高数微积分第四题
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原式=1/3 ∫0-1 f（x）d(x^3)
=1/3[f(x)x^3]0-1 - [1/3∫0-1 x^3 d(f(x)) ] 其中d(f(x)) =sinx/x^ 2. (2x+3)
=1/3(arc4-arc1) -[1/3∫0-1( 2x^2+3x)sinx dx ]
=1/3(arc4-arc1)-1/3(7sin1-cos1-4)
无语，这么简单的题多看看书行不，比照书上的例题我就不信有人做不出来
绛旓細绛変环鏃犵┓灏忔浛鎹紝鍒嗗瓙~1/2*x^2锛屽垎姣崀x^2锛屾墍浠ョ粨鏋滀负1/2锛岄夐」A姝ｇ‘
绛旓細=鈭(0鈫1)dy鈭(0鈫抷)e^(-y^2)dx =鈭(0鈫1)ye^(-y^2) dy =-1/2路e^(-y^2) |(0鈫1)=1/2路(1-1/e)
绛旓細浠=sec胃锛屽垯dx=d(sec胃)=sec胃*tan胃d胃 鍒欙紝x^2-1=sec^2 胃-1=tan^2 胃 鎵浠ワ紝鍘熷紡=鈭玔sec胃*tan胃/(sec^2 胃*tan胃)]d胃 =鈭(1/sec胃)d胃 =鈭玞os胃d胃 =sin胃+C 宸茬煡sec胃=x锛屽嵆1/cos胃=x 鎵浠ワ紝cos胃=1/x 閭ｄ箞锛宻in胃=鈭(1-x^2)/x 鎵浠ワ紝鍘熷紡=鈭(1-x...
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绛旓細鍒嗛儴绉垎娉曪細鍘熷紡=1/3 鈭0-1 f锛坸锛塪(x^3)=1/3[f(x)x^3]0-1 - [1/3鈭0-1 x^3 d(f(x)) ] 鍏朵腑d(f(x)) =sinx/x^ 2. (2x+3)=1/3(arc4-arc1) -[1/3鈭0-1( 2x^2+3x)sinx dx ]=1/3(arc4-arc1)-1/3(7sin1-cos1-4)...
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