《微积分》(中国商业出版社)课后习题答案四

合集下载

《微积分》(中国商业出版社_经管类)课后习2题答案九

《微积分》(中国商业出版社_经管类)课后习2题答案九

则 u ( x, y ) =
∫ (0 , 0) ( x + 2 y )dx + (2 x − 3 y)dy
( x , y)
1 2 3 x + 4x − y 2 = c 2 2 1 3 ∴ : x 2 + 4 xy − y 2 = c 2 2 y 3+5 3x + 5 y dy x 5. = =− y 4 x + 6 y dx 4+6 x y dy du 令 = ux.则 =u+x x dx dx = ⇒− 3 + 5u + 4u + 6u 2 du 3 2 + 3u =x ⇒ − (ln x + c) = 2 du 4 + 6u dx 2 2u + 3u + 1
2⎞ 1 解: y' +⎛ ⎜− ⎟y = 1+ ⎝ 3⎠
2y +1 ,且曲线 L 过点(1,0).试求曲线 L 的方 x
x

dy ⎛ 2 ⎞ + ⎜ − ⎟ y = 0 则 ln y = −2 ln x + c dx ⎝ 3 ⎠ c x2
则y=
则设通解为 y =
c( x ) x2
1 ⎡ ⎤ 1⎞ 2 1 ⎛ y (x ) = x 2 ⎢ ⎜1 + ⎟e x dx + c ⎥ = 3 x 2 − 2 x − 1 ⎢ ⎝ ⎥ 2 x⎠ ⎢ ⎥ ⎣ ⎦


(4) (x + 2 y )dx + (2 x − 3 y )dy = 0
̇ ⋅ y +1 2 dy 2y + x x = = dx 3 y − 2 x 3 ⋅ y − 2 x

微积分答案

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.。

微积分课后题答案习题详解

微积分课后题答案习题详解

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!nn =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。

大学一年级上学期-微积分课后练习及答案-4-2-微积分基本定理

大学一年级上学期-微积分课后练习及答案-4-2-微积分基本定理

∫ xtf (t )dt
7.设
f ( x) 为连续正值函数,证明:当 x
>
0 时, F ( x) =
0 x ∫
f (t)dt
为单调增加函数.
0
x
x
x
xf ( x)∫ f (t)dt − f ( x)∫ tf (t)dt f ( x)∫ ( x − t) f (t)dt
解: F ′( x) =
0
0
=
0
(1) lim 0
x→0
x
∫ xcos t 2dt 解: lim 0
= lim cos x 2
=1
x→0
x
x→0 1
1 ∫
e −t2 dt
(2)
lim
x→0
cos x
x2
1 ∫
e −t2 dt
解: lim cos x
= lim − e − cos2 x (− sin x) = e −1
=
1
x→0
x2
8.计算下列定积分.
(1) ∫ 3 x 3dx
1
解: ∫ 3 x 3dx = 1 x 4 3 = 1 (34 − 1) = 20
1
4 14
9
(2) ∫
x (1 +
x )dx
4
解: ∫ 9
4
x (1 +
x )dx
=
1
∫49( x 2
+
x)dx
=
(2 3
3
x2
+
1 2
9
x2)
=
271 6
=
45 1 6
4
x→0
2x

微积分4习题答案

微积分4习题答案

一、填空题1.=--→x x x e e x 0lim 212.=--→14arctan lim1x x x π21 3.=-→221)1(ln lim x xx 14.21xxy +=的单调区间是 增区间)1,1(-;减区间),1(),1,(+∞--∞ 5.函数xx y 82+=(0>x )的单调增加的区间是 ),2(∞+6.一质点作直线运动,其运动规律为t t t t S +-=246)(,则它速度开始增加的时刻为=t 1=t7.函数)0(542<-=x xx y 的最小值是 12 8.函数x x y 2⋅=的极小值点为 2ln 1-9.曲线16623-+=x x y 的拐点坐标是 )0,2(-10.函数xxx y ln +=的渐近线有 0=x 或x y =二、单项选择题1.如果函数b x ax y +-=sin 在),(+∞-∞上严格单调递增,则b a ,应满足 D A .1≥a 且0<b B .1>a 且0>b C .1<a 且0<b D .1>a 且b 为任意实数2.下列各命题中,正确的是 B A .若)(x f y =在0x x =处有0)("0=x f ,则),(00y x 一定是曲线)(x f y =的拐点 B .若可导函数)(x f y =在0x x =处取得极值,则0)('0=x fC .若)(x f y =在0x x =处有0)('0=x f ,则)(x f 在0x x =处一定取得极值D .极大值就是最大值3.已知函数bx ax x x f ++=23)(在点1=x 处取得极值2-,则 B A .0,3=-=b a 且点1=x 为函数)(x f 的极小值点 B .3,0-==b a 且点1=x 为函数)(x f 的极小值点 C .0,3=-=b a 且点1=x 为函数)(x f 的极大值点 D .3,0-==b a 且点1=x 为函数)(x f 的极大值点4.函数)(x f y =的导数)(''x f y =的图像如图所示,则下列结论正确的是 D A .在)1,(--∞内,曲线)(x f y =是凸的 B .在),(+∞-∞内,曲线)(x f y =是凸的 C .在),(+∞-∞内,曲线)(x f y =是直线 D .在),(+∞-∞内,曲线)(x f y =是凹的5.设函数)(x f 满足x x f x f =+2)]('[)(",且0)0('=f ,则 CA .)0(f 是)(x f 的极大值B .)0(f 是)(x f 的极小值C .))0(,0(f 是曲线)(x f y =的拐点D .)0(f 不是)(x f 的极大值,))0(,0(f 也不是曲线)(x f y =的拐点三、求下列极限1.1cos )1(lim 0--→x e x x x (=-2) 2.xxe e x x x cos 13lim 20----→(=3)3.1tan 1tan 1lim 0---+→x x e xx [解] 1tan 12sec tan 12sec lim 1tan 1tan 1lim 220000=---+---+→→=x x x x exx x x e x x 4.2)ln(lim nxm be a x x +++∞→(0,0>>n b ) (n 1=)5.2201)21(lim x x x x -+→ (=4) 6.x x x x x ln 1lim 1-→ (=1)7.)1ln(sin 2tan tan lim 2--→x x x (2sec 2=) 8.||ln 1lim 21x x x --→ (=2)9.⎪⎭⎫ ⎝⎛-+∞→x x x 1cos 1lim 2(21=) 10.)233(lim 112-+-∞→x x x x (2)3(ln =)11.⎪⎭⎫ ⎝⎛-→2201sin 1lim x x x (31=) 12.⎥⎦⎤⎢⎣⎡+-+→)21ln(3sin )21ln(3lim 220x x x x x (49=) 13.x x x +→0lim (=1) 14.xxx e x 1)(lim ++∞→ (=2e ) 15.)21ln(3lim x x x++∞→(3e )四、应用题1.求x x y ln 22-=的单调区间。

微积分课后题答案习题详解

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》各章习题及详细答案

《微积分》各章习题及详细答案

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(lim 22x x x x 。

3、0→x 时,x x sin tan -是x 的 阶无穷小。

4、01sin lim 0=→xx kx 成立的k 为 。

5、=-∞→x e xx 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(lim0 。

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xxx 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 。

(A)α是比β高阶的无穷小; (B)α是比β低阶的无穷小; (C )α与β是同阶无穷小; (D )βα~。

《微积分》上册部分课后习题答案

《微积分》上册部分课后习题答案

微积分上册 一元函数微积分与无穷级数第2章 极限与连续2.1 数列的极限1.对于数列n x ,若a x k →2(∞→k ),a x k →+12(∞→k ),证明:a x n → (∞→n ). 证. 0>∀ε, a x k →2 (∞→k ), Z K ∈∃∴1, 只要122K k >, 就有ε<-a x k 2; 又因a x k →+12(∞→k ), Z K ∈∃∴2, 只要12122+>+K k , 就有ε<-+a x k 12. 取{}12,2m ax 21+=K K N , 只要N n >, 就有ε<-a x n , 因此有a x n → (∞→n ). 2.若a x n n =∞→lim ,证明||||lim a x n n =∞→,并举反例说明反之不一定成立.证明: a x n n =∞→lim ,由定义有:N ∃>∀,0ε,当N n >时恒有ε<-||a x n又 ε<-≤-||||||a x a x n n对上述同样的ε和N ,当N n >时,都有ε<-||||a x n 成立 ∴ ||||lim a x n n =∞→反之,不一定成立.如取 ,2,1,)1(=-=n x nn显然 1||lim =∞→n n x ,但n n x ∞→lim 不存在.2.2 函数的极限1. 用极限定义证明:函数()x f 当0x x →时极限存在的充要条件是左、右极限各自存在且相等.证: 必要性. 若()A x f x x =→0lim , 0>∀ε, 0>∃δ, 当δ<-<00x x 时, 就有()ε<-A x f . 因而, 当δ<-<00x x 时, 有()ε<-A x f , 所以()A x f x x =+→0lim ; 同时当δ<-<x x 00时, 有()ε<-A x f , 所以()A x f x x =-→0lim .充分性. 若()A x f x x =+→0lim ,()A x f x x =-→0lim . 0>∀ε, 01>∃δ, 当100δ<-<x x 时, 就有()ε<-A x f , 也02>∃δ, 当200δ<-<x x 时, 有()ε<-A x f . 取{}21,m in δδδ=,则当δ<-<00x x 时, 就有()ε<-A x f . 所以()A x f x x =→0lim .2.写出下列极限的精确定义:(1)A x f x x =+→)(lim 0,(2)A x f x =-∞→)(lim ,(3)+∞=+→)(lim 0x f x x ,(4)-∞=+∞→)(lim x f x ,(5)A x f x =+∞→)(lim .解:(1)设R x U f →)(:0是一个函数,如果存在一个常数R A ∈,满足关系:0,0>∃>∀δε,使得当δ<-<00x x 时,恒有ε<-|)(|A x f ,则称A 是)(x f 当+→0x x 时的极限,记作A x f x x =+→)(lim 0或 )()(0+→=x x A x f . (2)设R f D f →)(:是一函数,其中0,),,()(>>--∞⊃αααR f D .若存在常数R A ∈,满足关系:0)(,0>∈∃>∀R X ε,使得当X x -<时,恒有ε<-|)(|A x f 成立,则称A 是)(x f 当-∞→x 时的极限,记作:A x f x =-∞→)(lim 或 A x f =)()(-∞→x .(3)设R x U f →)(:0是任一函数,若0>∀M ,0>∃δ,使得当δ<-<00x x 时,恒有M x f >)(,则称当+→0x x 时)(x f 的极限为正无穷大,记作+∞=+→)(lim 0x f x x 或 +∞=)(x f )(0+→x x . (4)设R f D f →)(:是一函数,其中R f D ∈>+∞⊃ααα,0),,()(,若存在常数R A ∈,满足关系:0>∀M ,0)(>∈∃R X ,使得当X x >时,恒有M x f -<)(则称当+∞→x 时)(x f 的极限为负无穷大,记作:-∞=+∞→)(lim x f x 或 -∞=)(x f )(+∞→x .(5)设R f D f →)(:是一函数,其中R f D ∈>+∞⊃ααα,0),,()(,若存在常数R A ∈,满足关系:0,0>∃>∀X ε,使得当X x >时,恒有ε<-|)(|A x f 成立,则称A是)(x f 当+∞→x 时的极限,记作:A x f x =+∞→)(lim 或 A x f =)()(+∞→x .2.3 极限的运算法则1.求∑=∞→+⋯++Nn N n 1211lim. 解. ()()⎪⎭⎫ ⎝⎛+-=+=+=+⋯++111212211211n n n n n n n⎪⎭⎫ ⎝⎛+-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-++⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛-=+⋯++∑=1112111312121122111N N N n Nn 21112lim 211lim1=⎪⎭⎫ ⎝⎛+-=+⋯++∴∞→=∞→∑N nN Nn N 2.求xe e xxx 1arctan11lim110-+→. 解. +∞=+→x x e 10lim , 0lim 10=-→xx e,,21arctan lim 11lim 1arctan11lim 0110110π=-+=-++++→--→→x ee x e e x xxx xxx ,21arctan lim 11lim 1arctan11lim 0110110π=-+=-+---→→→x e e x e e x x xx x x x 21arctan 11lim 110π=-+∴→x e e x xx3.设)(lim 1x f x →存在,)(lim 2)(12x f x x x f x →+=,求)(x f . 解:设 )(lim 1x f x →=A ,则A x x x f ⋅+=2)(2再求极限:A A A x x x f x x =+=⋅+=→→21)2(lim )(lim 211⇒ 1-=A∴ x x xA x x f 22)(22-=+=.4.确定a ,b ,c ,使 0)1(3)1()1(lim 2221=-+-+-+-→x x c x b x a x 成立.解:依题意,所给函数极限存在且 0)1(lim 21=-→x x∴ 0]3)1()1([lim 221=+-+-+-→x c x b x a x ⇒ 2=c∴ 上式左边=])32)(1(11[lim ))1(321(lim 21221++-+--+=-+-+-+→→x x x x b a x x x b a x x])32)(1(1)32([lim 221++---+++=→x x x x b a x同理有 0]1)32([lim 21=--++→x x b x ⇒ 21=b ∴ 163)23)(1(8)1(3lim )32)(1(1)32(21lim221221=++---=++---++-=→→x x x x x x xx a x x 故 2,21,163===c b a 为所求.2.4 极限存在准则1. 设1x =10,n n x x +=+61,( ,2,1=n ).试证数列{n x }的极限存在,并求此极限. 证: 由101=x , 4612=+=x x , 知21x x >. 假设1+>k k x x , 则有21166+++=+>+=k k k k x x x x . 由数学归纳法知, 对一切正整数n , 有1+>n n x x ,即数列{n x }单调减少. 又显然, () ,2,10=>n x n , 即{n x }有界. 故n n x ∞→lim 存在.令a x n n =∞→lim , 对n n x x +=+61两边取极限得a a +=6, 从而有062=--a a ,,3=∴a 或2-=a , 但0,0≥∴>a x n , 故3lim =∞→n n x2.证明数列 nn n x x x x ++=<<+3)1(3,3011收敛,并求其极限.证明:利用准则II ,单调有界必有极限来证明.∴301<<x ,由递推公式33312131213213)1(30111112=++<++=++=++=<x x x x x x∴ 302<<x 同理可证:30<<n x 有界又 03)3)(3(333)1(311112111112>++-=+-=-++=-x x x x x x x x x x∴ 12x x > 同理 23x x > ,… ,1->n n x x ∴数列 }{n x 单调递增,由准则II n n x ∞→lim 存在,设为A ,由递推公式有:AA A ++=3)1(3 ⇒ 3±=A (舍去负数)∴ 3lim =∞→n n x .3.设}{n x 为一单调增加的数列,若它有一个子列收敛于a ,证明a x n n =∞→lim .证明:设}{k n x 为}{n x 的一子列,则}{k n x 也为一单调增加的数列,且a x k k n n =∞→lim对于1=ε,N ∃,当N n >时有1||<-a x k n 从而||1||||||||a a a x a a x x k k k n n n +<+-≤+-=取|}|1|,|,|,max {|1a x x M N n n += ,对一切k n 都有 M x k n ≤|| 有界.由子列有界,且原数列}{n x 又为一单调增加的数列,所以,对一切n 有M x n ≤||有界,由准则II ,数列}{n x 极限存在且a x n n =∞→lim .2.5 两个重要极限1. 求]cos 1[cos lim n n n -++∞→.解: 原式 =21sin 21sin2lim nn n n n -+++-+∞→⎪⎪⎭⎫⎝⎛++=-+=-+-+-+++-=+∞→n n n n n n nn nn nn n 1110212121sin21sin2lim 2. 求)1sin(lim 2++∞→n n π.解. 原式=()()n nn n n nn n -+-=-+++∞→+∞→1sin 1lim )1sin(lim 22ππππ()()()()0111sin 1lim 222=-+⋅-+-+-=+∞→n nn n nnnn πππ3. 求x x xx )1cos 1(sinlim +∞→. 解. 原式=()[]()e t t t tttt tt xt =⎥⎦⎤⎢⎣⎡+=+=→→=22sin 2sin 10212012sin 1lim cos sin lim 令4. 设 ⎩⎨⎧+-=32)cos 1(2)(x x x x f 00≥<x x 求 20)(lim x x f x →. 解: 1lim )(lim 232020=+=++→→x x x x x f x x ,1)cos 1(2lim )(lim 2020=-=--→→x x x x f x x ∴ 1)(lim2=→xx f x .2.6 函数的连续性1. 研究函数()[]x x x g -=的连续性,并指出间断点类型. 解. n x =,Z n ∈ (整数集)为第一类 (跳跃) 间断点.2. 证明方程)0(03>=++p q px x 有且只有一个实根.证. 令()()()0,0,3>∞+<∞-++=f f q px x x f , 由零点定理, 至少存在一点ξ使得()0=ξf , 其唯一性, 易由()x f 的严格单调性可得.3.设⎪⎩⎪⎨⎧≤<-+>=-01),1ln(0 ,)(11x x x e x f x ,求)(x f 的间断点,并说明间断点的所属类型. 解. )(x f 在()()()+∞-,1,1,0,0,1内连续, ∞=-→+111lim x x e,0lim 111=-→-x x e, ()00=f , 因此,1=x 是)(x f 的第二类无穷间断点; (),lim lim 1110--→→==++e ex f x x x()()01ln lim lim 00=+=--→→x x f x x , 因此0=x 是)(x f 的第一类跳跃间断点.4.讨论nx nxn e e x x x f ++=∞→1lim )(2的连续性.解. ⎪⎩⎪⎨⎧<=>=++=∞→0,0,00,1lim)(22x x x x x e e x x x f nxnxn , 因此)(x f 在()()+∞∞-,0,0,内连续, 又()()00lim 0==→f x f x , ()x f ∴在()+∞∞-,上连续.5.设函数),()(+∞-∞在x f 内连续,且0)(lim=∞→xx f x ,证明至少存在一点ξ,使得0)(=+ξξf .证:令x x f x F +=)()(,则01]1)([lim )(lim>=+=∞→∞→x x f x x F x x ,从而0)(>xx F .由极限保号性定理可得,存在01>x 使0)(1>x F ;存在02<x 使0)(2<x F .)(x F 在],[12x x 上满足零点定理的条件,所以至少存在一点ξ使得0)(=ξF ,即0)(=+ξξf .6.讨论函数nnx x x x f 2211lim )(+-=∞→的连续性,若有间断点,判别其类型.解: ⎪⎩⎪⎨⎧-=101)(x f 1||1||1||>=<x x x ,显然 1±=x 是第一类跳跃间断点,除此之外均为连续区间.7.证明:方程)0,0(sin >>+=b a b x a x 至少有一个正根,且不超过b a +. 证明:设b x a x x f --=sin )(,考虑区间],0[b a +0)0(<-=b f ,0))sin(1()(≥+-=+b a a b a f ,当0))sin(1()(=+-=+b a a b a f 时,b a x +=是方程的根;当0))sin(1()(>+-=+b a a b a f 时,由零点定理,至少),0(b a +∈∃ξ使0)(=ξf ,即 0sin =--b a ξξ成立,故原方程至少有一个正根且不超过b a +.2.7 无穷小与无穷大、无穷小的比较1. 当0→x 时,下面等式成立吗?(1))()(32x o x o x =⋅;(2))()(2x o xx o =;(3) )()(2x o x o =. 解. (1)()()()002232→→=⋅x xx o x x o x , ()()()032→=⋅∴x x o x o x (2) ()()()0)(,00)()(2222→=∴→→=x x o x x o x x x o xxx o(3) ()2xx o不一定趋于零, )()(2x o x o =∴不一定成立(当0→x 时) 2. 当∞→x 时,若)11(12+=++x o c bx ax ,则求常数c b a ,,.解. 因为当∞→x 时,若)11(12+=++x o c bx ax , 所以01lim 111lim 22=+++=++++∞→+∞→c bx ax x x c bx ax x x , 故c b a ,,0≠任意.3.写出0→x 时,无穷小量3x x +的等价无穷小量.解: 11lim 1lim lim303630=+=+=+→→→x xx xxx x x x∴ 当0→x ,3x x +~6x第3章 导数与微分3.1 导数概念1. 设函数)(x f 在0x 处可导,求下列极限值. (1)hh x f h x f h )3()2(lim000--+→;(2)000)()(lim 0x x x xf x f x x x --→.解.(1) 原式()()()000000533)3(22)2(lim x f h x f h x f h x f h x f h '=⎥⎦⎤⎢⎣⎡⋅---+⋅-+=→(2) 原式()[]()()()()00000000)(limx f x f x x x x x x f x f x f x x x -'=----=→2.设函数R f →+∞),0(:在1=x 处可导,且),0(,+∞∈∀y x 有)()()(y xf x yf xy f += 试证:函数f 在),0(+∞内可导,且)1()()(f xx f x f '+='. 解:令1==y x ,由()()()y xf x yf xy f +=有()()121f f =得()01=f .()+∞∈∀,0x ,()()()()()()()()()()xx f f x x f xx f x x f x x f x f x x x x xf x x f x x x f x x f x x f x f x x x x +'=+∆-⎪⎭⎫⎝⎛∆+=∆-⎪⎭⎫ ⎝⎛∆++⎪⎭⎫ ⎝⎛∆+=∆-⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛∆+=∆-∆+='→∆→∆→∆→∆111lim 11lim 1lim lim 0000 故()x f 在()+∞,0内处处可导,且()()()xx f f x f +'='1. 3.设()f x 在(,)-∞+∞内有意义,且(0)0f =,(0)1f '=, 又121221()()()()()f x x f x x f x x ϕϕ+=+,其中22()cos xx x x e ϕ-=+, 求()f x '.解: ()()()()()()()()x x f x x f x x f x x f x x f x f x x ∆-∆+∆=∆-∆+='→∆→∆ϕϕ00lim lim()()()()()()()()()001lim 0lim 00ϕϕϕϕ'+'=∆-∆+∆-∆=→∆→∆x f x f xx x f x x f x f x x ()x e x x x 22cos -+==ϕ4.设函数0)(=x x f 在处可导,且21arctan lim )(0=-→x f x e x,求)0(f '.解:由已知,必有0]1[lim )(0=-→x f x e,从而0)(lim 0=→x f x ,而0)(=x x f 在连续,故0)0(=f .于是)0(1)0()(1lim )(lim 1arctan lim200)(0f xf x f x f x e x x x x f x '=-==-=→→→. 故21)0(='f .5.设)(x f 具有二阶导数,)(,sin )()2(lim )(2x dF t xx f t x f t x F t 求⎥⎦⎤⎢⎣⎡-+=∞→.解: 令t h 1=,则)(2 sin )()2(lim)(0x f x hhxh x f h x f x F t '=⋅-+=→.从而)(2)(2)(x f x x f x F ''+'=',dx x f x x f dx x F x dF )]()([2)()(''+'='=.6.设f 是对任意实数y x ,满足方程 22)()()(xy y x y f x f x f +++= 的函数,又假设1)(lim=→xx f x ,求:(1))0(f ;(2))0(f '; (3))(x f '. 解:(1)依题意 R y x ∈∀,,等式 22)()()(xy y x y f x f y x f +++=+ 成立令0==y x 有 )0(2)0(f f = ⇒ 0)0(=f(2)又 1)(lim=→x x f x ,即 )0(10)0()(lim 0f x f x f x '==--→,∴ 1)0(='f(3)xx f x x f x f x ∆-∆+='→∆)()(lim )(0x x f x x x x x f x f x ∆-∆⋅+∆⋅+∆+=→∆)()()()(lim 220 x x x x x x f x ∆∆⋅+∆⋅+∆=→∆220)()(lim ])([lim 20x x x xx f x ∆⋅++∆∆=→∆ ]1)0(22x x f +=+'=∴ 21)(x x f +='.7.设曲线)(x f y =在原点与x y sin =相切,试求极限 )2(lim 21nf nn ∞→. 解:依题意有 1)0()0(='='f y 且0)0(=f∴ 222)0()2(lim )2(lim 2121=⋅-⋅=⋅∞→∞→n nf n f n nf n n n .8.设函数)(x f 在0=x 处可导且0)0(,0)0(='≠f f ,证明1])0()1([lim =∞→nn f n f .证:n n n n f f n f f n f ])0()0()1(1[lim ])0()1([lim -+=∞→∞→.=10)0(11)0()01(lim )0()0()1(lim ===⋅-+-∞→∞→e ee f nf n f f f n f n n n .1.计算函数baxax xb ab y )()()(= (0,0>>b a )的导数.解. a xb bx a b a x xb a b a a x b a x a b x b x b a a x x b a b a b y )(1)()()()(ln )(121⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛+='-- ⎥⎦⎤⎢⎣⎡+-=x b x a a b a x x b a b b a x ln )()()( 2.引入中间变量,1)(2x x u +=计算1111ln 411arctan 21222-+++++=x x x y 的导数dx dy .解. 引入,1)(2x x u += 得11ln 41arctan 21-++=u u u y ,于是dxdudu dy dx dy ⋅=, 又 ()()4242422111111111141121x x x u u u u du dy +-=+-=-=⎪⎭⎫ ⎝⎛--+++=,21xx dx du +=, 则()22242121121xx x x x x x dx dy ++-=+⋅⎪⎭⎫⎝⎛+-= 3.设y y x +=2,232)(x x u +=,求dudy. 解. dudxdx dy du dy ⋅= , 又()()1223,12212++=+=x x x dx du y dy dx ,得121+=y dx dy , ()x x x du dx ++=21232, 则得()()xx x y du dy +++=2121232 4.已知 2arctan )(),2323(x x f x x f y ='+-=,求=x dx dy .解:22)23(12)2323arctan()2323()2323(+⋅+-='+-⋅+-'='x x x x x x x f y π43)23(12)2323arctan(02200=+⋅+-='=∴===x x x x x x y dxdy .1. 计算下列各函数的n 阶导数. (1) 6512-+=x x y ; (2) x e y xcos =. 解 (1)⎪⎭⎫⎝⎛+--=611171x x y ,()()()()()()⎥⎦⎤⎢⎣⎡+---=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+-⎪⎭⎫⎝⎛-=∴++1161117!1611171n n nn n n x x n x x y (2) ()⎪⎭⎫ ⎝⎛+=⎥⎦⎤⎢⎣⎡-=-='4cos 2sin 21cos 212sin cos πx e x x e x x e y x x x()⎪⎭⎫ ⎝⎛⋅+=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-⎪⎭⎫ ⎝⎛+=''42cos 24sin 4cos 22πππx ex x e y xx由此推得 ()()⎪⎭⎫ ⎝⎛⋅+=4cos 2πn x eyxnn2. 设x x y 2sin 2=, 求()50y .解 ()()()()()()()()()()"+'+=248250249150250502sin 2sin 2sin x x C x x C x x y⎪⎭⎫ ⎝⎛⋅+⋅⨯+⎪⎭⎫ ⎝⎛⋅+⋅+⎪⎭⎫ ⎝⎛⋅+=2482sin 2249502492sin 2502502sin 24950250πππx x x x xx x x x x 2sin 212252cos 2502sin 24950250⋅+⋅+-= ()[]x x x x 2cos 1002sin 212252249+-=3. 试从y dy dx '=1, 0≠'y , 其中y 三阶可导, 导出()322y y dy x d '''-=, ()()52333y y y y dy x d '''''-''= 解 y dy dx '=1 ,()()322211y y y y y dy dx y dx d dyx d '''-='⋅'-''=⋅⎪⎪⎭⎫ ⎝⎛'=∴ ()()()()()()52623333313y y y y y y y y y y y dy dx y y dx d dy x d '''''-''='⋅'''⋅'⋅''+''''-=⋅⎪⎪⎭⎫ ⎝⎛'''-=∴ 4. 设()x f 满足()()0 312≠=⎪⎭⎫⎝⎛+x xx f x f , 求()()()()x f x f x f n ,,'.解 以x 1代x ,原方程为()x x f x f 321==⎪⎭⎫ ⎝⎛,由()()⎪⎪⎩⎪⎪⎨⎧=+⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+x x f x f x x f x f 321 312,消去⎪⎭⎫⎝⎛x f 1,求得()x x x f 12-=,且得()212xx f +=',()()()()2!111≥-=++n x n x f n n n . 5.设()arcsin f x x =,试证明()f x 满足 (1)2(1)()()0x f x xf x '''--= (2) ,1,0,0)()()12()()1()(2)1()2(2==-+--++n x f n x xf n x f x n n n(3)求()(0)n f解 (1)()211x x f -=',()()()22221112211xx xx x x x f --=-⋅--='', ()()()012='-''-∴x f x x f x ,(2)上式两边对x 求n 阶导数得()()[]()()[]()()()()()()()()()()()()()()()[]x f n x xf x f n n x f x n x f x x f x x f x n n n n n nn⋅⋅+-⋅-⋅---+-='-''-=+++1221211021222即 ()()()()()()()()01212122=-+--++x f nx xf n x f xn n n 。

微积分课后题答案

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。

微积分详解习题答案

微积分详解习题答案

微积分详解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规律与性质。

在学习微积分的过程中,习题是不可或缺的一部分。

通过解答习题,我们可以更好地理解微积分的概念和原理。

本文将详解一些微积分习题的答案,帮助读者更好地掌握微积分知识。

首先,让我们来看一个简单的求导题目。

假设有一个函数f(x) = 3x^2 + 2x + 1,我们需要求出它的导数。

根据求导的定义,我们可以逐项对函数中的每一项进行求导。

对于3x^2,我们可以应用求导法则,得到它的导数为6x;对于2x,它的导数为2;对于常数1,它的导数为0。

因此,函数f(x)的导数为f'(x) = 6x+ 2。

接下来,我们来看一个求定积分的例子。

假设我们需要计算函数g(x) = x^2在区间[0, 2]上的定积分。

根据定积分的定义,我们可以将区间[0, 2]划分成无限小的小矩形,然后将这些小矩形的面积相加,即可得到定积分的值。

在本例中,我们可以将区间[0, 2]划分成无限小的宽度为Δx的矩形,然后将这些矩形的面积相加。

每个矩形的面积可以用函数g(x)在该点的函数值乘以Δx来表示。

因此,定积分的值可以表示为lim(Δx→0)∑(g(xi)Δx),其中∑表示求和,xi表示每个小矩形的中点。

通过计算,我们可以得到定积分的值为∫[0, 2]x^2dx =[x^3/3]0^2 = 8/3。

除了求导和定积分,微积分还涉及到一些其他的概念,比如极限和微分方程。

下面我们来看一个关于极限的习题。

假设有一个函数h(x) = (x^2 - 1)/(x - 1),我们需要求出它在x趋近于1时的极限。

当x趋近于1时,函数h(x)的分子和分母都会趋近于0。

我们可以应用洛必达法则来求解这个极限。

根据洛必达法则,我们可以对函数h(x)的分子和分母同时求导,然后再求极限。

对于函数h(x)的分子,它的导数为2x;对于函数h(x)的分母,它的导数为1。

因此,在x趋近于1时,函数h(x)的极限可以表示为lim(x→1)(2x) / 1 = 2。

《微积分》中国商业出版社_课后习题答案详解二

《微积分》中国商业出版社_课后习题答案详解二

n→∞ 5n − 4 81n8 + 2
n→∞ 5
2
− 4 81 +
n
n8
16.设 x → a 时 f (x) → ∞ , g(x) → ∞ ,则下列各式中成立的是(D).
= 0.
⎧3
2)因为
xn
=
xn
=
⎪ ⎨1
⎪⎩ 3
n为偶数 n为奇数
所以 xn 是发散的;
3)发散的.因为当
n


时,
1 n

0
;所以
xn
= 1g
1 n

−∞
;
4)因为
xn
=
⎧1 ⎨ ⎩− 1
n为偶数 n为奇数
所以 xn 是发散的;
5)收敛的.因为当 n → ∞ 时,
1 n
→ 0 ;所以 xn
= 3 + (−1)n
3)由于 f (x) − 0 = x +1 所以对于任意给定的 ξ > 0 ,存在 δ = ξ 2 当 0 < x + 1 < δ 时
恒有 f (x) − 0 < ξ 成立
故 lim x +1 = 0
x → −1+
4)对于任意给定的正数 ξ 要使 10x − 0 < ξ 成立即 x > 1gξ 成立
,要使
5 − n2 1 +
3n2 + 1 3

成立即 lim
x → xo
1 f (x)
=
+∞
n>
16 − 3ξ 成立
9ξ 2
所以对于正数 ξ ,存在 N =
16 - 3ξ 成立

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
微积分第一章课外习题参考答案
9
p4.3.证明 : { xn }有界, M 0, 使得 | xn | M , n 1,2,
n
.
0, lim yn 0, N ,当n N 时, | yn |

M | xn yn || xn || yn | , lim xn yn 0.
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
1
第一章 极限与连续
微积分第一章课外习题参考答案
2
预备知识(1-2)
p1. 一.1. { x | x 3且x 0} . 2. [1,1],[2k ,(2k 1) ], k Z . 1 x 3. 1 1 e x 1 1 x1 , x2 , 1 x1 1 e x 1 x 1 . x 1
x0 1 三. f [ g ( x )] 0 x0 1 x 0 e | x | 1 g[ f ( x )] 1 | x | 1 注意作图形. 1 | x | 1 e
微积分第一章课外习题参考答案 5
p2. 四 . 证明: f ( x ) f (2a x ) f (2b 2a x ) f [2(b a ) x ] 周期 T 2 | b a | . 五 . 证明 f ( x ) log a ( x x 1)
8
p4.
2.
解 :由题意,
n 2
1 1 1 ( 1) P1 Pn 1 2 3 2 2 2 2n 2 1 n 1 1 n 1 1 ( ) 2 2( ) 2 2 1 3 1 2 1 n 1 2 2( ) 2 2 lim P1 Pn lim n n 3 3

实用文档之《微积分》各章习题及详细答案

实用文档之《微积分》各章习题及详细答案

实用文档之"第一章 函数极限与连续"一、填空题1、已知x x f 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 kx 成立的k 为 。

5、=-∞→x e xx 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xxx 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 。

(A)α是比β高阶的无穷小; (B)α是比β低阶的无穷小; (C )α与β是同阶无穷小; (D )βα~。

微积分各章习题及详细答案

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 。

微积分第四版习题答案

微积分第四版习题答案

微积分第四版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。

对于学习微积分的学生来说,习题是巩固知识和提高能力的重要途径。

然而,对于微积分第四版习题的答案,很多学生可能会感到困惑。

在本文中,我将为大家提供微积分第四版习题的答案,希望能够帮助到大家。

第一章:函数与极限1.1 函数的概念与性质1.1.1 习题答案1. a) 函数的定义域是实数集,值域是实数集。

b) 函数的奇偶性与定义域无关,只与函数的表达式有关。

c) 函数的周期性与定义域无关,只与函数的表达式有关。

1.1.2 习题答案1. a) 函数的图像是一条抛物线,开口向上。

b) 函数的图像关于x轴对称,是一个偶函数。

c) 函数的图像关于y轴对称,是一个奇函数。

1.2 一元函数的极限1.2.1 习题答案1. a) 当x趋于无穷大时,函数的极限为无穷大。

b) 当x趋于无穷小时,函数的极限为0。

c) 当x趋于无穷小时,函数的极限不存在。

1.2.2 习题答案1. a) 函数的极限存在,且等于2。

b) 函数的极限不存在。

c) 函数的极限存在,且等于0。

第二章:导数与微分2.1 导数的概念与性质2.1.1 习题答案1. a) 函数在x=1处的导数为2。

b) 函数在x=0处的导数不存在。

c) 函数在x=2处的导数为1。

2.1.2 习题答案1. a) 函数在x=1处的导数为-1。

b) 函数在x=0处的导数不存在。

c) 函数在x=2处的导数为2。

2.2 函数的求导法则2.2.1 习题答案1. a) 函数的导数为f'(x) = 3x^2 - 2x + 1。

b) 函数的导数为f'(x) = 4x^3 - 6x^2 + 2x。

c) 函数的导数为f'(x) = 2x^2 + 4x - 2。

2.2.2 习题答案1. a) 函数的导数为f'(x) = 2x^3 + 3x^2 - 4x + 1。

b) 函数的导数为f'(x) = 3x^2 + 4x - 1。

微积分第四版课后习题答案

微积分第四版课后习题答案

微积分第四版课后习题答案微积分是数学中的一门重要学科,它研究的是变化和积分的关系。

对于学习微积分的同学来说,课后习题是巩固知识和提高能力的重要途径。

然而,有时候我们会遇到一些难题,没有答案或者不知道如何解答。

为了帮助大家更好地学习微积分,本文将为大家提供微积分第四版课后习题的一些答案和解析。

在微积分的学习中,函数是一个非常重要的概念。

函数是一种映射关系,它将一个自变量的值映射到一个因变量的值。

在微积分中,我们经常需要求解函数的导数和积分。

导数表示函数在某一点的变化率,积分则表示函数在一段区间上的累积效应。

对于求解导数的问题,我们可以使用导数的定义或者一些常用的求导法则。

例如,对于函数 f(x) = x^2,我们可以使用导数的定义来求解它的导数。

根据导数的定义,我们有:f'(x) = lim(h→0) [f(x+h) - f(x)] / h将函数 f(x) = x^2 代入上述公式,我们可以得到:f'(x) = lim(h→0) [(x+h)^2 - x^2] / h化简上述表达式,我们可以得到:f'(x) = lim(h→0) [2xh + h^2] / h继续化简,我们可以得到:f'(x) = lim(h→0) 2x + h由于 h 在趋于 0 的过程中,2x 是一个常数,所以我们可以得到:f'(x) = 2x因此,函数 f(x) = x^2 的导数为 f'(x) = 2x。

对于求解积分的问题,我们可以使用积分的定义或者一些常用的积分法则。

例如,对于函数 f(x) = 2x,我们可以使用积分的定义来求解它的积分。

根据积分的定义,我们有:∫f(x)dx = lim(n→∞) Σ[f(xi)Δx]将函数 f(x) = 2x 代入上述公式,我们可以得到:∫2xdx = lim(n→∞) Σ[2xiΔx]化简上述表达式,我们可以得到:∫2xdx = lim(n→∞) 2Σ[xiΔx]继续化简,我们可以得到:∫2xdx = 2lim(n→∞) Σ[xiΔx]由于 n 在趋于无穷大的过程中,Σ[xiΔx] 是一个常数,所以我们可以得到:∫2xdx = 2Σ[xiΔx]因此,函数 f(x) = 2x 的积分为∫2xdx = x^2 + C,其中 C 为常数。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。


(3) lim (1 − x) tan πx ;
x →1
2
(4) lim x1−x ;
x →1
(5)
⎛ xli→m0⎜⎜⎝1 +
1 x2
x
⎞ ⎟⎟ ⎠

1
(8) lim (cot x) ln x ;
x→0+
1
(6) lim ⎜⎛ π − arctan x ⎟⎞ ln x ;
x→+∞⎝ 2

(7) lim cot x⎜⎛
1+ x
解:(1) y ' = 4x3 − 4x = 0 ⇒ x1 = 0, x2 = 1, x3 = −1 所以,当 x ∈ (−1,0) ∪ (1, + ∞) 单调增 当 x ∈ (−∞, −1) ∪ (0,1) 调减
(2) y ' = 1 − ex = 0 ⇒ x = 0
所以,当 x < 0 单调增 当 x > 0 单调减
x→0
x→0
x + ex −1 x
x→0
x
sin 2 x 1
(10) lim (arcsin x)tan x = lim etan x ln arcsin x = lim e ln arcsin x = lim e arcsin x 1−x2 = e0 = 1
x→0+
x→0+
x→0+ cot x
x→0+
=
1+
cos
x
在 [0,
π ]
上验证柯西中值定理.
2
解:据柯西中值定理
f (π ) − 2 π
g( ) −
f (0) g (0)
=
f ' (ε ) g ' (ε )
2
cos ε
π
⇒ −1 =
⇒ε =
− sin ε
4
5.设 f (x) 在 [a ,b] 上有二阶导数 f "(x) ,且 f "(x) ≠ 0 ,证明 f (x) 在 (a,b) 中至多有一个驻点.
Pn(n−1) (x0 ) = (n −1)!
an−1 = f (n−1) (x0 ) ⇒ an−1 =
f (n−1) (x0 ) (n −1)!
依此类推,可得 a0 = f (x0), a1 = f ' (x0)
x2 sin 1
10.极限 lim
x 能否用洛必达法则计算?其值为何?
x→0 sin x
习题四 (A)
1.判断下列函数在给定区间上是否满足罗尔定理条件,若满足条件,求出相应的 ξ 值. (1) f (x) = x 6 − x在[0, 6]上; (2) f (x) = 3 x在[1 −, 1] 上.
解:(1) f ' (x) = 6 - x − x = 0
2 6−x
⇒ x = 4 ∈ [0, 6]
(5)
lim
x→∞
ln(x ln xa
x)
(a
>
0)

(4) lim ex − e−x − 2x ;
x→0 x − sin x
(6) lim esin3 x − 1 ;
x→0 x(1 − cos x)
(7) lim
h→0
f
(a
+
h) +
f
(a − h2
h) − 2 f
(a)
(设
f
"(x) 在)
x
=
a
解:∵ f (a) = f (b) = 0 ,f (c) > 0
∴存在 a ∈ (a,c),β∈ (c ,b)
使得 f ' (a) > 0 , f ' (β) < 0
且 x ∈ ( α ,β), f ' (x)递减
∴ Eε ∈ (a ,b) 使 f "(ε ) < 0
9.设 f (x) 在 (a,b) 中有 n +1 阶导数,求 (x − x0 ) 的 n 次多项式
1

1
⎞ ⎟

x→0 ⎝ sin x x ⎠
1
(9) lim (x + ex ) x ;
x→0+
(10) lim (arcsin x)tan x .
x→0+
1
解:(1) lim⎜⎛ x −
x→1⎝ x −1
1 ⎟⎞ = ln x ⎠
lim xln x − x +1 x→1 ln x(x −1)
=
lim x→11 −
x
⎞ ⎟⎟ ⎠
=
⎛ 1⎞
lim
x
e
ln⎜1+ ⎝
x2
⎟ ⎠
x→0
lim
ln⎜1+ ⎝
x2
⎟ ⎠
x→0 1
=e
x
2x2 lim
= ex→0 x3 +x
=1
1
⎛ ln⎜
π
−arctan
x
⎞ ⎟
⎝2

(6)
lim
⎛π ⎜

arctan
x
⎞ ⎟
hx
=
lim
e
hx
= lim e
x
1 =
x→∞⎝ 2

x→∞
x→∞
∴ 满足
(2)
f (x)
=
1
x

2 3

0
3
∴ 不满足
2.下列函数在给定区间上是否满足拉格朗日定理条件?如满足,求出相应的 ξ 值
(1) f (x) = x3 + 2x在[0, a] 上;
(2)
f
(x)
=
sin x在⎢⎣⎡0,
π 2
⎥⎦⎤上

解:(1) f ' (x) = 3x + 2
f (0) = 0 f(a) = a3 + 2a
2
− sin = lim
2
+
cos 2
2
(1 − x)
=
2
x →1
2
x→1
πx cos
x→1
π πx − sin
π
2
22
1
hx
ln x
1
(4) lim x1−x
=
lim e1−x
= e xli→m11− x
lim
= e x→1 x
=
1
x →1
x →1
e
⎛ 1⎞
(5) xli→m0⎜⎜⎝⎛1 +
1 x2
f (a) − f (0) = f ' (ε ) ⇒ a2 + 2 = 3ε 2 + 2 ⇒ ε = 3 a
a
3
∴ 满足
(2) f (x) = cosx
f (0) = 0
f
π (
)
=
1
2
f
(π ) − 2
f
(0)
=
f
' (ε )

2
=
cosε

ε
=
arccos
2
∈ (0,π )
π
π
π
2
2
∴ 满足
3.证明:
Pn (x) = a0 + a1(x + x0) + a2(x − x0)2 + ⋯ + an (x − x0)n
使得
Pn(k) (x0) = f (k) (x0) (k = 0,1,2,⋯, n)
解: Pn(n)(x0 ) = n!
an = f (n) (x0 ) ⇒ an =
f (n) (x0 ) n!
⎛π ⎜
− arctan x ⎟⎞(1+ x2)
e
⎝2

(7)
lim cot
x→0
x⎜⎛ ⎝
1 sin x

1 x
⎞ ⎟ ⎠
=
lim
x→0
cos
x(x − sin x) sin2 x
=
⎜⎜⎛1 − lim ⎝
x→0
x2 2
⎟⎞ ⎟
⎜⎛ ⎜
x

⎠⎝
x3
x
+
x3 3!
⎟⎞ ⎟ ⎠
=
1 6
1 (− csc 2 x)
点邻近连续).
解:(1) lim x4 −16 lim 4x3 = 32
x→2 x − 2 x→2 1
(2) lim (1 + x)x −1 = lim π (1 + x)x−1 = π
x→0
x
x→0
(3)
lim
ln cot x =
lim
1 (−csc2 x)
cot x
= lim
− 2x
= −1
x→0+ ln x x→0+
13.设
f
(
x)
=
⎪⎧ ⎨
sin x
x
,
⎪⎩ ax + b,
在 x = 0 点可导,求 a, b
x > 0, x≤0
sin x
lim f (x) = lim
=1
解: x→0+
x→0+ x
⎫ ⎪ ⎬⇒ b =1
lim f (x) = lim (ax + b) = b⎪
x→0−
x→0−

f
' (0+ )
=
lim
相关文档
最新文档