5.3定積分的計算

1、第三節(jié)第三節(jié)定積分的計算定積分的計算教學目的：使學生熟練掌握定積分換元積分法與分部積分法教學目的：使學生熟練掌握定積分換元積分法與分部積分法教學重點：定積分換元積分法教學重點：定積分換元積分法一、換元積分法一、換元積分法定理定理假設(shè)函數(shù)f(x)在區(qū)間[a?b]上連續(xù)?函數(shù)x??(t)滿足條件?(1)?(??)?a??(?)?b?(2)?(t)在[???](或[???])上具有連續(xù)導數(shù)?且其值域不越出[a?b]?則有?dtttfdxxfb

2、a)()]([)(????????這個公式叫做定積分的換元公式?證明由假設(shè)知?f(x)在區(qū)間[a?b]上是連續(xù)?因而是可積的?f[?(t)]??(t)在區(qū)間[???](或[???])上也是連續(xù)的?因而是可積的?假設(shè)F(x)是f(x)的一個原函數(shù)?則?F(b)?F(a)?dxxfba)(?另一方面?因為F[?(t)]??F?[?(t)]??(t)?f[?(t)]??(t)?所以F[?(t)]是f[?(t)]??(t)的一個原函數(shù)?從而?F

3、[?(?)]?F[?(?)]?F(b)?F(a)?dtttf)()]([??????因此?dtttfdxxfba)()]([)(????????例1計算(a0)???adxxa022解?????20sin022coscos?tdtatadxxataxa令?????2022022)2cos1(2cos??dttatdta?220241]2sin21[2atta?????提示??dx?acost?當x?0時t?0?當x?a時?tataaxa

4、cossin22222????2??t例2計算?xdxxsincos520??解令t?cosx?則xxdxdxxcoscossincos520520????????????????aaaadxxfdxxfdxxfxf00)(2)(2)]()([討論?若f(x)在[?a?a]上連續(xù)且為奇函數(shù)?問？???aadxxf)(提示?若f(x)為奇函數(shù)?則f(?x)?f(x)?0?從而?0)]()([)(0???????aaadxxfxfdxxf例

5、6若f(x)在[0?1]上連續(xù)?證明(1)????2020)(cos)(sin??dxxfdxxf(2)???????00)(sin2)(sindxxfdxxxf證明(1)令?則tx??2?dttfdxxf)]2[sin()(sin0220??????????????2020)(cos)]2[sin(???dxxfdttf(2)令x???t?則??????00)][sin()()(sin????dttftdxxxf???????????

6、?00)(sin)()][sin()(dttftdttft???????00)(sin)(sindtttfdttf????????00)(sin)(sindxxxfdxxf所以???????00)(sin2)(sindxxfdxxxf例7設(shè)函數(shù)?計算?????????????01cos110)(2xxxxexfx??41)2(dxxf解設(shè)x?2?t?則????????????200121412cos11)()2(dttedttdttfd