设f(x)=∫(1,x^3)sint/tdt,求∫(0,1)x^2f(x)dx (若f(x)=∫(1,x^n)sint/
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设f(x)=∫(1,x^3)sint/tdt,求∫(0,1)x^2f(x)dx (若f(x)=∫(1,x^n)sint/tdt,则∫(0,1)x^(x-1)f(x)dx又为什么
显然f(1)=0;由微积分基本定理知道f'(x)=sin(x^3)/x^3 *3x^2=3sin(x^3)/x.
于是∫(0,1)x^2f(x)dx
=∫(0,1)f(x)d(x^3/3)
=x^3*f(x)/3|上限1下限0-∫(0,1)x^3*f‘(x)/3dx
=-∫(0,1) x^2sin(x^3)dx
=cos(x^3)/3|上限1下限0
=(cos1-1)/3.
下一问类似来做即可.结果是(cos1-1)/n.
于是∫(0,1)x^2f(x)dx
=∫(0,1)f(x)d(x^3/3)
=x^3*f(x)/3|上限1下限0-∫(0,1)x^3*f‘(x)/3dx
=-∫(0,1) x^2sin(x^3)dx
=cos(x^3)/3|上限1下限0
=(cos1-1)/3.
下一问类似来做即可.结果是(cos1-1)/n.
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