2个回答
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1.设x-t=u,代入得:
∫(0,x)tf(x-t)dt=-∫(x,0)(x-u)f(u)du=∫(0,x)(x-u)f(u)du
=x∫(0,x)f(u)du-∫(0,x)uf(u)du
代入并求导得:3x^2=xf(x)+∫(0,x)f(u)du-xf(x)
所以:∫(0,x)f(u)du=3x^2
∫(0,1)f(2x-3)dx=(1/2)∫(0,1)f(2x-3)d(2x-3)=(1/2)∫(-3,-1)f(u)d(u)
=(1/2)(∫(0,-1)f(u)d(u)-∫(0,-3)f(u)d(u))
=(1/2)(3-27)=-12
2.ln(1+f(x)/sinx)与f(x)/sinx等价,2^x-1与xln2等价
所以:3=lim[f(x)/sinx]/(xln2)=(1/ln2)limf(x)/xsinx)=(1/ln2)limf(x)/x^2)
极限=3ln2
∫(0,x)tf(x-t)dt=-∫(x,0)(x-u)f(u)du=∫(0,x)(x-u)f(u)du
=x∫(0,x)f(u)du-∫(0,x)uf(u)du
代入并求导得:3x^2=xf(x)+∫(0,x)f(u)du-xf(x)
所以:∫(0,x)f(u)du=3x^2
∫(0,1)f(2x-3)dx=(1/2)∫(0,1)f(2x-3)d(2x-3)=(1/2)∫(-3,-1)f(u)d(u)
=(1/2)(∫(0,-1)f(u)d(u)-∫(0,-3)f(u)d(u))
=(1/2)(3-27)=-12
2.ln(1+f(x)/sinx)与f(x)/sinx等价,2^x-1与xln2等价
所以:3=lim[f(x)/sinx]/(xln2)=(1/ln2)limf(x)/xsinx)=(1/ln2)limf(x)/x^2)
极限=3ln2
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