第二类)；(3) x=0(第一类；(4) x=2(第二类)；x=-2(可去)，定义f(-2)=0;(5) x=0(可去)，定义f(0)=0．６．f(x)=sgnx，x=0（第一类)，f(x)∈C［（－∞，０）∪（０，＋∞）］７．（１）１２；（２）３；（３）０；（４）π3；(5) 1．习题3-1１．２９．２．－１x２０．３．４x-y-4=0,8x-y-16=0４．（１）-f′(x0);(2) -f′(x0);(3) 2f′(x0)５．（１）１２x；（２）－２３x－５３；（３）１６x－５6．６．连续但不可导．８．（１）f′(2) f′12，f′９．f′(x)＝cosx，１，x＜０，x≥０．１０．a=2,b=-1．１１．（１）在x=0处连续，不可导；(2) 在x=0处连续且可导；(3) 在x=1必连续，不可导．１３．(1) -０．７８m/s；(2) 10-gt;(3) 10g（s)．１４．dＱdtt=t0．１５．（１）limΔＴ→０Q(T＋ΔＴ）－Ｑ（Ｔ）ΔＴ；（２）a＋２bＴ．习题3-2１．(1) 3t；(2) xx+12xlnx；(3) 2xsin２x－２xsinx＋cosx－x２cosx－sin２x＋x２sin２x．(4)1-sinx-cosx(1-cosx)2；(5)sec2x；（６）xsecxtanx-secxx2-3secx²tanx；(7)１x１－２ln10＋３ln２；(8)－1+2x(1+x+x2)2．２．（１）２４１＋π２；（２）f′(0)=３２５，f′（２）＝１７１５；（３）f′（１）＝５．３．略．４．(1) 3e3x；(2) 2x1+x4；(3) 12x+1e2x+1；(4) 2xln(x+1+x２）＋１＋x２；(5)２x²sin１x２－２xcos１x２；(6)－3ax２sin２ax３；(7)xx２²x２－１；(8)２arcsinx２４－x２；(9)lnxx²1+ln２x；(10)nsinn-1x²cos(n+1)x;(11)１1-x2+1-x２；(12)－１（1+x）２x(1-x)；(13)-thx；（１４）a２-x2．５．１３．６．2x+3y-3=0; 3x-2y+2=0; x=-1; y=0．７．(1) 2xf′(x2);(2) sin2x［f′（sin２x）-f′(cos2x)］．８．（１）-x2-ayy2-ax；(2) 1-yx(lnx+lny+1)；(3) -ey+yexxey+ex;(4)x+yx-y；(5)ex+y－yx-ex+y．９．（１）x+2(3-x)4(x+1)512(x+2)－43-x-5x+1；(2) sinxcosxcos２xsinx-sinxln sinx；(3) e2x(x+3)(x+5)(x-4)2+1x+1-12(x+5)-12(x-4)．１０．（１）sinat+cosbtcosat-sinbt;(2)cosθ-θsinθ1-sinθ-θcosθ．１１．３－２．习题3-3１．f(n)(x)=(-1)n-1(n-1)!(1+x)n．２．y(n)=(-1)n²an²n!²(ax+b)-(n+1)．f(n)(x)=(-1)n2·n!·1(x-1)n+1-1(x+1)n+1３．(1)0;(2)4e,8e;(3)7200,720．４．(1) -b4a2y3;(2) e2y(3-y)(2-y)3;(3) -2csc2(x+y)cot3(x+y);(4)2x2y［3(y2+1)2+2x4(1-y2)］(y2+1)3．５．(1) -1a(1-cost)2;(2) 1f″(t)．６．(1) 4x２f″(x2)+2f′(x2);(2) f″(x)f(x)-［f′(x)］2f．习题3-4１．(1) sint;(2)－１ωcosωt;(3)ln(1+x);(4) -12e-2x;(5)2x;(6)13tanx;(7) ln2x2;(8)-1-x2．２．（１）０．２１，０．２，０．０１；（２）０．０２０１，０．０２，０．０００１．３．（１）（x+1)exdx;(2)1-lnx〖〗x2dx;(3)-12xsinxdx;(4)2ln5²5ln tanx²1sin2xdx;(5)-12cscx2dx;(6)8［xx(1+lnx)-12e2x］dx;(7) 121-x2arcsinx+2arctanx1+x2dx．４．(1) ey1-xeydx;(2)-b2xa2ydx;(3) 22-cosyds;(4)1-y21+2y²1-y2dx．５．(1)２．００８３；（２）－０．０１；（３）０．７９５４．习题3-5１．（１）１．１；（２）６５０；（３）６５０－５０１２９．２．（１）９６．５６；（２）是，提高2．３．（１）a,axax+b,aax+b;(2)abebx,bx,b;(3) axa-1,a,ax．４．提高8%；提高16%．５．５．９．习题4-1１．ξ＝π２．２．（１）满足，有ξ＝０；（２）不满足第二个条件，没有；(3) 不满足第一和第三个条件，有ξ＝π２．３．有分别位于区间(1，2)，(2，3)，(3，4)内的三个根．４．ξ＝３３．习题4-2１．（１）－３５；（２）１２；（３）mnam-n；（４）１a（５）０；（６）０；（７）1；(8) 32；(9) e；（１０）e－２π；（１１）１e；（１２）∞（１３）１３；（１４）e－１２．２．m=-4,n=3４．f″（x）；习题4-3１．xex=x+x2+x32!+…+xn(n-1)!+1(n+1)!(n+1+θx)eθxxn+1（０＜θ＜１）．２．1x=-1-(x+1)-(x+1)2-…-(x+1)n+(-1)n+1(x+1)n+1［-1+θ(x+1)］n+2（０＜θ＜１）．３．f(x)=-56+21(x-4)+37(x-4)2+11(x-4)3+(x-4)4．４．(1) １６(提示：只要将sinx展开成三次多项式即可)．(2) 12(提示：令u=１x，再将ln(1+u)展开成二次多项式）．习题4-4１．（１）(-∞，-1)和(3，+∞)为增区间，(-1，3)为减区间，f(-1)=3为极大值，f(3)=-61为极小值．(2) (1，+∞)为增区间，(0，1)为减区间，f(1)=1为极小值．（３）（－∞，2)为增区间，(2，+∞)为减区间，f(2)=1为极大值．（４）（－∞，０）和(0，2)为增区间，(2，+∞)为减区间，f(2)=-4为极大值．５．当a=2时，f(x)在x=π３取极大值3．习题4-5１．15元２．x=αcPQ１１－α３．（１）Ｑ＝３；（２）ＭＣ＝＝６４．(1) 1000件；(2) ６０００件５．(1) ４３１．３２５吨(2) 12次(3) ３０．４５２天(4) 13６６４３．９元６．α＝２３（３－６）π．７．t=１４r２．８．v=３２００００≈２７．１４（km/h）习题4-6１．（１）在-∞，１３下凸，１３，＋∞上凸，拐点13，227；(2) 在(-∞，－１）上凸，(-1，1)下凸，(1，+∞)上凸，拐点(-1，ln２)及(1，ln ２）；（３）在(-∞，－２)上凸，(-2，+∞)下凸，拐点(-2，-2e－２)；（４）在(-∞，+∞)下凸，无拐点；(5) 在(-∞，-3)上凸，(-3，6)上凸，(6，+∞)下凸，拐点6，227；(6) 在-∞，12上凸，12，+∞下凸，拐点12，earctan１２．３．a=-32，b=９２．４．（１）垂直渐近线x=0;(2) 水平渐近线y=0;(3) 水平渐近线y=0，垂直渐近线x＝3；(4) 垂直渐近线x=12，斜渐近线y=12x+1〖〗4．５．（１）定义域(-∞，+∞)，极大值f(1)=12，极小值f(-1)=-12,拐点3，34，-3，-34，渐近线y=0；(2) 定义域(-∞，+∞)，极大值f(-1)=π2-1，极小值f(1)=1-π２，拐点(0，0)，渐近线y=x+π,y=x-π;(3) 定义域(0，+∞)，极大值f(1)=2e，拐点，２，４e2，渐近线y=0．习题5-1１．（１）２７x７〖〗２-103x3２+C；(2) 2x-43x3２+25x5２+C；(3) 3xex１＋ln３＋Ｃ；（４）x＋sinx２＋Ｃ；（５）２x－５２３xln２－ln３＋C；（６）－（cotx＋tanx）＋C．２．（１）y＝x2-2x+1;(2) cosx+C;(3)x-sinx;(4)Q=1000１３Ｐ习题5-2１．(1) 1a；(2) 17；(3)110；(4) -12；(5) 112；(6) 12；(7) -2；(8) 15；(9) -1；(10) -1；(11) 13；(12) 12；(13) -1；(14) 32．２．（１）１５e５t＋Ｃ；（２）－１８（３－２x）４＋Ｃ；（３）－１２ln１－２x＋Ｃ；（４）－１２（２－３x）２３＋Ｃ；（５）－２cost＋Ｃ；（６）lnlnlnx＋Ｃ；（７）１１１tan１１x＋Ｃ；（８）－１２e-x２＋Ｃ；（９）lntanx＋Ｃ；（１０）－lncos１＋x２＋Ｃ；（１１）arctanex＋Ｃ；（１２）－１３（２－３x２）１２＋Ｃ；（１３）－３４ln１－x４＋Ｃ；（１４）１２cos２x＋Ｃ；（１５）１２arcsin２x３＋１４９－４x２＋Ｃ；（１６）x２２－９２ln（x２＋９）＋Ｃ；（１７）１２２ln2x-12x+1+C；(18) 13lnx-2x+1+C；(19) t2+14ωsin２（ωt＋φ）＋Ｃ；（２０）－１３ωcos３（ωt＋φ）＋Ｃ；（２１）１２cosx－１１０cos５x＋Ｃ；（２２）１３sin３x２＋sinx２＋Ｃ；（２３）１４sin２x－１２４sin１２x＋Ｃ；（２４）１３sec３x－secx＋Ｃ；（２５）（arctanx）２＋Ｃ；（２６）－１arcsinx＋Ｃ；（２７）１２（lntanx）２＋Ｃ；（２８）－1xlnx＋Ｃ；（２９）a２２（arcsinxa－xa２a２－x２）＋Ｃ；（３０）x１＋x２＋Ｃ；（３１）x９－９－３arccos３x＋Ｃ；（３２）１２（arcsinx＋lnx＋１－x２）＋Ｃ；（３３）arcsinx－x１＋１-x２＋Ｃ；（３４）arcsinxa－a２－x２＋Ｃ；（３５）－４－x２x－arcsinx２＋Ｃ；（３６）ln１＋x＋x２＋２x－２xx２＋２x＋Ｃ；（３７）－１１＋tanx+Ｃ；（３８）x＋lnx１＋xex＋Ｃ．习题5-3１．（１）-xcosx+sinx+C;(2) -(x+1)e－x +C;(3) xarcsinx+１－x２+C;(4) sinx-cosx2e-x+C;(5)-217e-2xx2＋4sinx2+C;(6) -12x2+xtanx+lncosx+C;(7) -t2+14e－2t+C;(8)x(arcsinx)2+21-x2arcsinx-2x+C;(9) 12-15sin2x-110cos2x）ex+C;(10) 3e3x(３x2-23x+2+C;(11)x2(coslnx+sinlnx)+C;(12) -12x2-32cos2x+x2sin2x+C;(13) 12（x２-1)ln(x-1)-14x2-12x+C;(14) x36+12x2sinx+xcosx-sinx+C;(15) -1x(ln3x+3ln2x+6lnx+6)+C;(16) -14xcos2x+18sin2x+C;(17) -12xcot2x-12x-12cotx+C;(18) 12x2ex2+C;(19)xlnlnx+C;(20) (1+ex)ln(1+ex)-ex+C;(21) 12tanxsecx-12lnsecx+tanx+C;(22) -ln(x+1+x22(1+x2)+x22+x2+C;(23) ex1+x+C;(24) x-121+x2earctanx+C．习题5-4(1) lnx+1x2-x+1+3arctan2x-13+C;(2) x3３+x２2+x+8lnx-3lnx-1-4lnx+1+C；（３）x-tanx+secx+C；（４）14lntanx2-18tan2x2+C．习题6-1１．１３（b3-a3)+b-a．２．（１）１；（２）14πa2．３．(1)∫10x2dx较大；(2) ∫10exdx较大．４．(1)6≤∫41(x2+1)dx≤51;(2)π9≤∫3１3xarctanxdx≤23π；(3)2ae－a2＜∫a-ae-x2dx＜2a；(4)-2e2≤∫02ex2-xdx≤-2e-1〖〗4．习题6-2１．（１）２x１+x4;(2)x5e-3x;(3)(sinx-cosx)cos(πsin2x);(4) sinx-xcosxx2．2．(1)-12;(2) 6;(3) 2．３．cosxsinx-1．4．当x=0时．5．(1)23(8-33);(2) 16;(3) 1+π8；(4) 203．６．－３２．习题6-3１．（１）０；（２）５１５１２；（３）１６；（４）１４；（５）π６-３８；（６）２（３－１）；（７）２－２３３；（８）π２；（９）１２ln３２；（１０）ln２－１３ln５；（１１）７ln２－６ln（６２＋１）；（１２）４３．２．（１）０；（２）０；（３）３２π．习题6-4２．（１）１－２e;(2)14（e２+1);(3) 4(2ln2-1);(4) 14-133π+12ln32;(5)15(eπ-2);(6)2-34ln2；(7) π3６－π4；(8) 12(esin1-ecos1+1);(9) ln2-12;(10)12-38ln3．３．０．习题6-5１．（１）１；（２）２；（３）４３；（４）７６；（５）１２＋ln２；（６）１６；（７）e＋１e-２；（８）b-a．２．（１）Vy=2π;(2) Vx=1287π,Vy=12．8π;(3) Vy=310π;(4) Vx=pa2π;(5)Vy=4π2．３．（１）a=1e，(x0,y0)=(e2,1);(2)S=16e2-12．4．12ln2提示：f(x)=0,x1+x2,x≥０x＜０．5．a=-4,b=6,c=0．６．50;100．7．(1) Q=2．5,L=6．25;(2) 0．25．８．９６．７３习题6-6１．（１）１３；（２）发散；(3) １a；（４）发散；(5) 发散；(6) π；（７）８３；（８）１；（９）π２；（１０）－１；（１１）发散；(12) 1．２．当k＞１时收敛于1(k-1)(ln2)12-1；当k≤1时发散；当k＝１－１lnln２时取得最小值．３．n!．４．（１）π4；(2) π2５．In=-(2n)!!(2n+1)!!=22n(n!)2〖〗(2n+1)!(n=0,1,2,…）．６．（１）1nΓ1n;(2) Γ(α+1);(3)1nΓm+1n;(4)12Γn+12．习题7-1１．略．２．(1) (a,b,-c),(-a,b,c),(a,-b,c);(2) (a,-b,-c),(-a,b,-c),(-a,-b,c);(3) (-a,-b,-c)．３．坐标面：(x0,y0,0),(0,y0,z0),(x0,0,z0)；坐标轴：(x0,0,0),(0,y0,0),(0,0,z0)．４．x轴：34，y轴：41，z轴：5．５．（０，１，－２）．６．略．习题7-2１．ＭＡ→＝－１２（a+b);MB→＝１２（a-b);MC→=12(a+b);MD →=12(b-a)．２．略．３．（２，１，１）．４．（１６，０，－２０）．５．Ｍ１Ｍ２→＝（１，－２，－２），Ｍ１Ｍ２→＝３．１３，－２３，－２３或－１３，２３，２３．习题7-3１．（１）１；（２）４；（３）２８．２．（１）３，５i+j+7k;(2) -18，10i+2j+14k；(3) -10i－２j－１４k．３．－３２．４．±（６２，８２，０）．５．１４．６．略．７．４５j－３５k或－４５j＋３５k．８．∠Ａ＝７６°２２′，∠Ｂ＝７９°２′，∠Ｃ＝２４°３６′．习题7-4１．３x-2y+5z-22=0．２．2x+9y-6z=121．３．略．４．x+z-1=0．５．x+y+z-2=0．６．2x+3y+z-6=0．７．(1)x=2；(2)x+3y=0；(3)x-y=0．８．１３，２３，－２３．９．（１）互相垂直；(2) 互相平行；(3) 斜交(相交但不垂直)．习题7-51．（１）x-23=y-31=z-11；(2)x-31=y-4２=z+4-1；(3)x-21＝y-20=z+1〖〗0；(4)x2=y-31=z+23．２．x+3-5=y=z-25，[JB(｛〗x=-3-5t，y=t，z=2+5t．３．x-2＝y-23=z-4〖〗1．４．x-21=y+22=z3．５．x-10=y+37=z+2〖〗16．６．４６１，６６１，－３６１．７．Ｂ＝１，Ｄ＝－９．８．x-3-1=y-31=z1．９．φ＝arcsin1310．１０．４x-y-2z-1=0．１１．y-z+3=0，x-y-z+1=0．１２．５．１３．（１）垂直，(2) 平行，(3) 重合．习题7-6１．（x+1)2+(y+3)2+(z-2)2=32．２．以点(1，-2，-1)为球心，半径等于6的球面．３．(1) x23+y24+z24=1;x23+y24+z23＝１；(2)x2-y2-z2=1;x2+y2-z2=1．４．（１）母线平行于z轴的椭圆柱面；(2) 母线平行于x轴的抛物柱面；(3) 椭圆锥面；(4) 旋转椭球面；(5) 双叶双曲面；(6) 圆锥面．５．3y2-z2=16,3x2+2z2=16６．x2+y2+(1-x)2=9，z=0；(1-z)2+y2+z2=9，x=0；x+z=1，y=0．７．（１）椭圆；(2) 双曲线；(3) 抛物线．８．略．习题8-1１．（１）（x,y)x2a2+y2b2≤１；(2)｛(x,y)x＞y,且x-y≠１｝；(3)(x,y)-1≤yx≤１，且x≠０＝｛x＞０，－x≤y≤x；x＜０，x≤y≤－x｝；（４）｛(x,y)x≥y，x2+y2≤１，y≥０｝．２．（１）３１；（２）１x３－４xy＋１２y２；（３）（x+y）3－２（x２－y２）＋３（x－y）２．３．f(x)＝（x＋２）x，Ｆ（x,y）＝y＋x－１．４．略习题8-2１．（１）不存在，(2) 存在．２．（１）０，（２）１，（３）２，（４）０．３．｛（x,y)y２＝２x，x∈Ｒ｝．习题8-3１．（１）z′x=y(1+x)y-1,z′y=(1+x)yln(1+x);(2) z′x=-yx２cotyx²sec2yx，z′y=1xcotyx²sec2yx；（３）z′x=-yx2+y2，z′y=xx2+y2；(4)u′x=-zlnyx2²yzx，u′y=zx²yzx-1，u′z=１xyzx²lny．２．－１，２．３．１，１＋π６．４．略．５．偏导数存在．６．α＝π４．７．Δz=-0．12,dz=-0．1．８．（１）du＝dx-dy；（２）dz=-xy（x２＋y２）３／２dx＋xy（x２＋y２）３／２dy．习题8-4１．（１）２e2cost＋３t２［３t-sint］；（２）３－４t－３＋３２t１２sec23t+2t2+t32．２．（１）z′u=(2xy-y2)cosv+(x2-2xy)sinv；(2)z′v=-(2xy-y2)usinv,z′y=euvx２+y２（ux+vy）．３．（１） u x=1yf′1， u y＝－xy２f′１＋１zf′２， u z＝－yz２f′２；（２） z x＝２xf′， zy＝２yf′；（３） u x＝f′1+yf′2+yzf′3， u y＝xf′2+xzf′３， u z＝xyf′2．４．略．５．（１）dz=(x２+y２)sin（２x＋y）２sin（２x＋y）x２＋y２（xdx＋yd y）＋cos（２x＋y）ln（x２＋y２）（２dx＋dy）；（２）du＝１f（x２＋y２－z２）dy－yf′（x２＋y２－z２）f（x２＋y２－z２）（２xd x＋２ydy－２zdz）．６．（１）z′x＝ex＋y＋yzez－xy，z′y＝ex＋y＋xzez－xy；（２） z x＝zx＋z，z y＝z２y（x＋z）．７．略．８． z x＝（vcosv－usinv）e－u， z y＝（ucosv+vsinv)e-u．９．dudx＝f′x+y２f′y1-xy+zf′zxz-x．习题8-5１．（１） ２z x２＝１２x２－８y２， ２z y２＝１２y２－８x２， ２z x y＝－１６xy；（２） ２z x２＝２xy（x２＋y２）２， ２z y２＝－xy（x２＋y２）２， ２z xy＝y２－x２（x２＋y２）２；（３） ２z x２＝yxln２y，２zy２＝x（x－１）yx-2； ２z xy＝yx－１（１＋xlny）；（４） ２z x＝１x， ２z y２＝－xy２， ２z x y＝１y．２．（１） ２z x２＝４xf″（x２＋y２）＋２f′（x２＋y２），２z y２＝４yf″（x２＋y２）＋２f′（x２＋y２）；（２） ２z x２＝y２f″１１＋２yf″１２＋f″２２， ２z y２＝x２f″１１＋４xf″１２＋４f″２２，２z x y＝xyf″１１＋２yf′１２＋f′1+xf″２１＋２f″２２．３． ２z x２＝z（２x－２－z２）x２（z－１）３， ２z y２＝z（２z－２－z２）y２（z－１）３， ２z x y＝－zxy（z－１）３．习题8-6１．１＋２３．２．２３．３．α＝π４时取得最大值2；α＝５π４时取得最小值-2；α＝７π４时，方向导数为零．习题8-7１．（１）极大值f(0,0)=3；(2) 极小值f１２，－１＝－e２；（３）极大值fa３，a３＝a３２７（a＞０)，极小值fa３，a３＝a３２７（a＜０）．２．极大值z（４，１）＝７，最小值z４３＋２２３，－１≈-１１．６７．３．极小值z(2,2)=4．４．a≥１２，最小距离为a－１４；a≤１２，最小距离为a．５．a的分法是三等分时，乘积最大为a３２７．６．x=100,y=25,f(100,25)=1250．７．x=70,y=30,λ＝－７２，Ｌ＝１４５（万元)．习题8-8１．（１）∫１－１dx∫３－３f(x,y)dy, ∫３－３dy∫１－１f(x,y)dx;（２）∫４０dx∫２xxf(x,y)dy,∫４０dy∫y１４y２f(x,y)dx;（３）∫r－rdx∫r２－x２０f(x,y)dy,∫r０dy∫r２－y２－r２－y２f(x,y)dx．２．（１）∫１０dx∫xx２f（x，y）dy；（２）∫a０dy∫a＋a２－y２a－a２－y２f （x，y）dx;（３）∫１０dy∫２－yyf（x，y）dx．３．（１）e－１e２；（２）２９１５；（３）－１２；（４）２３；（５）１－２π；（６）２πＲ２２＋Ｒ３；（７）３６４π２；（８）２－π２．４．５１４４．５．π．６．８π．７．ＳＤ＝１２e－１，ＶＤ＝１２e２－e－１２．习题9-1１．（１）a＞１收敛；0＜a≤１发散；(2) 发散；(3) 发散；(4) 收敛；(5) 发散；(6) 发散；(7) 发散；(8) 发散．２．（１）收敛，s＝３２；（２）收敛，s＝１４；（３）发散；(4) 发散．习题9-2１．（１）收敛；(2) 发散；(3) 发散；(4) 收敛；(5) a＞１，收敛；0＜a≤１发散；(6) 发散；(7) 发散；(8) 收敛；(9) 发散；(10) 发散；(11) 收敛；(12) 收敛；(13) 收敛；(14) 收敛；(15) 收敛；(16) 收敛．习题9-3１．（１）条件收敛；(2) 绝对收敛；(3) 绝对收敛；(4) 绝对收敛；(5) 绝对收敛；(6) 条件收敛；(7) 绝对收敛；(8) 条件收敛．习题9-4１．（１）（-∞，+∞)；(2) (-e，e）；(3) (-2，2)；(4) (-1，1)；(5) (-4，0)；(6) １２，３〖〗２．２．（１）－ln（１＋x）；x＜１；（２）２x（１－x２）２，x＜１；（３）当x≠０且x＜１时，s(x)＝１＋１x－１ln（１－x）；当x＝０时，s(x)＝０；（４）１＋x（１－x）２，x＜１．３．（１）１５３２；（２）１２ln（１＋２）；（３）１０９；（４）４．习题9-5１．（１）１－x２２²２！＋x４２²４！－…＋（－１）nx２n２²（２n）！＋…（－∞＜x＜＋∞）；（２）∑∞n=1（－１）n-1（２n－１）！x２２n－１（－∞＜x＜＋∞）；（３）∑∞n=1（－１）n-1x2n-1〖〗（n－１）！（－∞＜x＜＋∞）；（４）∑∞n=0x2n,x＜１；（５）２２∑∞n=0（－１）nx2n（2n）！＋x2n+1（２n+1）！（－∞＜x＜＋∞）．２．（１）∑∞n=0１２n+1（x－１）n（－１＜x＜３）；（２）∑∞n=0[ＪＢ（（〗（－１）n２²x－π３２n（２n）！＋（－１）n＋１３２x－π２２n＋１（２n＋１）！（－∞＜x＜＋∞）；（３）∑∞n=0(－１）n１２n＋２－１２２n＋３（x－１）n（－１＜x＜３）；（４）∑∞n=0（－１）n３n＋１（x－３）n（０＜x＜６）．３．（１）２．７１８２８；（２）０．２５０４９．习题10-1１．（１）一阶，(2) 二阶，(3) 三阶，(4) 一阶．２．略．３．y′=y-xx．４．y′＝y－x＋１．习题10-2１．（１）（１－x)(1+y)=Ｃ（Ｃ为任意常数，以下Ｃ，Ｃ１，Ｃ２…均为任意常数)；(2) １－x２＝lny＋Ｃ；（３）y２＝Ｃ（１－x２）－１；（４）secx＋tany＝Ｃ；（５）２y３＋３y２－２x３－３x２＝５；（６）（y＋１）e－y＝１２（１＋x２）；（７）ey＝１２（e２x＋１）．２．Ｔ＝Ｔ０e－kt＋α（１－e－kt），k为比例系数．３．（１）y+x２+y2=Ｃx2;(2) y=2xarctan(Ｃx);(3) x3+y3=Ｃx2;(4) y=2x1+x2;(5) y=xe1-x;(6) (x+3)２+(y+1)2=Ｃe-arctanyx;(7)x+3y+2ln2-x-y=Ｃ．４．（１）y=Ｃex－12（sinx+cosx);(2) y=xn(Ｃ+ex);(3) x=2(y-1)+Ｃe-y;(4) x=y+Ｃcosy；(5) y=(x+1)ex;(6) y=2(1+x3)3(1+x2);(7) y=2lnx-x+2;(8) y=(1+sinx-xcosx)²e－x2；(9) y3=Ｃx3+3x4;(10)１x2=1-y2+Ｃe-y2．５．y′=３yx2－2²yx，y-x=-x3y．６．x=ab+x0-abe-bt．７．f(x)=-2e-3x-1．８．Ｃ(x)=(x+1)［Ｃ０+ln(x+1)］．９．x＝ab（Ｃ０x０－a）１b＋１²x０．习题10-3１．（１）y=(x-3)ex+12Ｃ1x2+Ｃ2x+Ｃ3;(2) y=xarctanx-12ln（１＋x２）＋C1x＋C2；(3) y=C１arctanx+C2;(4) y=-lnx+c１+c2；(5) 1+C1x2=(C2t+C2)2；（６）lny=C1(y-x)+C2．２．（１）y=16x3lnx-1136(x3-1);（２）y=lnx+12ln２x;(3) y=x．３．C１＋C２ex＋x．４．（１）y=(Ｃ１＋Ｃ２x)e２x；（２）y＝Ｃ１e－x＋Ｃ２e２x；（３）y＝９e－２x－８e－３x；（４）y＝－１３exxcos３x．５．（１）y＝（１－１２x）e－２x＋Ｃ１e－５x＋Ｃ２e２x；（２）y＝（x＋１）２＋Ｃ１e２x＋Ｃ２e４x；（３）y＝１１８cosx＋４sinx－１８cos３x；（４）y＝x＋１２x２e４x．６．f（x）＝２（ex－x）．７．a=-3,b=2,α=-1;y=Ｃ１ex＋Ｃ２e２x＋xex．８．φ（x）＝１２（sinx＋cosx＋ex）．９．y＝２３e２x－２３e－x－xe－x．１０．y＝－７e－２x＋８e－x＋（３x２－６x）e－x１１．s＝mgkt－m２gk２（１－e－kmt）．习题10-4１．C（x)=3ex（１＋２e３x）－１．２．Ｒ＝abs０（ebt－１），Ｓ（t）＝s０e－bt．３．Ｙ（t）＝Ｙ０eγt，Ｄ（t）＝αＹ０γeγt＋βt＋Ｄ０－αＹ０γ，limt→＋∞Ｄ（t）Ｙ（t）＝α〖〗γ．４．（１）Ｙ（t）＝（Ｙ０－Ｙe）eμt＋Ｙe，Ｙe＝b１－a，μ＝１－aka，Ｃ（t)=a(Ｙ０－Ｙe）eμt＋Ｙe，Ｉ（t)=(1-a)(Ｙ０－Ｙe）eμt；(2) limt→＋∞Ｙ（t）Ｉ（t）＝１１－a．５．y(6)=５０００１＋１１．５e－３（ln１１．５－ln８）．习题11-1１．（３），（４）．２．（１）一阶；(2) 五阶；(3) 三阶；(4) 六阶；(5) 二阶．３．（１）Δ２yt＝２；（２）Δyt＝（e－１）２et；（３）Δ２yt＝６（t＋１），Δ３yt＝６；（４）Δ２yt＝lnt２＋４t＋３t２＋４t＋４．４．略．习题11-2１．yＡ(t)＝Ａ１＋Ａ２t＋１（Ａ１，Ａ２为任意常数．以下A，A１，Ａ２…均为任意常数)．２．a(t)＝－１＋１５，f(t)＝１－１t²２t．３．略．４．（１）yt＝Ａ－１３t＋１；（２）yt＝Ａ－１２t＋７９＋１３t ；（３）yt＝Ａ（－１）t＋１３²２t；（４）yt＝Ａ－１３²２tcosπt．５．（１）yt＝０．１³３８t＋０．１；（２）yt＝１２t－２＋t；（３）yt＝２t－t＋４；（４）yt＝（－４）t＋sinπt．６．yt＝Ａ（－a）t＋b１＋a．７．（１）略；(2) yt＝１y０－bＣ－aaＣt＋bＣ－a）－１，１y０＋bat－１，当Ｃ≠a时，当Ｃ=a时．（３）yt＝１２t＋１＋３２－１．习题11-3１．（１）yＡ(t)＝Ａ１（－１）t＋Ａ２１２t；（２）yＡ（t）＝（３）t（Ａ１cosωt＋Ａ２sinωt），tanω＝－２；（３）yＡ（t）＝（Ａ１＋Ａ２t）²４t；（４）yＡ（t）＝Ａ１cosπ３t＋Ａ２sinπ３t；（５）yＡ（t）＝Ａ１（１．８）t＋Ａ２（２．１）t；（６）yＡ（t）＝Ａ１［２（a＋１）＋２a＋１）］t＋Ａ２［２（a＋１）－２a＋１］t．２．（１）yt＝Ａ１５＋１７２t＋Ａ２５－１７２t－１；（２）yt＝２tＡ１cosπ３t＋Ａ２sinπ３t＋１３（a＋bt）；（３）yt＝Ａ１＋Ａ２²２t＋１４³５t；（６）yt＝Ａ１（－２）tt＋Ａ２²３t²t１１５t－２２５．３．（１）t＝２５t２＋１２５t＋６４１２５＋１８６１２５（－４）t；（２）t＝４t＋４３（－２）t－４３；（３）t＝１９５１３０－２０〖〗１３０（－４）t－９２６１３t；（４）t＝４＋３２１２t＋１２－７２t．习题11-4１．Ｙt＝（Ｙ０－Ｙe）αt＋Ｙe，Ｙe＝１＋β１－α；Ｃt＝（Ｙ０－Ｙe）αt＋αＩ＋β１－α．２．Ｙt＝（Ｙ０－Ｙe）²λt＋Ｙe，其中λ＝１＋r（１－α），Ｙe＝β１－α；Ｃt＝α（Ｙ０－Ｙe)λt＋Ｙe；Ｉt＝（１－α）（Ｙ０－Ｙe）λt．３．Ｙt＝Ｙ０＋βα²λt－βα，其中λ＝δrδr－α；Ｓt＝（αＹ０＋β）²λt；Ｉt＝１δ（αＹ０＋β）²λt．４．Ｄn（t）＝Ａ１λt１＋Ａ２λt２，其中λ１，２＝２［（ab＋１）±１＋２ab］．。
为什么这么多人推荐初学者看《托马斯微积分》_
作为一个在国内学过一点微积分，现为的留学狗，我以个人角度比较下托马斯微积分（国外）和国内微积分吧（以下用托马斯和同济简便代替）。
托马斯是一本比较注重基础的书，就单从第一章节（都是基础的函数）来看的话就会发现很多习题要比高中的习题还要简单，好的基础才是未来前进的关键。
如果你不是对数学有良好敏感度的话（国内的书都是只看的懂字，看不懂在讲啥），托马斯非常适合理解微积分，这本书的每一个定理公式都会用最简单的例子娓娓道来，其格局之大之严谨，足以让我们感慨并仔细品尝！托马斯以步步为营讲解的方式建立微积分体系，有言道不成体系的知识都是碎片！托马斯的牛叉之处就在此。
反观国内的数学。
我在国外老师是华人会说中文，他有次对我们中国留学生说在中国学数学最好了。
他的这句话非常耐人寻味，而事实却是如此，对比国外的微积分教学理念，国内的数学学习更像是中医，即总结经验学，比如托马斯这本书在第六，七章（不同的版本导致章节变动）单独讨论了超越函数（对数和指数函数等）的微积分，这在国内的教材是难以看到的，国内老师要求我们要把所有的微分和积分公式记得滚瓜烂熟，然后直接做题。
还比如，我们在学微分的时候面对复合函数都要用替换法，令一个函数等于u～～，而托马斯却在超越函数这一章讲指数函数e的复合时用了一种更为拧巴的方法我看了半天才明白！（实际还是设u）所以国内善于总结，在做题的时候将困难问题简单化，但这需要你花很多精力来理解微积分。
我们的体系建立是记忆，通过做题慢慢理解，然后熟能生巧。
此外，国内汉语托马斯是第十版，2002年左右出的距今已有十几年了，里面有很多打印错误我都不想说什么了，还请大家注意！并且很多翻译都是直接硬翻的，非常难懂，还不如看英文版的，毕竟英语每个句子只有一个动词。
第二，托马斯微积分自从2002年的第十版到2014的第十三版，已经变更了4版！他的每一次更新都是名义上的改革，宣称比上一版好的不知多少，但实际上是换汤不换药，就是把内容换下位置并无太多变化，甚至连题目都若干年不变，如果要是在同一时间读了不同版本的托马斯，是会被这颠来倒去的顺序搞晕的。
[托马斯微积分.第11版.(附带习题答案)].review1
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tan x
(d)
dx
(ln cos x)2
√
3
2
(e)
arctan 2t dt
0
3 dx (f )
0 x−1
x
(g)
dx
(x − 3)2
ln(ln 3)
(h)
eexex dx
ln(ln 2)
1 + 4x
(i) √
dx
1 + x + 2x2
4
1 (j) x2 − 4x + 8 dx
0
1
√ x
(k)
√ dx
∞1 (u) −∞ ex + e−x dx
4
1
(v) 1 x2 − 5x + 6 dx
dx (w)
x(ln x)2
π
2
sin x
(x)
dx
1 + cos x
0
sin(e−2x)
(y)
e2x dx
4
√
ex
(z) √
√ dx
x(1 + e x)
1
Math 222—Exam 1 Review
2
2. Determine whether the following integrals converge or diverge.
5. A chemical manufacturing company has a 1000-gallon holding tank which it uses to control the release of pollutants into a sewage system. Initially the tank has 360 gallons of fluid containing 2 pounds of pollutant per gallon. Fluid containing 3 pounds of pollutant per gallon enters the tank at the rate of 80 gallons per hour and is uniformly mixed with the fluid already in the tank. Simultaneously, fluid is released from the tank at the rate of 40 gallons per hour. Determine how many pounds of pollutant is present in the tank after one hour.
托马斯微积分第一章节与第二章节练习题
Assignment-11.Find the domain and range of the function y =tan(2x −π).2.If f (x )=2−x and g (x )=3√x +1,find(1)(f ◦g )(−1)(2)(g ◦f )(2)(3)(f ◦f )(x )(4)(g ◦g )(x )3.Write formulas for (f ◦g )and (g ◦f )and find the domain and range of each.f (x )=√x ,g (x )=√1−x .4.Sketch the graphs of f and (f ◦f ),wheref (x )=x +1,−2≤x <0x −1,0≤x ≤212Assignment-1 .5.Describe how each graph is obtained from the graph of y=f(x)(a).y=f(x−5)(b).y=f(4x)(c).y=f(−3x)(d).y=f(2x+1))−4(e).y=f(x3(f).y=−3f(x)+146.ABC is a right triangle with the right angle at C.The sides opposite angle A,B and C are a,b andc respectively.a.Find a and b if c=2,B=π/3b.Find a and c if b=2,B=π/3c.Express sin A in terms of a and cd.Express sin A in terms of b and c7.Find the average rate of change of the function over the given intervals.h(t)=cot t, a.[π/4,3π/4] b.[π/6,π/2]3.8.Find (a)the slope of the curve at the given point P,and (b)an equation of the tangent line at P .(1).y =x 2−4x ,P :(1,−3)(2).y =2−x 3,P :(1,1)9.Explain why the limit does not exist limx →0x|x |10.If f (1)=5,must lim x →1f (x )exist?If it does,then must lim x →1f (x )=5?Can we conclude anything about lim x →1f (x )?Explain.11.Find the limits.1).lim h →0√5h +4−2h4Assignment-12).lim y →05y 3+8y 23y 4−16y 212.Suppose that lim x →−2p (x )=4,lim x →−2r (x )=0,and lim x →−2s (x )=−3.Finda.lim x →−2(r (x )+p (x )+s (x ))b.lim x →−2p (x )·r (x )·s (x )c.lim x →−2(−4p (x)+5r (x ))s (x )13.Let G (x )=(x +6)/(x 2+4x −12).a.Make a table of the values of G at x =−5.9,−5.99,−5.999and so on.Then estimate lim x →−6G (x ).What estimate do you arrive at if you evaluate G at x =−6.1,−6.01,−6.001,···instead?b.Support your conclusion in part(a)by graphing G and using Zoom and Trace to estimate y −valueson the graph as x →−6.c.Find lim x →−6G (x )algebraically.14.If lim x →0f(x )x 2=1,finda.lim x →0f (x )b.lim x →0f (x )x。
《微积分》课后答案（复旦大学出版社（曹定华李建平毛志强著））第1章
第一章习题1-11.用区间表示下列不等式的解2(1)9;(2)1;1(3)(1)(2)0;(4)00.011 x x x x x ≤>--+<<<+解(1)原不等式可化为(3)(3)0x x -+≤,其解为33x -≤≤,用区间表示是[-3,3].(2)原不等式可化为11x ->或11x -<-,其解为2x >或0x <,用区间表示是(-∞,0)∪(2,+∞).(3)原不等式的解为21x -<<,用区间表示是(-2,1).(4)原不等式可化为0.0110.0110x x -<+<⎧⎨+≠⎩即 1.010.991x x -<<-⎧⎨≠⎩用区间表示是(-1.01,-1)∪(-1,-0.99).2.用区间表示下列函数的定义域:1(1)(2)arcsin(1)lg(lg );1(3).ln(2) y y x x xy x ==-+=-解(1)要使函数有意义,必须2010x x ≠⎧⎨-≥⎩即011x x ≠⎧⎨-≤≤⎩所以函数的定义域为[-1,0)∪(0,1].(2)要使函数有意义,必须111lg 0x x x -≤-≤⎧⎪>⎨⎪>⎩即0210x x x ≤≤⎧⎪>⎨⎪>⎩所以函数的定义域是12x <≤,用区间表示就是(1,2].(3)要使函数有意义,必须2650ln(2)020x x x x ⎧--≥⎪-≠⎨⎪->⎩即6112x x x -≤≤⎧⎪≠⎨⎪<⎩所以函数的定义域是-6≤x <1,用区间表示就是[-6,1).3.确定下列函数的定义域及求函数值f (0),f),f (a )(a 为实数),并作出图形(1)1,0,2,011,12x x y x x x ⎧<⎪⎪=⎨≤<⎪⎪<≤⎩；(2)y＝211,12x x x ⎧≤⎪⎨-<<⎪⎩解(1)函数的定义域(){|0}{|01}{|12}{|112}(,1)(1,2]或D f x x x x x x x x x =<≤<<≤=<<≤=-∞10(0)200,1,()201112a af ff a a a a ⎧<⎪⎪=⨯===⎨≤<⎪⎪<≤⎩,图1-1图1-2(2)函数的定义域(){|1}{|12}{|2}(2,2)D f x x x x x x =≤<<=<=-221(0)1,11,()112a f ff a a a ≤===-==-<<⎪⎩4.设1,1()1,1x f x x ⎧≤⎪=⎨->⎪⎩,求f (f (x )).解当|x |≤1时,f (x )=1,f (f (x ))=f (1)=1;当|x |>1时,f (x )=-1,f (f (x ))=f (-1)=1,综上所述f (f (x ))=1(x ∈R ).5.判定下列函数的奇偶性：(1)f (x )＝21cos x x-；(2)f (x )＝(x 2＋x )sin x ；(3)f (x )＝1e ,0e 1,0x xx x -⎧-≤⎨->⎩解(1)∵221()1()()cos()cos x x f x f x x x----===-∴f (x )是偶函数.(2)∵222()[()()]sin()()(sin )()sin ()f x x x x x x x x x x f x -=-+--=--=--≠且()()f x f x -≠-,∴f (x )是非奇非偶函数.(3)当x <0时,-x >0,()1(1)()ee xx f x f x ---=-=--=-;当x ≥0时,-x ≤0,()()11(1)()ee e x x xf x f x ---=-=-=--=-,综上所述,x ∀∈R ,有f (-x )=-f (x ),所以f (x )是奇函数.6.设f (x )在区间(-l ,l )内有定义,试证明：(1)f (-x )+f (x )为偶函数；(2)f (-x )-f (x )为奇函数.证(1)令()()()F x f x f x =-+(,)x l l ∀∈-有()[()]()()()()F x f x f x f x f x F x -=--+-=+-=所以()()()F x f x f x =-+是偶函数;(2)令()()()F x f x f x =--,(,)x l l ∀∈-有()[()]()()()[()()]()F x f x f x f x f x f x f x F x -=----=--=---=-所以()()()F x f x f x =--是奇函数.7.试证：(1)两个偶函数的代数和仍为偶函数；(2)奇函数与偶函数的积是奇函数.证(1)设f (x ),g (x )均为偶函数,令()()()F x f x g x =±则()()()()()()F x f x g x f x g x F x -=-±-=±=,所以()()f x g x ±是偶函数,即两个偶函数的代数和仍为偶函数.(2)设f (x )为奇函数,g (x )为偶函数,令()()()F x f x g x =⋅,则()()()()()()F x f x g x f x g x F x -=-⋅-=-=-,所以()()f x g x ⋅是奇函数,即奇函数与偶函数之积是奇函数.8.求下列函数的反函数:22(1)2sin 3;(2);212101,(3)()2(2)1 2.xx y x y x x f x x x ==+-≤≤⎧=⎨--<≤⎩解(1)由2sin 3y x =得1arcsin 32yx =所以函数2sin 3y x =的反函数为1arcsin (22)32xy x =-≤≤.(2)由221x x y =+得21xy y =-,即2log 1y x y =-.所以函数221xx y =+的反函数为2log (01)1x y x x=<<-.(3)当01x ≤≤时,由21y x =-得1,112yx y +=-≤≤;当12x <≤时,由22(2)y x =--得22x y =<≤;于是有1112212y y x y +⎧-≤≤⎪=⎨⎪<≤⎩,所以函数22101()2(2)12x x f x x x -≤≤⎧=⎨--<≤⎩的反函数是1112()212xx f x x +⎧-≤≤⎪=⎨⎪<≤⎩.9.将y 表示成x 的函数,并求定义域:222(1)10,1;(2)ln ,2,sin ;(3)arctan ,().为实数u v y u x y u u v x y u u v a x a ==+======+解(1)211010ux y +==,定义域为(-∞,+∞);(2)sin ln ln 2ln 2sin ln 2vxy u x ====⋅定义域为(-∞,+∞);(3)arctan y u ===(a 为实数),定义域为(-∞,+∞).习题1-21.下列初等函数是由哪些基本初等函数复合而成的?(1)y=；(2)y =sin 3ln x ;(3)y =tan 2x a；(4)y =ln ［ln 2(ln 3x )］.解(1)令arcsin x u a =,则y =再令x v a =,则arcsin u v =,因此y =是由基本初等函数arcsin ,xy u v v a ===复合而成的.(2)令sin ln u x =,则3y u =,再令ln v x =,则sin u v =.因此3sin ln y x =是由基本初等函数3,sin ,ln y u u v v x ===复合而成.(3)令2tan u x =,则uy a =,再令2v x =,则tan u v =,因此2tan x y a=是由基本初等函数2,tan ,u y a u v v x ===复合而成.(4)令23ln (ln )u x =,则ln y u =,再令3ln(ln )v x =则2u v =,再令3ln w x =,则ln v w =,再令ln t x =,则3w t =,因此23ln[ln (ln )]y x =是由基本初等函数2ln ,,ln ,y u u v v w ===3,ln w t t x ==复合而成.2.设f (x )的定义域为［0,1］,分别求下列函数的定义域：(1)f (x 2)；(2)f (sin x );(3)f (x +a ),(a ＞0)；(4)f (e x +1).解(1)由f (x )的定义域为[0,1]得0≤x 2≤1,于是-1≤x ≤1,所以f (x 2)的定义域为[-1,1].(2)由f (x )的定义域为[0,1]得0≤sin x ≤1,于是2k π≤x ≤(2k +1)π,k ∈z ,所以f (sin x )的定义域为[2k π,(2k +1)π],k ∈Z .(3)由f (x )的定义域为[0,1]得0≤x+a ≤1即-a ≤x ≤1-a 所以f (x+a )的定义域为[-a ,1-a ].(4)由f (x )的定义域为[0,1]得0≤e x +1≤1,解此不等式得x ≤-1,所以f (e x +1)的定义域为(-∞,-1].3.求下列函数的表达式：(1)设ϕ(sin x )=cos 2x +sin x +5,求ϕ(x );(2)设g (x -1)=x 2+x +1,求g (x );(3)设1()f x x +=x 2+21x,求f (x ).解(1)法一:令sin t x =,则222cos 1sin 1x x t =-=-,代入函数式,得:22()156t t t t t ϕ=-++=+-,即2()6x x x ϕ=++.法二:将函数的表达式变形得:22(sin )(1sin )sin 56sin sin x x x x xϕ=-++=+-令sin t x =,得2()6t t t ϕ=+-,即2()6x x x ϕ=+-.(2)法一:令1t x =-,则1x t =+,将其代入函数式,得22()(1)(1)133g t t t t t =++++=++即2()33g x x x =++.法二:将函数表达式变形,得22(1)(21)(33)3(1)3(1)3g x x x x x x -=-++-+=-+-+令1x t -=,得2()33g t t t =++,即2()33g x x x =++.(3)法一:令1x t x +=,两边平方得22212x t x++=即22212x t x+=-,将其代入函数式,得2()2f t t =-,即2()2f x x =-.法二:将函数表达式变形,得222111222f x x x x x x ⎛⎫⎛⎫⎛⎫=-=-++++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭令1x t x+=,得2()2f t t =-,即2()2f x x =-.4.设f (x )为奇函数,证明：若f (x )在x =0有定义,则f (0)=0.证∵f (x )为奇函数,且f (x )在x =0处有定义,∴(0)(0)f f -=-又(0)(0)f f -=于是(0)(0)f f =-即2(0)0,(0)0f f =∴=.5.证明：狄利克雷函数是周期函数,任何一个正有理数均是它的周期,但无最小正周期.证狄利克雷函数1,,()0,当为有理数时当为无理数时.x D x x ⎧=⎨⎩设T 是任一正有理数,x ∀∈R ,当x 为有理数时,x+T 为有理数,于是()1D x T +=,又()1D x =,所以()()D x T D x +=;当x 为无理数时,x+T 为无理数,于是()0D x T +=,又()0D x =,所以()()D x T D x +=.综上所述,x ∀∈R 有()()D x T D x +=,所以()D x 是周期函数,任何一个正有理数均是它的周期,又设P 是任一无理数,x P ∃=-∈R ,使()(0)1D x P P +==,而()0D x =,故()()D x P D x +≠,即无理数不是()D x 的周期;因为不存在最小的正有理数,所以()D x 无最小正周期.习题1-31.设销售商品的总收入是销售量x 的二次函数,已知x =0,2,4时,总收入分别是0,6,8,试确定总收入函数TR(x ).解设2()TR x ax bx c =++,由已知(0)0,(2)6,(4)8TR TR TR ===即04261648c a b c a b c =⎧⎪++=⎨⎪++=⎩解得1240a b c ⎧=-⎪⎪⎨=⎪⎪=⎩所以总收入函数21()42TR x x x =-+.2.设某厂生产某种产品1000吨,定价为130元/吨,当一次售出700吨以内时,按原价出售；若一次成交超过700吨时,超过700吨的部分按原价的9折出售,试将总收入表示成销售量的函数.解设销售量为x ,实际每吨售价为P 元,由题设可得P 与x 间函数关系为1307001177001000x P x ≤⎧=⎨<≤⎩,总收入130700()130700(700)1177001000TR x x x x x ≤⎧=⎨⨯+-⨯<≤⎩,即130700()91001177001000TR xx x x x ≤⎧=⎨+<≤⎩.3.已知需求函数为105QP =-,成本函数为C =50+2Q ,P 、Q 分别表示价格和销售量.写出利润L 与销售量Q 的关系,并求平均利润.解由题设知总收入2()105Q R Q PQ Q ==-,则总利润()221()()()8505021055Q L Q R Q C Q Q Q Q Q ⎛⎫=-=-=--+- ⎪⎝⎭,平均利润()150()85L Q AL Q Q Q Q==--.4.已知需求函数Q d 和供给函数Q s ,分别为Q d =100233P -,Q s =-20+10P ,求相应的市场均衡价格.解当d s Q Q =时供需平衡,由d s Q Q =得1002201033P P -=-+,解得5P =所以市场均衡价格5P =.。
托马斯微积分第13版第三章答案
lim
2
( x3 3 x 2 h 3 xh 2 h3 3 x 3h ) ( x3 3 x ) h h 0
2 2 h 3 3h lim 3 x h 3 xh h
h 0
lim (3x 3xh h 3) 3 x 3; 3x 3 0 x 1 or x 1. Then f (1) 2 and f (1) 2 (1, 2)
h 0
x 2. Then f (2) 4 8 1 5 (2, 5) is the point on the graph where there is a horizontal tangent.
26. 0 m lim
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1 1 ( x h ) 1 x 1
( x 1) ( x h 1)
( x h) x x x h m lim x h lim x h x h x lim lim 1 . 28. 1 4 h h 2 x x h x h 0 h x h x h 0 h x h x h 0 h 0 x 1 1 1 x 2 x 4 y 2. The tangent line is y 2 4 ( x 4) 4 1. Thus, 4 2 x f (2 h ) f (2) h h 0
P (5) lim
h 0
24. (a) From t 0 to t 3, the derivative is positive. (b) At t 3, the derivative appears to be 0. From t 2 to t 3, the derivative is positive but decreasing. 25. At a horizontal tangent the slope m 0 0 m lim
微积分习题集带参考答案
微积分习题集带参考答案2（2），求圆的面积为1时，面积变量S 相对于周长l 的变化率。
解 此时S 是l 的函数 πππ4222l l S =⎪⎭⎫ ⎝⎛=。
于是S 对周长l 的变化率为 π2l dl dS =。
当1=S 时π2=l ，此时ππ12==l dl dS 。
5(2). 设ax y ||=，在0=x 点可导，求α的取值范围。
解 设ax x f ||)(=。
当0≤α时，0=x 是函数的间断点，此时函数不可导。
只讨论0>α。
考虑左导数 ⎪⎩⎪⎨⎧>=<∞===---+→1,0111,0)0()(lim10ααααa x x xxx f x f ， 考虑右导数 ⎪⎩⎪⎨⎧>=-<∞=--=-=----→1,0111,)()(0)0()(lim10ααααa x x x x x f x f ， 因此该函数当1>α时在0=x 点可导，导数为0.6. 设⎪⎩⎪⎨⎧≥+-<≤+<-=1,1)1sin(10,0,1)(x x b x a x x e x f x 。
求b a ,使得)(x f 在1,0=x 可导。
解法1 因可导必连续，则 a f x f x ===-→)0(0)(lim 0，则0=a 。
这样在1=x 处)(x f 也连续。
此时 110)0()(lim )0(0=-=--='-→-x e x f x f f x x ，1lim 0)0()(lim )0(00==--='+→+→+xxx f x f f x x ，。
1111)1()(lim)1(1=--=--='-→-x x x f x f f x ，b x x b x f x f f x x =--=--='+→+→+1)1sin(lim 1)1()(lim )1(11。
若)1('f 存在，则应有b =1。
此时1)1('=f 。
解法2 同理可得0=a 。
《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第7章
M 3 (0, 0,5) ，则点 M (4, 3,5) 到 x 轴，y 轴，z 轴的距离分别为：
d x | MM 1 | (4 4) 2 (3 0) 2 (5 0) 2 34. d y | MM 2 | (4 0) 2 (3 3) 2 (5 0) 2 41. d z | MM 3 | (4 0) 2 (3 0) 2 (5 5) 2 5.
5． 在 yＯz 面上，求与三个已知点 A(3，1，2)，B(4，-2，2)和 C(0，5，1)等距离的点． 解：设所求点 P (0, b, c) ,则 | PA || PB || PC | 即 9 (b 1) (c 2) 16 (b 2) (c 2)
2 2 2 2
| AB || AC | ,且 | AB |2 | AC |2 | BC |2
所以 ABC 是等腰直角三角形. 习题 7-2 1．在平行四边形 ABCD 内，设 AB a , AD b ，M 为对角线的交点，试用向量 a 和 b 表示向量 MA, MB, MC 和 MD ． 解： （如图） DC AB a, BC AD b,
2．试用向量证明：如果平面上一个四边形的对角线互相平分，则该四边形是平行 四边形． 证： （如上题图） ，依题意有 AM MC , DM MB. 于是 AB AM MB MC DM DC. 故 ABCD 是平行四边形. 3．已知向量 a=i-2j+3k 的始点为(1，3，-2)，求向量 a 的终点坐标． 解：设 a 的终点坐标为( x, y, z ),则 源自
a ( x 1)i ( y 3) j ( z 2)k ,
而 a i 2 j 3k , 从而有