英文版微积分试卷答案(1)
微积分英文版第九版课后练习题含答案
微积分英文版第九版课后练习题含答案微积分是数学中重要的一部分，其关注的是一条曲线或曲面的局部性质。
美国著名数学家Stewart所著的微积分英文版第九版是微积分学习的重要教材之一，本文将介绍其课后练习题，并提供答案供大家进行自我学习和测试。
课后练习题微积分英文版第九版的课后练习题共分为两部分，其中Part 1是选择题，Part 2是填空题和证明题，共计约1700道题目。
Part 1中包含了大量的选择题，这些题目主要考察读者对微积分理论的掌握和应用。
大多数题目都要求读者用所学知识推理或计算来获得正确答案。
这些题目中难易程度参差不齐，有一些比较简单，但也有一些比较困难。
Part 2的题目类型较多，包括了填空题、证明题、计算题等。
这些题目内容繁杂、难度较大，需要读者花费很多时间和精力来解答。
这些题目主要是为了检测读者对所学知识的深层次理解和应用能力，考察读者的逻辑思维和推理能力。
答案微积分英文版第九版提供了相应的课后练习答案，可以帮助读者检验自己的答案是否正确，同时也可以帮助读者更好地理解和掌握所学知识。
答案分为Part 1和Part 2两部分，且每部分分别包含了选择题和非选择题的答案。
这些答案详细、准确，提供了完整的解题思路和方法，帮助读者更好地理解题目的解法，并弥补了部分教材中的不足之处。
读者可以通过该教材的官方网站或者其他渠道获得课后练习题的答案。
结论微积分是数学中非常重要的一门学科，对于各个领域的科学研究、技术发展和社会进步都有着举足轻重的作用。
而微积分英文版第九版课后练习题则是培养和检验读者对微积分学习的深刻理解和应用能力的重要途径。
通过对这些题目的研究和答案的掌握，可以帮助读者更好地掌握微积分学科，提升自己的学术能力和科研能力。
AP 微积分 AB 2007 (Form B) 真题与答案
AP® Calculus AB2007 Scoring GuidelinesForm BThe College Board: Connecting Students to College SuccessThe College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns.© 2007 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation.Permission to use copyrighted College Board materials may be requested online at:/inquiry/cbpermit.html.Visit the College Board on the Web: .AP Central is the official online home for the AP Program: .Question 1Let R be the region bounded by the graph of 22x x y e −= and thehorizontal line and let S be the region bounded by the graph of2,y =22x x y e −= and the horizontal lines and 1y =2,y =as shown above. (a) Find the area of R . (b) Find the area of S .(c) Write, but do not evaluate, an integral expression that gives thevolume of the solid generated when R is rotated about the horizontal line 1.y =222x x e −= when 0.446057,1.553943x =Let and 0.446057P = 1.553943Q =(a) Area of ()2220.5Qx x P14dx −=−=⌠⌡⎪⎩R e3 : ⎪⎨ 1 : integrand1 : limits 1 : answer ⎧(b) when 2221x x e −=0,x =Area of S e Area of R()22201x x dx −=−−⌠⌡− Area of2.06016= 1.546R =OR()()()222220110.219064 1.1078860.219064 1.546Px x x x Qe dx Q P e d −−−+−⋅+−=++=⌠⌠⌡⌡1x⎪⎩3 : ⎪⎨ 1 : integrand 1 : limits 1 : answer ⎧(c) Volume ()()2222121Qx x P e d π−⎛⎞−−⎜⎟⎝⎠⌠⎮⌡x =−3 : {2 : integrand1 : constant and limitsQuestion 2A particle moves along the x -axis so that its velocity v at timeis given by The graph of v is shown abovefor 0t ≥()()2sin .v t t=0t ≤≤ The position of the particle at time t is ()x t and its position at time is 0t =()05x =.(a) Find the acceleration of the particle at time t 3.=(b) Find the total distance traveled by the particle from time 0t =to t 3.==(c) Find the position of the particle at time t 3.(d) For 0t ≤≤ find the time t at which the particleis farthest to the right. Explain your answer.Question 3The wind chill is the temperature, in degrees Fahrenheit ()F ,° a human feels based on the air temperature, in degrees Fahrenheit, and the wind velocity v , in miles per hour ()mph . If the air temperature is 32 then the wind chill is given by and is valid for 56F,°()0.1655.622.1W v v =−0.v ≤≤ (a) Find ()20.W ′ Using correct units, explain the meaning of ()20W ′ in terms of the wind chill.(b) Find the average rate of change of W over the interval 560.v ≤≤ Find the value of v at which theinstantaneous rate of change of W is equal to the average rate of change of W over the interval 560.v ≤≤ (c) Over the time interval hours, the air temperature is a constant 32 At time the windvelocity is mph. If the wind velocity increases at a constant rate of 5 mph per hour, what is the rate of change of the wind chill with respect to time at 0t ≤≤4 F.°0,t =20v =3t = hours? Indicate units of measure.Question 4Let f be a function defined on the closed interval 55x −≤≤ with ()13f=. The graph of ,f ′ the derivative of f , consists of two semicircles and two line segments, as shown above.(a) For −< find all values x at which f has arelative maximum. Justify your answer.5x 5,<5,<(b) For −< find all values x at which the graph of fhas a point of inflection. Justify your answer.5x (c) Find all intervals on which the graph of f is concave upand also has positive slope. Explain your reasoning.(d) Find the absolute minimum value of ()f x over the closed interval 5x 5.−≤≤ Explain your reasoning.Question 5Consider the differential equation 11.2dy x y dx =+−(a) On the axes provided, sketch a slope field for the given differential equationat the nine points indicated.(Note: Use the axes provided in the exam booklet.)(b) Find 2d ydxin terms of x and y . Describe the region in the xy -plane inwhich all solution curves to the differential equation are concave up.(c) Let ()y f x = be a particular solution to the differential equation with theinitial condition ()01f =. Does f have a relative minimum, a relative maximum, or neither at Justify your answer. 0?x =(d) Find the values of the constants m and b , for which y mx b =+ is asolution to the differential equation.y −1Question 6Let f be a twice-differentiable function such that ()2f 5= and ()52f .= Let g be the function given by ()()().g x f f x =(a) Explain why there must be a value c for 25c << such that () 1.f c =−′(b) Show that Use this result to explain why there must be a value k for 2 such that()()2g g =′′5.5k <<()0.g k =′′ (c) Show that if ()0f x =′′ for all x , then the graph of g does not have a point of inflection. (d) Let ()().h x f x x =− Explain why there must be a value r for 25r << such that ()0.h r =。
英文版微积分考试样题3
Problems 37 - 39: Solve the problem. Assume projectile is ideal, launch angle is measured from the horizontal, and launch is over a horizontal surface, unless stated otherwise. 24) An ideal projectile is launched from level ground at a launch angle of 26° and an initial speed of 48 m/sec. How far away from the launch point does the projectile hit the ground? A) ≈ 60 m B) ≈ 230 m C) ≈ 290 m D) ≈ 185 m 25) A projectile is fired with an initial speed of 528 m/sec at an angle of 45°. What is the greatest height reached by the projectile? Round answer to the nearest tenth of a meter. A) 7111.8 m B) 76.2 m C) 69,696.0 m D) 28,447.3 m Find the unit tangent vector of the given curve. 26) r(t) = (5 + 2t7 )i + (4 + 10t7 )j + (8 + 11t7 )k A) T(t) = C) T(t) = 2 2 11 i+ j+ k 15 3 15 2 2 11 i+ j+ k 225 45 225 B) T (t) = 14 14 77 i+ j+ k 15 3 15
微积分试题及答案-1
《微积分》试题 第1页（共8页）微积分试题及参考答案与评分标准一、填空题（本大题共10小题，每小题2分，共20分）1、函数21()arcsinln(1)3x f x x -=+-的定义域为 ； 2、2lim(1)x x x→∞-= ；3、若22(sin )cos ,f x x '=则()f x = ；4、设2sin(1)(),1x f x x -=-则x = 是()f x 的可去间断点; 5、若0(32)(3)1limh f h f h→--=,则(3)f '= ；6、1(arctan )x =d ；7、可导函数()f x 是偶函数，若(1)3,f '=则(1)f '-= ;8、函数()f x =[0, 3]上满足罗尔定理条件，结论中的=ξ ； 9、曲线C :2ln 1xy x =-的垂直渐近线是 ； 10、设某商品的需求函数是402Q p =-，则需求价格弹性15|p η== 。
二、单项选择题（本大题共5小题，每小题2分，共10分）1、函数)(x f 在0x 处连续是)(x f 在0x 处可导的（ ）．（A ）必要但非充分条件； （B ）充分但非必要条件；（C ）充分必要条件； （D ）无关条件．《微积分》试题 第2页（共8页）2、当0→x 时，2x 是x cos 1-的（ ）无穷小．（A ）等价； （B ）同阶但不等价； （C ）高阶； （D ）低阶．3、设函数1)(1+=xe xf ，则0=x 为)(x f 的间断点类型是（ ）． （A ）跳跃间断点； （B ）可去间断点； （C ）振荡间断点； （D ）无穷间断点.4、设()f x 的一个原函数是2x ，则2(1)xf x x -=⎰d （ ） （A ）222(1)x C -+； （B ）222(1)x C --+；（C ）221(1)2x C -+； （D ）221(1)2x C --+.5、函数1sin ,0()0,0x x f x xx ⎧≠⎪=⎨⎪=⎩在0x =处（ ）． （A ）不连续； （B ）极限不存在；（C ）连续且可导； （D ）连续但不可导．三、计算题（本大题共8小题，每小题7分，共56分）1、求极限+01lim(1)xx x→+.2、求极限11lim()1ln x x x x→--.《微积分》试题 第3页（共8页）3、设ln(x y e =，求,y y '''.4、求曲线C :2ln(1)y x =+的凹凸区间与拐点.5、求曲线C :1x y xy e ++=在0x =对应点处切线的方程.《微积分》试题 第4页（共8页）6、求函数2()1xf x x =+的单调区间和极值.7、求不定积分()112ln dx x x +⎰.8、求不定积分⎰.《微积分》试题 第5页（共8页）四、应用题（本大题共1小题，共8分）设某产品的总成本函数为：2()5,C x x =+需求函数为：120.5,x p =-其中x 为产量，p 为价格，求（1）收益最大时的产量和价格；（2）利润最大时的产量和价格。
英文版-微积分试卷答案-(1)
（考点：经济函数，课本4.8节。
此题型为常考题，属于送分题，大家可以做相应的4.8节的练习加以巩固）9、State the second derivative test theorem testing maximum and prove it.Suppose is continuous near c''f If and , then has a local maximum at c .'c f （)=0''c f （)< 0Proof: Because near c and so is concave downward near ''c f （)< 0f c . This means that the graph of lies below its horizontal tangent at c and so has a local maximum at c .10、Show that the equation has at most one root in the 3150x x c -+=interval.[2,2]- Suppose the equation has two root in 3150x x c -+=1212,and x x x x <the interval .12,[2,2]x x ∈-Let then there exist 3F()15x x x c =-+12F()F()=0x x =Using the Rolle’s Theorem we can know that there exist one number c can satisfy12F'(c)0(,)c x x =∈2F'()3150[2,2]x x when x =-<∈-But , Therefore, the suppose is wrong. 2F'()3150[2,2]x x when x =-<∈-Then the equation has at most one root in the interval3150x x c -+=.[2,2]（考点：罗尔地理。
微积分B(1)第3次习题课(Stolz定理、函数极限)答案
( A − ε )(1 −
即
( A − ε )(−
bN +1 a a b a ) + N +1 < n < ( A + ε )(1 − N +1 ) + N +1 bn bn bn bn bn
因为
n →∞
lim bn = +∞
，所以 lim bb
n →∞
bN +1 a a b a ) + N +1 − ε < n − A < ( A + ε )( − N +1 ) + N +1 + ε bn bn bn bn bn
n →∞
+ 2m + ⋯ + n m n m +1
，其中 m 为自然数．
lim 1m + 2m + ⋯ + nm (n + 1) m = lim n →∞ n →∞ ( n + 1) m +1 − n m +1 nm +1
= lim
( n + 1) m 1 = n →∞ (m + 1)m m −1 m +1 m (m + 1) n + n +⋯ +1 2
1 1
．
（3）求
n− 2 2n −1 2 2 2n−1 22 2 lim 2 3 ⋯ n n →∞ 2 − 1 2 −1 2 −1
1
．
解：令
所以
n−2 2n−1 2 2 2n−1 22 2 an = 2 3 ⋯ n 2 −1 2 −1 2 −1
英文版微积分
Infinite Sequences and Series
一、Determine whether the series is convergent or divergent
2 2 n !
(1) n 1 n n
(2)
(
n )n
n1 2n 1
2n2 3n1
(3)
(4)
1
n1 3n7 n2 2
Determine whether the series
an
n0
is convergent or divergent. And find the sum of this series.
Example
(19) 求幂级数
1 n1 x2n1
n1 n 2n 1
的收敛域及和函数
(9) 若级数
求幂级数
1 (1)n x2n (x1)
n1
2n
的和函数f(x)及其极值.
True or false
lim a 0 a 1. If
,then
n n
is convergent.
n
2.
a If
is divergent ,then
divergent. n
is
an
3.
a b If
for all ,and
z
z .
x y
x3 y3 2.If f(x,y)x2 y2
calculate
and 0
x2 y2 0 x2 y2 0
fx (0,0)
f y (0,0).
Vectors and Geometry of Space
If is the angle between the nonzero
英文版微积分试卷答案（1）
1、(1) sin 2limx x x→∞=0 . (2)d(arctan )x =21d 1+x x (3) 21d sin x x =⎰-cot +Cx x (4).2()()x n e =22nxe (5)412d x x +=⎰26/3 2、(6) The right proposi on in the following proposi ons is ___A_____.A. If lim ()x af x →exists and lim ()x ag x →does not exist then lim(()())x af xg x →+does not exist.B. If lim ()x af x →,lim ()x ag x →do both not exist then lim(()())x af xg x →+does not exist. C. If lim ()x af x →exists and lim ()x ag x →does not exist then lim ()()x af xg x →does not exist. `D. If lim ()x a f x →exists and lim ()x ag x →does not exist then()lim ()x a f xg x →does not exist.(7) The right proposi on in the following proposi ons is __B______.A. If lim ()()x af x f a →=then ()f a 'exists. B. If lim ()()x a f x f a →≠then ()f a 'does not exist. C. If ()f a 'does not exist then lim ()()x af x f a →≠.D. If ()f a 'does not exist then the cure ()y f x =does not have tangent at (,())a f a .(8) The right statement in the following statements is ___D_____. A. sin lim1x xx x→∞= B. 1lim(1)xxx e→∞+=C. 11d 1x x x Cααα+=++⎰ D. 5511d d 11bb a a x yx y =++⎰⎰(9) For con nuous func on ()f x , the erroneous expression in the following expressions is ____D__./A.d (()d )()d ba f x x fb b =⎰ B. d (()d )()d ba f x x f a a =-⎰C. d (()d )0d ba f x x x =⎰ D. d (()d )()()d ba f x x fb f a x=-⎰(10) The right proposi on in the following proposi ons is __B______.A. If ()f x is discon nuous on [,]a b then ()f x is unbounded on [,]a b .B. If ()f x is unbounded on [,]a b then ()f x is discon nuous on [,]a b . C. If ()f x is bounded on [,]a b then ()f x is con nuous on [,]a b .D. If ()f x has absolute extreme values on [,]a b then ()f x is con nuous on [,]a b . 3、Evaluate 2011lim()xx e x x →-- 201=lim()xx e x x →--01=lim()2xx e x →-01=lim =22xx e →（考点课本节洛比达法则，每年都会有一道求极限的解答题，大多数都是用洛比达法则去求解，所以大家要注意节的内容。
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1、 (1) sin 2limx xx→∞= 0 .(2) d(arctan )x = 21d 1+x x(3)21d sin x x =⎰ -cot +C x x(4).2()()x n e = 22n x e .(5)x =⎰26/32、(6) The right proposition in the following propositions is ___A_____.A. If lim ()x af x →exists and lim ()x ag x →does not exist then lim(()())x af xg x →+does notexist.B. If lim ()x af x →,lim ()x ag x →do both not exist then lim(()())x af xg x →+does not exist.C. If lim ()x af x →exists and lim ()x ag x →does not exist then lim ()()x af xg x →does notexist.D. If lim ()x af x →exists and lim ()x ag x →does not exist then ()lim()x af xg x →does not exist. (7) The right proposition in the following propositions is __B______.A. If lim ()()x af x f a →=then ()f a 'exists.B. If lim ()()x af x f a →≠ then ()f a 'does not exist.C. If ()f a 'does not exist then lim ()()x af x f a →≠.D. If ()f a 'does not exist then the cure ()y f x =does not have tangent at (,())a f a .(8) The right statement in the following statements is ___D_____.A. sin lim 1x xx→∞= B. 1lim(1)x x x e →∞+=C. 11d 1x x x C ααα+=++⎰ D. 5511d d 11b b a a x y x y =++⎰⎰ (9) For continuous function ()f x , the erroneous expression in the followingexpressions is ____D__. A.d (()d )()d b a f x x f b b =⎰ B. d (()d )()d baf x x f a a =-⎰C. d (()d )0d b a f x x x =⎰D. d (()d )()()d baf x x f b f a x =-⎰(10) The right proposition in the following propositions is __B______.A. If ()f x is discontinuous on [,]a b then ()f x is unbounded on [,]a b .B. If ()f x is unbounded on [,]a b then ()f x is discontinuous on [,]a b .C. If ()f x is bounded on [,]a b then ()f x is continuous on [,]a b .D. If ()f x has absolute extreme values on [,]a b then ()f x is continuous on [,]a b .3、Evaluate 2011lim()x x e x x →-- 201=lim()x x e x x →--01=lim()2x x e x →-01=lim =22x x e →（考点课本节洛比达法则，每年都会有一道求极限的解答题，大多数都是用洛比达法则去求解，所以大家要注意节的内容。
注意洛比达法则的适用范围。
）4．Find 0d |x y =and (0)y ''if 20x x x y y t e +=+⎰.2'()'x x x y y t e +=+⎰（）1'2()'2()1x x y x y x e y x y x e +=⋅+⇒=⋅+-0(20(0)1)0x dy y e dx dx==⋅⋅+-=''(2()1)'2()2'()x xy x y x e y x xy x e =⋅+-=++200-(0)0-01x x y y t e x y e =+⇒=+=⎰0''02(0)20'(0)=3y y y e =+⋅+（） （考察微积分基本定理与微分，书上节）5、 Find 22arctan d (1)xx x x +⎰=22221)arctan d (1)x x x x x x +-+⎰（22arctan arctan =d d (1)x x x x x x -+⎰⎰-12311=-arctan +d arctan +2x x x x x x -⎰22-1221++1=-arctan +d arctan 1+2x x x x x x x x -⎰（） -12211=-arctan +d d arctan 1+2x x x x x x x x --⎰⎰（）-12211=-arctan +In In 1+arctan 22x x x x x ---121=-arctan arctan +C 2x x x - （凑微分求不定积分，积分是微积分的重点及难点，大家一定要掌握透彻。
）6、 Given that 22()1x f x x =+.(1) Find the intervals on which ()f x is increasing or decreasing.22’22212()1x x x x f x x +-⋅=+（）（）2221x x =+（）When ’()00f x x >⇒>’()00f x x <⇒< Therefore, the increasing interval is ()0+∞，, the decreasing interval is ()0-∞，(2) Find the local maximum and minimum values of ()f x’()00f x x =⇒= The function is increasing in interval ()0+∞，, decreasing ininterval ()0-∞，, therefore, the function exist the local minimum value, it is ()0f x =(3)Find the intervals of concavity and the inflection points.'22224222242422181642''()111x x x x x x f x x x x +-+--+⎛⎫=== ⎪+++⎝⎭（）（）（）（）（）4224642''()0133x x f x x x x --+=>⇒><-+（）4224642''()0001x x f x x or x x --+=<⇒<<<<+（）1''(4f f =Therefore, the concave upward interval are 3⎛⎫-∞- ⎪ ⎪⎝⎭，,3⎛⎫+∞ ⎪ ⎪⎝⎭, the concave downward intervalare -03⎛⎫⎪⎪⎝⎭,03⎛⎫ ⎪ ⎪⎝⎭， and the inflection pointsare 134⎛⎫ ⎪ ⎪⎝⎭，,134⎛⎫⎪ ⎪⎝⎭， (4) Find the asymptote lines of the cure ()y f x =2221lim =1111+x x x x→∞=+ T herefore, the liney = 1 is a horizontal asymptote（考点：节，、节。
近几年经常会考一道作图题。
这种题目应该在注意的点主要包括函数的定义域，对称性，增减区间，极值点，凹凸性，拐点，以及渐近线等。
大家参照课本的节进行作图）7、Let R be the region bounded by the curve 1y x=, and the line y x = and 2x =.(a)Evaluate the area of the region R. R=211x dx x -⎰2211=In 2x x -2211=2In21In122⋅--⋅+3=In22- (b)Find the volume of the solid generated by revolving the R about they -axis .V=212121214)?4)y dy dy y ππ-+-⎰⎰（（12311211443x y y y π⎛⎫ ⎪=-++ ⎪⎝⎭33111142241141423312π⎛⎫=⋅-⋅-⋅+⋅+⋅+-⋅- ⎪⎝⎭83π=（考点：求面积以及体积，课本、节。
这类题目是常考题，较简单。
望同学一定要做相应的题目加以巩固。
）8、 Determine the production level that will maximize the profit for a company with cost and demand functions23()1450360.580.001C x x x x =+-+and ()600.01p x x =-.Solution 2()(600.01)600.01R x x x x x=-=-223()()()600.01(1450360.580.001)P x R x C x x x x x x =-=--+-+320.0010.57241450x x x =-++-'2()0.003 1.142P x x x =-++Let '()040020P x x or x =⇒==- since x>0, then x=400''()0.006 1.14P x x =-+When x=400 ''()0.006400 1.14 1.260P x =-⋅+=-<32(400)0.0014000.5740024400145035350P =-⋅+⋅+⋅-=Therefore, when the production level is 400 that will maximize the profit 35350 （考点：经济函数，课本节。