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当x→0+时,与∫x0sint2Dt等价的无穷小量是( )A...

此题用罗必塔法则比较简单,lim(x--0) f(x)/g(x)=lim(x--0)f'(x)/g'(x) =lim(x--0)[sin(sinx)^2]cosx/(3x^2+4x^3) =(等价无穷小代换) lim(x--0)x^2/3x^2=1/3,所以选B.

这是0/0型,用两次洛必达法则,得到极限是2/3,求导的时候注意到(\int_0^x f(t)dt)' = f(x)

令x-t=u;则:dt=-du;ddx∫x0sin(x?t)2dt=ddx∫0x(?sinu2)du=ddx∫x0sinu2du=sinx2.故本题答案为:sinx2.

先求∫(0->x)sin(x-t)^2dt =∫(0->x)(1-cos(2x-2t)/2 dt =1/2∫(0->x)dt-1/2∫(0->x)cos(2x-2t)dt =x/2+1/4∫(0->x)cos(2x-2t)d(2x-2t) =x/2+1/4sin(2x-2t)|(0->x) =x/2+1/4(sin(2x-2x)-sin(2x-2*0) =x/2+sin2x/4 所以 d/dx∫(0->x)sin(x-t)^2dt =d(x/...

x-t=u t=x-u t=x u=0 t=0 u=x d/dx ∫(0,x) sin(x-t)²dt=d/dx ∫(0,x) sinu²du =sinx²

[∫(0→x) sint^2 dt =[∫(0→x) (1-cos2t)/2dt =1/2t-1/4sin2t.把t换成x =1/2x-1/4sin2x

这样形式的积分应该在极限中出现的,所以只有求导化简

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