微积分习题答案经济数学微积分主编张建梅马庆华

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微积分答案经济数学微积分主编张建梅马庆华科学出版社广外

y2
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7. (2) 计算积分 I
y x

1 2 1 4
dy 1 e dx 1 dy
2 2
y
y x
1
y y
e dx .
y x
解 e d x 不能用初等函数表示
先改变积分次序.
原式
y x
y x2
I

1
1 2
1
1 2
dx
x x
2
e dy
x
y x
dxdy ,
解由对称性，可只考虑第一象限部分，
D 4 D1
D1

D
注意：被积函数也要有对称性. 2 2 4 sin( x y )
x2 y2

2
dxdy

D1
sin(
2
x2 y2 ) x y
2
dxdy
4 d
0
2
1
sin r rd r 4. r
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6.
(1)
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6(2)
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6(3) 改变积分

1 0
dx
2 x x2 0
f ( x , y )d y d x
1
2
2 x 0
f ( x , y )d y 的次序.
解积分区域如图
y 2 x
y 2x x2
原式

1 0
dy
xy
t 解: 两个隐函数方程两边对 x 求导, 得
u

微积分答案

International Monetary FundMoldova and the IMF Press Release:IMF Executive Board Completes Second Review Under the Extended Credit Facility and the Extended Fund Facility Arrangements with Moldova, Approves US$79 Million Disbursement April 7, 2011Country’s Policy Intentions DocumentsE-Mail Notification Subscribe or Modify your subscription Moldova: Letter of Intent, Supplementary Memorandum of Economic and Financial Policies, and Technical Memorandum of UnderstandingMarch 24, 2011M OLDOVA:L ETTER OF I N TE N TChişinău, March 24, 2011 Mr. Dominique Strauss-KahnManaging DirectorInternational Monetary Fund700 19th Street NWWashington, DC 20431 USADear Mr. Strauss-Kahn:The economic program supported by the IMF is playing a crucial role in restoring stability and rebuilding confidence in Moldova. With growth significantly exceeding projections in 2010, GDP has broadly recovered to pre-crisis levels. Inflation is under control, and the fiscal deficit has narrowed substantially. These remarkable results were achieved notwithstanding the challenges that the economy faces: fiscal adjustment and promotion of export-led growth require profound structural reforms; rising international food and fuel prices rekindle inflation pressures; job creation lags behind and unemployment still exceeds pre-crisis levels.The program is broadly on track. All quantitative performance criteria for end-September and most indicative targets for end-December 2010 were observed. However, the difficult political environment of 2010 and unforeseen technical complications have taken their toll, and several structural benchmarks under the program were delayed. In the coming period, we will move expeditiously to implement these measures, as well as the new reforms set forth in our agreement with the IMF. The 2011 fiscal budget consistent with the program objectives will be adopted as a prior action for completion of this review. In addition, we have prepared the Annual Progress Report on the implementation of our National Development Strategy and circulated it to the IMF Executive Board for information.In consideration of our strong record of program implementation, we request the completion of the second review of the program supported by the Extended Credit Facility and the Extended Fund Facility arrangements and the associated disbursement of SDR 50 million. As the Executive Board consideration of our request falls in early April 2011, we also request waivers of applicability of the relevant end-March performance criteria. The third program review, assessing performance based on end-March 2011 performance criteria and relevant structural benchmarks, is envisaged for June 2011. Moldova remains committed to improving the well-being of the population through reforms that promote sustainable growth and reduce poverty. In the period ahead, our program will focus on maintaining the targeted pace of fiscal adjustment; reining in inflation pressures; strengthening financial stability of the banking sector; restructuring the energy sector; rolling out the long-awaitededucation and other structural reforms that would support Moldova’s reorientation toward export-led growth.We believe that the policies set forth in the attached Supplementary Memorandum of Economic and Financial Policies (SMEFP) are adequate to achieve these objectives but will take any additional measures that may become appropriate for this purpose. We will consult with the IMF on the adoption of such additional measures in advance of revisions to the policies contained in the SMEFP, in accordance with the Fund’s policies on such consultation. We will provide the Fund with the information it requests for monitoring progress during program implementation. We will also consult the Fund on our economic policies after the expiration of the arrangement, in line with Fund policies on such consultations, while we have outstanding purchases in the upper credit tranches. Sincerely yours,/s/Vladimir FilatPrime MinisterofRepublicMoldovatheGovernmentof/s/ /s/NegruţaVeaceslavValeriu LazărFinanceofDeputy Prime Minister MinisterEconomyMinisterof/s/Dorin DrăguţanuGovernorNational Bank of MoldovaAttachment: Supplementary Memorandum of Economic and Financial PoliciesUnderstandingofMemorandumTechnicalS UPPLEME N TARY M EMORA N DUM OF E CO N OMIC A N D F I N A N CIAL P OLICIESMarch 24, 20111.The present document supplements and updates the Memoranda of Economic and Financial Policies (MEFPs) signed by the authorities of the Republic of Moldova on January 14, 2010 and June 30, 2010. It accounts for recent macroeconomic developments and introduces policy adjustments, as well as additional policies necessary to achieve the objectives of the program. We remain determined to meeting our commitments made previously under the program.I. M ACROECO N OMIC D EVELOPME N TS A N D O UTLOOK2.Growth outperformed expectations in 2010, and the economic expansion is set to continue. Real GDP rebounded by 6.9 percent in 2010, more than offsetting the economic contraction of 6 percent recorded in 2009. We expect the economic growth to return to its sustainable pace of 4½-5 percent in 2011 and thereafter. Expansion of domestic demand, exports, and investment are expected to drive activity in the near term, with tailwinds from trade liberalization reforms, a more favorable external environment, and improving competitiveness.3.Barring severe external shocks, disinflation should continue in 2011-12. Despite adjustment of energy tariffs, depreciation of the leu, and higher excise rates, inflation remained under control at around 8 percent in 2010, while core inflation declined below 5 percent. Under our baseline assumptions for international food and energy prices, we expect that inflation will decline further to 7½ percent in 2011 and about 5 percent by end-2012, the medium-term target set by the NBM. However, we recognize the risk that further surges in international food and energy prices and faster than expected rebound in domestic demand can temporarily push headline inflation above the projected path.4.Strong economic recovery boosted budget revenues and helped improve the fiscal position. In 2010, revenue significantly exceeded the program projections in nominal terms, but underperformed as percent of GDP, mainly due to high contribution to growth of the largely untaxed agriculture. Expenditure targets were also comfortably met, albeit largely due to under-spending of the capital budget caused by capacity constraints. As a result, the cash budget deficit narrowed to 2½ percent of GDP in 2010, far below the program target of5.4 percent of GDP.5.After a sharp drop to single digits in 2009, the external current account deficit widened in 2010 and will remain elevated in 2011. Rising demand for consumer and investment goods has pushed the current account deficit to an estimated 12¾ percent of GDP in 2010. The same demand factors, along with higher costs of energy imports, will likely propel the deficit even higher in 2011. The elevated deficit in 2011 will be largely financed by official assistance, private capital flows, and FDI. As the economy’s borrowing space is filling up quickly, we realize that further external borrowing should proceed at a more measured pace. We expect that from 2013, thanks to our exportpromotion efforts and economic recovery in trading partners, higher exports will more than offset the rise in imports, and the current account deficit would decline towards 10 percent of GDP.6.The situation in the financial sector has improved as well, with domestic credit rebounding and nonperforming loans declining. After the decline of 2009, domestic bank credit expanded by about 13 percent in 2010, and interest rates have declined. Meanwhile, the share of nonperforming loans declined to 13.3 percent, in part reflecting write-offs. Moreover, banks maintain large liquidity and capital buffers, remaining resilient to potential risks.II. R EVISED P OLICY F RAMEWORK FOR 2011-12A. Fiscal Policy7.Building on the better-than-expected fiscal outcome in 2010, the structural fiscal adjustment will stay on course in 2011-12. Our goal is to bring down the structural fiscal deficit excluding grants—the fiscal deficit adjusted for the effects of economic cycles—from 5½ percent of GDP at end-2010 through 4½ percent of GDP in 2011 to 3½ percent of GDP by 2012. This would largely rid the budget from its dependency on exceptional foreign aid and make public finances more resilient to macroeconomic risks. In this context, we will continue to contain the unaffordable public sector wage bill and low priority current spending, while strengthening revenue through selected tax policy measures and improved tax administration. Using the created fiscal space to increase infrastructure investment and provide well-targeted social assistance to the most vulnerable will allow us to achieve our broader development goals.8.As a next step, we will adopt a 2011 budget with a deficit of 1.9 percent of GDP as a prior action. We project that the budget revenue will amount to 37¾ percent of GDP in 2011, on account of continued progress in the tax administration reform, increased excise rates on tobacco and hard liquor—in line with our EU Association agenda—and updates of selected local taxes and fees. Implementation of various structural reforms, described below, will allow us to reduce current expenditure by 1½ percent of GDP to 34½ percent of GDP. At the same time, priority social assistance spending will be safeguarded, and capital expenditure will increase to 5¼ percent of GDP. We will seek to maintain the targeted structural fiscal adjustment in case the economic outlook and budget revenue deviate from our current projections.9.With immediate fiscal pressures easing, structural reforms will help contain the large public sector wage bill while creating space for poverty reduction actions. The significant optimization efforts in the education sector (¶19) will help finance the increase of teachers’ wages planned for September 2011. During 2011, other public wage restraints will remain in place as described in Law 355, as amended in October 2009. The only exception will be made for low-income auxilliary personnel in the budget sector (with salaries below MDL 1500), whose wages will be indexed by 8.5 percent on average from July 1, 2011 to alleviate the impact of higher than expected food and fuel prices and to avoid disincentives to labor market participation. Moreover, public sectoremployment will be capped at 212,000 positions by end-2011, reflecting the effects of the education reforms, while all vacant positions in excess of that level will be eliminated in 2011.10.Greater emphasis will be placed on synchronizing fiscal consolidation efforts at the central and local levels. The local governments will be granted greater control over local tax rates and fees to allow better revenue planning. In particular, by end-March 2011, we will ensure parliamentary passage of the necessary legal amendments to remove ceilings on existing local taxes and fees. This would allow the Chişinău municipality to raise at least MDL 100 million in additional revenues to finance, among other things (discussed in ¶21), its program of granting wage supplements and heating assistance in 2011. The practice of granting these payments will be discontinued at end-2011. The Ministry of Finance will verify compliance with these commitments.11.Going forward, we will continue trimming down current spending while creating sufficient space for the large public investment needs. In 2012, we aim to reduce the budget deficit further to ¾ percent of GDP, mainly through further rationalization of current spending (1 percent of GDP), sustained by structural reforms (¶¶19-22) that will commence in 2011 and bear fruit over the medium term. Ensuring sustainability of public finances in the medium term will also require implementation of the following measures:∙To reduce spending on goods and services, we will persevere with our procurement reform, assisted by the World Bank. The reform, to be phased in during 2011, will lower the budget costs by automating the bids for delivery of goods and services in the government’scentralized procurement agency.∙To improve control over budget planning and execution, we have drafted a law on public finance and accountability which will introduce a rule-based fiscal framework, enhance fiscal discipline, and improve transparency. We expect the law to be passed by Parliament by end-September 2011 and used in the preparation of the 2012 budget.∙To ensure the most effective allocation of capital expenditure, we will review the list of existing and envisaged capital projects, with a view to prioritize execution on the basis oftheir viability and economic growth potential. The review will also take into account pastexecution rates and capacity for implementation.∙To ensure implementation of the recently approved tax compliance strategy, by April 30, 2011, the State Tax Service (STS) will put in place operational plans for the strategyimplementation, including audit, collection of arrears, and taxpayer service activities(structural benchmark). In addition, by September 30, 2011, we will draft and submit toParliament legislation to allow indirect assessment of individuals’ income based on theirassets and other indicators as specified in the compliance strategy. On this basis, byDecember 31, 2011, we will prepare operational plans to strengthen audit, enforcement,outreach to, and education of high-wealth individuals regarding their tax compliance.∙We will reform the outdated mechanism for sick leave benefits. By March 31, 2011, we will amend legislation to assign the financial responsibility for the first day of sick leave to theemployee and the second day to the employer, effective July 1, 2011 (structural benchmark for end-April). Further legal amendments—to accompany the passage of the 2012 budget—will increase the number of sick leave days covered by employers to 3 in 2012, 4 in 2013, and6 in 2014.∙Early retirement privileges will be gradually phased out. By March 31, 2011, we will adopt legislation that, starting July 1, 2011, would raise the statutory retirement age of civilservants, judges, and prosecutors by six months every year until it reaches the regularretirement age (structural benchmark for end-April). This legislation will also extend the requirement to pay social contributions to all persons employed in Moldova in line withbilateral treaties. Another related piece of legislation, also to be passed by March 31, 2011,will put in place a policy of increasing the years of contribution required for full pensioneligibility from 30 to 35 years (and from 20 to 25 years for military and police personnel), by6 months every year, starting July 1, 2011.∙Building on the findings and recommendations of the recent IMF TA mission, we will implement measures to rationalize the use of health care. In particular, from January 1, 2012 we will introduce a copayment of 20 lei for primary care visits for uninsured patients, tomotivate them to enroll into the health insurance system. From January 1, 2013, we willintroduce small copayments for each doctor and hospital visit (5 lei for primary care, 10 leifor specialists, and 20 lei for hospital admissions) for all other categories of patients,including those who currently receive medical services free of charge. This policy will raise revenue and deter the use of unnecessary care, thus reducing the burden on the system. Tothis end, by end-April 2011 we will prepare an action plan detailing needed legislativechanges, technical preparations, and public information campaign.B. Monetary and Exchange Rate Policies12.The N BM’s monetary policy will be focused on achieving its end-2012 inflation objective of 5 ± 1½ percent. Given the fast economic recovery, closing output gap, and inflation pressures from rising international food and energy prices, the NBM’s monetary policy stance will gradually shift from supporting the recovery to addressing inflation risks. Specifically, it should focus on anchoring expectations—thereby countering the second-round effects from surging food and energy prices—and preventing excessive credit expansion. In this context, the NBM’s recent tightening measures—the 100 basis points hike in the policy interest rate and the increase in required reserve ratio from 8 percent to 11 percent— adequately address current inflation concerns. Further tightening should be conditional on marked acceleration of credit growth or rising inflation expectations.13.At the same time, the N BM will continue to strengthen the operational and legal aspects of its monetary policy framework. Consistent with the transition to inflation targeting, theindicative target for reserve money under the program will be discontinued after March 2011. Nevertheless, the NBM will continue to monitor money growth closely as an indicator of the state of domestic demand and sharp sustained moves may warrant policy action. In parallel, the NBM will continue to further enhance its communication, research, and forecasting capacities. As regards the legal framework, by end-September 2011, the NBM will propose amendments to the central bank law to strengthen its independence in line with the international best practice and establish appropriate mechanisms of internal control over NBM’s corporate governance.14.Alongside, the N BM’s exchange rate policies will remain consistent with program objectives. Specifically, NBM interventions in the foreign exchange market will continue to aim at smoothing erratic movements, but not resist sustained depreciation pressures. Should capital inflows exceed program projections, the NBM will accelerate the pace of reserve accumulation to ensure adequate buffers against the still high external vulnerabilities.C. Financial Sector Policy15.To strengthen financial stability, we will address the quasi-fiscal liabilities stemming from recent crisis management efforts. The Government’s decision to shield from losses the depositors of Investprivatbank (IPB) that failed in 2009 was a necessary step to avoid potential panic and deposit runs. However, paying out these deposits by means of a loan from the majority state-owned Banca de Economii (BEM) to IPB—in turn, enabled by a liquidity-providing loan from the NBM—has created a burden on BEM’s balance sheet that is now inhibiting its development. To address this problem, by end-May 2011 the Government will issue to BEM a long-term bond equal to the residual face value of BEM’s loan to IPB by either purchasing this loan or—subject to agreement of BEM’s minority shareholders—recapitalizing the bank. Meanwhile, the NBM will consider a limited extension of its loan to BEM to mitigate the attendant liquidity risk, and will work with BEM and the IPB liquidator to accelerate the sale of IPB assets. The Deposit Guarantee Fund will assume the responsibility for the net cost of the payout to IPB depositors and may introduce an extraordinary deposit insurance premium to gradually reimburse the Government for the cost of the bond issued to BEM.16.To handle future risks better, we aim to put in place the remaining elements of our contingency planning framework. Recent strengthening of the bank resolution framework and the establishment of a high-level Financial Stability Committee (FSC) were followed by signing of a memorandum of understanding (MoU) between key institutions involved in responding to financial emergencies. As a next step, we aim to put in place specific contingency plans for each MoU participant by end-June 2011. These plans will establish a contingency framework based on a clear set of instruments, division of roles, responsibilities, as well as coordination channels between the involved parties.17.Looking ahead, as credit growth picks up speed, the N BM will need to strengthen its bank supervision framework by improving data collection and reducing scope for regulatoryarbitrage. To this end, the NBM, based on best international practices, will develop a new reporting system for commercial banks allowing a more detailed analysis of financial sector data. In addition, by end-September 2011, the NBM and the National Commission for Financial Markets, with assistance from the World Bank, will explore options and make proposals to consolidate all credit institutions—including banks, leasing companies, savings and credit associations, and microfinance institutions—as well as insurance companies and pension funds under a common supervisory framework. Finally, by end-September 2011, the NBM in cooperation with the World Bank will evaluate the feasibility of establishing a public credit bureau to promote information exchange and prudent lending policies by banks.18.Despite earlier delays, measures to strengthen the debt restructuring and contract enforcement frameworks are being developed and will be implemented in the coming months. The NBM has already allowed faster reclassification of restructured loans into lower-risk categories. We will now ensure by end-September 2011 parliamentary passage of the legal amendments described in the SMEFP of June 30, 2010 (¶15), to enhance the speed and predictability of collateral execution by banks and to strengthen incentives for banks to restructure nonperforming loans (structural benchmark). Furthermore, with technical assistance from the World Bank and in consultation with the IMF staff, we will seek to strengthen and simplify other aspects of the insolvency framework. Specific draft legal amendments in this area will be adopted by the Government by March 2012.D. Structural ReformsRaising Efficiency of the Public Sector19.In the coming months, we will roll out the comprehensive reform of the oversized education sector. Its main goals are to eliminate excess capacity, create a leaner and better-equipped education system with adequately trained and paid staff, and provide education that meets demands of the modern economy. The reform will seek class, school, and employment consolidation. A large part of the eventual budget savings and financial assistance from the World Bank will be used to improve school quality, secure transportation for students, and repair school bus routes. Nevertheless, the reform will save about 0.5 percent of GDP on a net permanent basis from 2013 on. Our reform strategy is based on the following elements:∙Class size optimization. By September 1, 2012, we will increase class size to 30-35 students in large schools and 25-30 students in the rest. For this purpose, we will pass legalamendments to eliminate the existing norms prescribed in the Law on Education by end-July 2011. This would reduce the number of teaching positions by 1,736, including 390 positions in 2011, and lead to estimated annual savings of about MDL 94 million.∙Optimization of the school network. Gradual consolidation of the school network through closure of schools with low enrollment and securing transportation of students to nearby“hub” schools will commence this year. Its full implementation during 2011-13 would reducethe number of teaching and non-teaching positions by 2,661 and 1,426 respectively and, when completed, will generate savings of about MDL 136 million a year. We will aim to limit the attendant transportation costs to MDL 61 million per year, and will seek grant assistance from the international financial community to defray this cost.∙Reduction of non-teaching personnel and vacant positions. As a first step, we will immediately freeze hiring of non-teaching staff and eliminate 2,400 vacant positions in thesector. Alongside, we will include in the budget law for 2011 a provision establishing wage bill ceiling for education sector, resulting in all rayons reducing personnel in educationinstitutions on average by 5 percent from their level of end 2010 (5,300 positions nationwide) before academic year 2011/12. These measures would provide savings of MDL 175 million on a full-year basis.∙Increasing flexibility of labor relations in the sector. Local authorities also need support and more flexibility to be able to consolidate schools and classes. By end-July 2011, we willadopt legal amendments to the Labor Code and other enabling legislation to (i) make fixed-term (one year) contracts mandatory for teachers beyond retirement age; and (ii) allow school principals’ hiring and dismissal decisions to be based on business need and performancerather than tenure. Estimated annual savings from this measure amount to MDL 48 million. ∙Rollout of a per-student financing system. Following successful implementation of per-student financing in the pilot rayons of Cauşeni and Rişcani, the system will be expandedstarting January 1, 2012 to 9 additional rayons, as well as municipalities of Chişinău andBalţi. The system will create strong incentives to optimize schools’ financial performance. Its nationwide implementation will take place in 2013.∙Putting social protection costs in education on a means-tested basis. By end-June 2011, in consultation with the World Bank and other partners, we will conduct a thorough review ofall social expenditure in the education budget (scholarships, dormitory assistance, schoolmeals, etc.) to explore options for better targeting of such assistance to the most vulnerablegroups.In consultation with the World Bank, the Government will develop and, by end-March 2011, adopt a detailed action plan to implement this reform.20.We will reform the civil service in a way that increases efficiency without destabilizing the fiscal position. To this end, we have developed descriptions of new job functions and responsibilities for staff in central government administration along with a merit- and performance-based wage system for civil servants. Implementation of this reform will start in October 2011, and will ensure that the reform does not affect the aggregate public sector wage bill as a ratio to GDP. 21.As regards the energy sector, we will strive to achieve a stable framework for payments of current bills, pending a comprehensive sector restructuring strategy to be finalized and implemented in cooperation with the World Bank and other partners. To ensure a stablefunctioning of the sector, the Ministry of Economy, the Chişinău municipality authorities, and the key participants in the energy sector will seek to negotiate in good faith a MoU with the following key elements: (i) a monthly schedule of payments to energy suppliers that is consistent with typical collection lags in Termocom’s receivables during the heating season, (ii) full repayment of current arrears by Termocom before the following heating season; (iii) a mechanism for covering the cash gap arising from collection lags in Termocom or a bank guarantee from the Chişinău municipality backing Termocom’s adherence to the agreed payment schedule; (iv) creditors’ commitment to abstain from blocking bank accounts as long as the MoU is observed. In this context, the Chişinău municipality will budget for and pay in full its remaining debt to Termocom of MDL 64 million by end-March 2011.22.Meanwhile, we will adopt a number of legal and regulatory amendments which would help ensure cost recovery in the heating sector. By end-August 2011, we will adopt the necessary legal and/or regulatory amendments to raise the heating fee for apartments disconnected from central heating from 5 percent to 20 percent of the average heating bill. This increase is in line with regional practices and would mostly affect consumers with relatively high incomes. At the same time, the Ministry of Regional Development and Construction, the Chişinău municipality, Termocom, and the water distributor Apă Canal will seek to put an end to persistent losses caused by under-billing for hot and cold water delivery; other municipalities will seek to resolve this issue as well. And to facilitate timely collection of heating bills, by end-August 2011, we will adopt the necessary legal and/or regulatory amendments introducing a minimum payment of 40 percent of the monthly bill and setting August 1 as the deadline for settling all heating bills for the past heating season.23.With the international investment climate gradually improving, the government will accelerate the efforts to divest its noncore assets. In the first half of 2011 the government, with assistance from IFC, will put in place an advisor to review various options for private sector participation in Moldtelecom. At the same time, by mid-2011, the government will expand the list of state assets subject to privatization. This will pave the way for privatization of other large public companies. By end-September 2011, the government will approach various international financial institutions, seeking an advisor to explore options to divest Air Moldova as soon as possible. Also by end-September 2011, we shall develop a roadmap for the privatization of Banca de Economii, and, if need be, resume the engagement of the privatization advisor.Improving the Business Environment and Removing Barriers for Trade24.The wheat export ban introduced in response to dwindling grain stocks in early 2011 will be abolished as soon as possible, and we will not introduce any new barriers to trade. We plan to abolish this ban by end-April 2011, provided that domestic and regional grain shortages are alleviated. Moreover, we shall refrain from introducing any new tariff or non-tariff barriers to exports. In addition, by end-May 2011 we will conduct an assessment of the existing tariff and non-tariff barriers to trade and their consistency with Moldova’s WTO commitments with regard to market access, and will develop roadmap for their gradual elimination.。

微积分教材习题答案

第一章初等函数习题答案练习题1.11. 数3.1415926是有理数。

2.是无理数。

3. 数4. 有限区间有4个，分别为(),a b [],a b [),a b (],a b 无限区间有5个，分别为(],b -∞ (),b -∞(),a +∞ [),a +∞ (),-∞+∞5. 实数集合{}||2|1,x x x R -<∈用区间表示为()1,36. 实数集合{}||1|2,x x x R -<∈可以认为是1为中心，长度为4的开区间。

7. 实数集合{}||1|2,x x x R -<∈可以称为1的邻域。

8. 以点3为中心，区间长度为1的邻域表示为{}||3|1,x x x R -<∈ 练习题1.21. 函数的三种表示法分别为公式法，图像法，列表法。

2. 单调增函数的是y x =，3y x =，xy e =，ln y x =，lg y x =，tan y x =，arcsin y x = arctan y x = ；单调减函数的是xy e -=， cot y x = 分区间的增减函数是2y x =， sin y x =，cos y x =. 3. 函数2()ln f x x =和()2ln g x x =不是相同函数。

由于2()ln f x x =的定义域是0x ≠；()2ln g x x =的定义域是0x >。

4. 函数()f x x =和()g x =不是相同函数。

由于()f x x =的值域是()f x R ∈，()g x 的值域是()0g x ≥。

5. 求下列函数的定义域：（1）y =解：[)(]1,,1+∞⋃-∞-（2）21()1f x x =- 解：[)2,-+∞且1x ≠± （3）()ln(1)f x x =+ 解：()1,-+∞（4）()lg(1)f x x =- 解：(][),22,-∞-⋃+∞6. 判断下列函数的奇偶性：（1）33y x x =+ 解：奇函数。

微积分第三版上册课后练习题含答案

微积分第三版上册课后练习题含答案微积分是数学的一个分支，它主要研究函数、极限、连续、导数、积分等概念和它们之间的关系。

微积分是自然科学、工程技术和经济管理等领域中不可或缺的数学工具。

本文将介绍微积分第三版上册的课后练习题，以及它们的答案和解析。

章节列表微积分第三版上册共分为12章，分别是：1.函数与极限2.导数及其应用3.曲线图形的相关概念4.定积分5.定积分应用6.不定积分7.不定积分的应用8.微分方程初步9.空间解析几何10.空间直线与平面11.空间曲面12.重积分每一章都包含了大量的练习题，这些题目是对每个章节中理论知识点的考察和巩固，同时也能够帮助读者构建更深入的理解。

练习题样例下面是微积分第三版上册第一章的一组练习题样例：1.1节练习1.求函数$f(x)=\\frac{x-1}{x+1}$在点x0=2处的导数。

2.求极限$\\displaystyle\\lim_{x \\to +\\infty}(\\sqrt{x^2+3x}-\\sqrt{x^2-5})$。

3.求函数$f(x)=\\sqrt{1+x}-1$的二阶导数。

1.2节练习1.求$f(x)=\\frac{1}{x}$的导函数和导数。

2.已知函数f(x)=x3+3x2+1，求它的单调区间和极值点。

3.求函数f(x)=x4−8x2的导函数和导数。

课后练习题答案微积分第三版上册的课后练习题答案可以在教材的补充练习答案中找到，答案涵盖了书中各章节的所有练习题。

下面是上述练习题的答案和解析。

1.1节练习答案1.$f'(2)=\\frac{2}{9}$2.$\\displaystyle\\lim_{x \\to +\\infty}(\\sqrt{x^2+3x}-\\sqrt{x^2-5})=+\\infty$3.$f''(x)=\\frac{1}{4(x+1)^{\\frac{3}{2}}}$1.2节练习答案1.$f'(x)=-\\frac{1}{x^2}$，$f''(x)=\\frac{2}{x^3}$2.f(x)在$(-\\infty,-1)$上单调递减，在$(-1,+\\infty)$上单调递增。

微积分课后题答案习题详解

微积分课后题答案习题详解IMB standardization office【IMB 5AB- IMBK 08- IMB 2C】第二章习题2-11. 试利用本节定义5后面的注（3）证明：若lim n →∞x n =a ,则对任何自然数k ，有lim n →∞x n +k =a .证：由lim n n x a →∞=，知0ε∀>，1N ∃，当1n N >时，有取1N N k =-，有0ε∀>，N ∃，设n N >时（此时1n k N +>）有由数列极限的定义得 lim n k x x a +→∞=.2. 试利用不等式A B A B -≤-说明：若lim n →∞x n =a ,则lim n →∞∣x n ∣=|a|.考察数列x n =(-1)n ，说明上述结论反之不成立.证：而 n n x a x a -≤- 于是0ε∀>，,使当时，有N n N ∃>n n x a x a ε-≤-< 即 n x a ε-<由数列极限的定义得 lim n n x a →∞=考察数列 (1)nn x =-，知lim n n x →∞不存在，而1n x =，lim 1n n x →∞=，所以前面所证结论反之不成立。

3. 利用夹逼定理证明：(1) lim n →∞222111(1)(2)n n n ⎛⎫+++ ⎪+⎝⎭=0； (2) lim n →∞2!n n =0.证：（1）因为222222111112(1)(2)n n n n n n n n n n++≤+++≤≤=+ 而且 21lim0n n →∞=，2lim 0n n→∞=，所以由夹逼定理，得222111lim 0(1)(2)n n n n →∞⎛⎫+++= ⎪+⎝⎭. （2）因为22222240!1231n n n n n<=<-，而且4lim 0n n →∞=，所以，由夹逼定理得4. 利用单调有界数列收敛准则证明下列数列的极限存在.(1) x n =11n e +，n =1,2,…；(2) x 1x n +1,n =1,2,…. 证：（1）略。

微积分第一章课外习题参考答案

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微积分1第三版习题答案

微积分1第三版习题答案微积分是数学中的重要分支，它研究的是函数的变化规律和求解各种问题的方法。

微积分1是大学数学课程中的一门基础课程，通过学习微积分1，我们可以掌握函数的基本概念、极限、导数和积分等重要知识。

而《微积分1第三版》是一本经典的教材，它提供了丰富的习题用于巩固和加深对微积分1知识的理解。

在本文中，我将为大家提供《微积分1第三版》习题的答案，希望能对大家的学习有所帮助。

在开始提供答案之前，我想强调一点，那就是习题的答案只是一种参考，通过自己思考和解答习题，我们才能真正理解微积分的概念和方法。

因此，在查看答案之前，我建议大家先尝试自己解答习题，再对照答案进行对照和讨论。

下面是《微积分1第三版》部分习题的答案：1. 求函数f(x) = 2x^3 + 3x^2 - 12x + 5在x = 2处的导数。

答案：f'(2) = 2(2)^3 + 3(2)^2 - 12(2) + 5 = 16 + 12 - 24 + 5 = 92. 求函数f(x) = sin(x) + cos(x)在x = π/4处的导数。

答案：f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 03. 求函数f(x) = ln(x^2 + 1)在x = 1处的导数。

答案：f'(1) = 2(1)/(1^2 + 1) = 1/1 = 14. 求函数f(x) = e^x在x = 0处的导数。

答案：f'(0) = e^0 = 15. 求函数f(x) = 3x^2 + 2x + 1的不定积分。

答案：∫(3x^2 + 2x + 1)dx = x^3 + x^2 + x + C，其中C为常数。

6. 求函数f(x) = 2sin(x) + 3cos(x)的不定积分。

答案：∫(2sin(x) + 3cos(x))dx = -2cos(x) + 3sin(x) + C，其中C为常数。

微积分--课后习题答案

习题1—1解答1．设y x xy y x f +=),(，求),(1),,(),1,1(),,(y x f y x xy f y x f y x f -- 解yxxy y x f +=--),(；x xy y y x f y x y x xy f x y xy y x f +=+=+=222),(1;),(;1)1,1(2．设y x y x f ln ln ),(=，证明：),(),(),(),(),(v y f u y f v x f u x f uv xy f +++=),(),(),(),(ln ln ln ln ln ln ln ln )ln )(ln ln (ln )ln()ln(),(v y f u y f v x f u x f v y u y v x u x v u y x uv xy uv xy f +++=⋅+⋅+⋅+⋅=++=⋅=3．求下列函数的定义域，并画出定义域的图形：（1）;11),(22-+-=y x y x f（2）;)1ln(4),(222y x y x y x f ---=（3）;1),(222222cz b y a x y x f ---=（4）.1),,(222zy x z y x z y x f ---++=解（1）}1,1),{(≥≤=y x y x D（2）{y y x y x D ,10),(22<+<=（3）⎫⎩⎨⎧++=),(22222b y a x y xD（4）{}1,0,0,0),,(222<++≥≥≥=z y x z y x z y x D4．求下列各极限：（1）22101limy x xy y x +-→→=11001=+- （2）2ln 01)1ln(ln(lim022)01=++=++→→e yx e x y y x（3）41)42()42)(42(lim 42lim000-=+++++-=+-→→→→xy xy xy xy xy xy y x y x（4）2)sin(lim )sin(lim202=⋅=→→→→x xy xy y xy y x y x5．证明下列极限不存在：（1）;lim 00yx y x y x -+→→ （2）2222200)(lim y x y x y x y x -+→→ （1）证明如果动点),(y x P 沿x y 2=趋向)0,0( 则322lim lim0020-=-+=-+→→=→x x xx y x y x x x y x ；如果动点),(y x P 沿y x 2=趋向)0,0(，则33lim lim0020==-+→→=→y yy x y x y y x yx所以极限不存在。

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］．。

微积分第一章课外习题参考答案

微积分第一章课外习题参考答案
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微积分课外习题参考答案
微积分第一章课外习题参考答案
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第一章极限与连续
微积分第一章课外习题参考答案
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微积分第一章课外习题参考答案 5
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微积分各章习题及详细答案

《微积分》各章习题及详细答案(总42页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--第一章函数极限与连续一、填空题1、已知x xf cos 1)2(sin +=，则=)(cos x f 。

2、=-+→∞)1()34(lim22x x x x 。

3、0→x 时，x x sin tan -是x 的阶无穷小。

4、01sin lim 0=→xx k x 成立的k 为。

5、=-∞→x e x x arctan lim 。

6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续，则=b 。

7、=+→xx x 6)13ln(lim 0 。

8、设)(x f 的定义域是]1,0[，则)(ln x f 的定义域是__________。

9、函数)2ln(1++=x y 的反函数为_________。

10、设a 是非零常数，则________)(lim =-+∞→xx ax a x 。

11、已知当0→x 时，1)1(312-+ax 与1cos -x 是等价无穷小，则常数________=a 。

12、函数x xx f +=13arcsin )(的定义域是__________。

13、lim ____________x →+∞=。

14、设8)2(lim =-+∞→xx ax a x ，则=a ________。

15、)2)(1(lim n n n n n -++++∞→=____________。

二、选择题1、设)(),(x g x f 是],[l l -上的偶函数，)(x h 是],[l l -上的奇函数，则中所给的函数必为奇函数。

（Ａ）)()(x g x f +；（Ｂ）)()(x h x f +；（C ）)]()()[(x h x g x f +；（D ）)()()(x h x g x f 。

2、xxx +-=11)(α，31)(x x -=β，则当1→x 时有。

微积分部分习题及答案 (1)

(2) 1(x2 4x 4) dx 0
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11.用牛顿莱布尼茨公式计算下列定积分
5
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35
14.用分部积分法计算下列定积分
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1x
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14.用分部积分法计算下列定积分
6
22
11.用牛顿莱布尼茨公式计算下列定积分

微积分课后习题答案

微积分课后习题答案微积分课后习题答案微积分是数学中的一门重要学科，它研究的是函数的变化和极限。

在学习微积分的过程中，课后习题是非常重要的一环。

通过做习题，我们可以巩固课堂上所学的知识，提高自己的解题能力。

然而，有时候我们可能会遇到一些难题，无法找到正确的解答。

因此，本文将为大家提供一些微积分课后习题的答案，希望能够帮助大家更好地理解微积分的知识。

一、函数的极限1. 求函数f(x) = (3x^2 + 2x + 1)/(2x^2 + x - 3)当x趋近于2时的极限。

解答：将x代入函数f(x)的表达式中，得到f(2) = (3(2)^2 + 2(2) + 1)/(2(2)^2 +2 - 3) = 13/9。

因此，当x趋近于2时，函数f(x)的极限为13/9。

2. 求函数f(x) = (x^2 - 4)/(x - 2)当x趋近于2时的极限。

解答：将x代入函数f(x)的表达式中，得到f(2) = (2^2 - 4)/(2 - 2) = 0/0。

此时，函数f(x)的极限不存在。

二、导数与微分1. 求函数f(x) = 3x^2 - 4x的导数。

解答：根据导数的定义，导数f'(x) = lim(h→0) [(f(x + h) - f(x))/h]。

将函数f(x)代入该定义中，得到f'(x) = lim(h→0) [(3(x + h)^2 - 4(x + h) - (3x^2 - 4x))/h]。

化简后可得f'(x) = 6x - 4。

2. 求函数f(x) = x^3 - 2x^2 + 3x - 4的微分。

解答：微分df(x) = f'(x)dx。

将函数f(x)的导数f'(x)代入该定义中，得到df(x) =(3x^2 - 4x)dx。

三、定积分1. 求函数f(x) = 2x在区间[1, 3]上的定积分。

解答：根据定积分的定义，定积分∫[1, 3] f(x)dx = lim(n→∞) Σ[i=1到n] f(xi)Δx，其中Δx = (b - a)/n，xi为区间[a, b]上的任意一点。

微积分第四版习题答案12

微积分第四版习题答案12微积分是数学中的一门重要学科，它研究的是函数的变化规律以及与之相关的概念和方法。

而习题则是学习微积分过程中的重要组成部分，通过解答习题可以帮助我们巩固所学的知识，提高解决问题的能力。

在微积分第四版中，有许多习题需要我们去解答，下面我将为大家提供一些习题的答案。

1. 习题：计算函数f(x) = x^2 + 3x - 2在x = 2处的导数。

答案：我们知道，函数在某一点处的导数等于该点的切线斜率。

因此，我们可以通过求函数在x = 2处的切线斜率来计算导数。

首先，我们需要计算函数在x = 2处的斜率。

利用导数的定义，我们有：f'(2) = lim(h->0) [f(2+h) - f(2)] / h代入函数f(x) = x^2 + 3x - 2，我们得到：f'(2) = lim(h->0) [(2+h)^2 + 3(2+h) - 2 - (2^2 + 3(2) - 2)] / h化简后，我们得到：f'(2) = lim(h->0) [4h + h^2 + 6h] / h继续化简，得到：f'(2) = lim(h->0) (h^2 + 10h) / h再次化简，得到：f'(2) = lim(h->0) (h(h + 10)) / h最后，化简为：f'(2) = lim(h->0) (h + 10) = 10所以，函数f(x) = x^2 + 3x - 2在x = 2处的导数为10。

2. 习题：计算函数g(x) = 3x^3 + 2x^2 - 5x + 1的不定积分。

答案：不定积分是求函数的原函数，即反向求导的过程。

对于给定的函数g(x) = 3x^3 + 2x^2 - 5x + 1，我们需要找到它的原函数F(x)，使得F'(x) = g(x)。

根据不定积分的性质，我们可以逐项对函数g(x)进行积分。

微积分习题答案 经济数学微积分 主编 张建梅 马庆华

微积分答案 经济数学微积分 主编张建梅 马庆华 科学出版社 广外

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微积分习题答案经济数学微积分主编张建梅马庆华

微积分答案经济数学微积分主编张建梅马庆华科学出版社广外