ImageVerifierCode 换一换
格式:PPTX , 页数:16 ,大小:420.60KB ,
文档编号:5897834      下载积分:15 文币
快捷下载
登录下载
邮箱/手机:
温馨提示:
系统将以此处填写的邮箱或者手机号生成账号和密码,方便再次下载。 如填写123,账号和密码都是123。
支付方式: 支付宝    微信支付   
验证码:   换一换

优惠套餐
 

温馨提示:若手机下载失败,请复制以下地址【https://www.163wenku.com/d-5897834.html】到电脑浏览器->登陆(账号密码均为手机号或邮箱;不要扫码登陆)->重新下载(不再收费)。

已注册用户请登录:
账号:
密码:
验证码:   换一换
  忘记密码?
三方登录: 微信登录  
下载须知

1: 试题类文档的标题没说有答案,则无答案;主观题也可能无答案。PPT的音视频可能无法播放。 请谨慎下单,一旦售出,概不退换。
2: 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。
3: 本文为用户(momomo)主动上传,所有收益归该用户。163文库仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知163文库(点击联系客服),我们立即给予删除!。
4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
5. 本站仅提供交流平台,并不能对任何下载内容负责。
6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

版权提示 | 免责声明

1,本文(《高数双语》课件section 5-3.pptx)为本站会员(momomo)主动上传,163文库仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。
2,用户下载本文档,所消耗的文币(积分)将全额增加到上传者的账号。
3, 若此文所含内容侵犯了您的版权或隐私,请立即通知163文库(发送邮件至3464097650@qq.com或直接QQ联系客服),我们立即给予删除!

《高数双语》课件section 5-3.pptx

1、Integration by substitution and by parts in definite integrals1Integration by Substitutions for definite integrals2,()baf x dx(),()F xCf x dx When we want to evaluate the value of a definite integral if we can find the corresponding indefinite integral then we can use the Newton-Leibniz formula to o

2、btain the value immediately.()()()()()dbbbaaaf x dxF bF aF xF x Integration by Substitutions for definite integrals3(Integration by substitution for definite integrals)Suppose that the function x=(t)is continuously differentiable on the interval,()=a,()=b,and that f is continuous in the range of ,I.

3、Then()()()baf x dxftt dt Integration by Substitutions for definite integrals4Proof()()()baf x dxF bF a ()()(),dFtfttdt ()()()ftt dtFt Let F be an antiderivative of f on the interval I,thenSince we haveTherefore,the formula holds.()()FF ()()F bF aIntegration by Substitutions for definite integrals5si

4、n,xt cos.dxtdt 0t 0;x 2t 1;x 1222001cosx dxtdt 1201.x dx Evaluate Let Then when when thus by the formula,we haveSolution2011sin222tt.4 Integration by Substitutions for definite integrals60202sinsin22nnxdxt dt 2200sincosnnxdxxdx Prove (n is any positive integer).Proof02cosntdt 20cosntdt 20cos.nxdx In

5、tegration by Substitutions for definite integrals7 Find 2001sinlimd.xxxtttx Solution Let,xtu then,utx dd.utx 0t 0;u2.uxtx Then0sindxxttt 20sinxu duuxx 20sindxuuu 2022000sind1sinlimdlimxxxxuuxtutxtx So 00220sin2lim2xxxxx 1.Integration by Parts for indefinite integrals 8When u and v are differentiable

6、 functions of x,the Product Rule for differentiation tell us that()ddvduuvuvdxdxdxIntegrating both sides with respect to x and rearranging leads to the integral equation()dvdduudxuvdxvdxdxdxdx.duuvvdxdx Integration by Parts for indefinite integrals When u and v are differentiable functions of x on t

7、he interval ,a b.Then bbbaaaudvuvvdu 92.dxtdt 42200022xttedxetdttdext Let,then 2xt,Thus Solution 22002tttee dt 22(1).eExample Evaluate 40.xedx Integration by Parts for indefinite integrals 10 Evaluate 120arcsin.xdx Solution Letarcsin,ux,dvdx then2,1dxdux ,vx 120arcsin xdx 120arcsinxx 12201xdxx 12 6

8、1222011(1)21dxx 12 12201x31.122 bbbaaaudvuvvduIntegration by Parts for indefinite integrals bbbaaaudvuvvdu11 Evaluate 40.1 cos2xdxx Solution21cos22cos,xxQ401cos2xdxx 4202cosxdxx 40tan2xdx 401tan2xx 401tan2xdx 401lnsec82x ln2.84 Integration by Parts for indefinite integrals bbbaaaudvuvvdu12 Evaluate

9、120ln(1).(2)xdxx Solution120ln(1)(2)xdxx 101ln(1)2x dx 10ln(1)2xx 101ln(1)2dxx ln23 101121dxxx 1112xx 10ln2ln(1)ln(2)3xx 5ln2ln3.3Integration by Parts for indefinite integrals 13201)sinnnIxdx bbbaaaudvuvvdu Compute the following integrals:4222001)sin;2)sincos.nnIxdxxxdx Solution(1)20sin(cos)nxdx 222

10、0(1)(1sin)sinnnxxdx 22200(1)sin(1)sinnnnxdxnxdx 2220(1)cossinnnxxdx 2(1)(1).nnnInI 1222200sincos(1)cossinnnxxnxxdx 21(2,3,)nnnIInn Integration by Parts for indefinite integrals 144222001)sin;2)sincos.nnIxdxxxdx Solution(continued)bbbaaaudvuvvdu241132nnnnnnIIInnn 113 2,2 3nnInn 241132nnnnnnIIInnn If

11、n is odd013 1.2 2nnInn If n is even210sin1,Ixdx 200,2Idx Since134 2,ifis odd;25 3133 1,ifis even.24 2 2nnnnnnInnnnn we obtain Compute the following integrals:Integration by Parts for indefinite integrals 15424222002)sincossin(1sin)xxdxxx dx134 2,ifis odd;25 3133 1,ifis even.24 2 2nnnnnnInnnnn Comput

12、e the following integrals:4222001)sin;2)sincos.nnIxdxxxdx Solution(continued)bbbaaaudvuvvdu462200sinsinxdxxdx3 15 3 14 2 26 4 2 2.32 Quadrature Problems for elementary fundamental functions16By the previous examples,we have seen that quadratures are much more difficult than differentiations.When int

13、egrands are continuous,their integral must exist,but their computation sometimes requires skill,and sometimes may nor even be expressible by elementary functions.For instance,the integrals:24sin,1xxdxe dxdxxx seems very simple,and the integrands are all continuous.All of these integrals exist,but we can not express them in terms of elementary functions.In general,we have known that for any rational function and any rational trigonometric function,their integrals can be expressed by elementary functions.

侵权处理QQ:3464097650--上传资料QQ:3464097650

【声明】本站为“文档C2C交易模式”,即用户上传的文档直接卖给(下载)用户,本站只是网络空间服务平台,本站所有原创文档下载所得归上传人所有,如您发现上传作品侵犯了您的版权,请立刻联系我们并提供证据,我们将在3个工作日内予以改正。


163文库-Www.163Wenku.Com |网站地图|