lim(X→0)【[∫上限X下线0(sin2)tdt]/xcosx】要过程

定积分:
∫上限9下限4√x(1+√x)dx
∫上限2下限-1(x^2-1)dx
∫上限2下限1(1/x+x)dx
∫上限π/4下限0(sinx+cosx)dx
∫上限2下限1(√x+1/x^2)dx

lim[x→0] {∫[0,x] sin²t/dt} / (xcosx),应用洛必达法则
= lim[x→0] sin²x / (cosx - xsinx)
= (0)² / (1 - 0)
= 0

∫[4,9] √x(1+√x) dx
= ∫[4,9] (√x+x) dx
= (2/3)x^(3/2) + x²/2
= [(2/3)(9)^(3/2) + 9²/2] - [(2/3)(4)^(3/2) + 4²/2]
= 271/6

∫[-1,2] (x²-1) dx
= x³/3 - x
= [2³/3 - 2] - [(-1)³/3 - (-1)]
= 0

∫[1,2] (1/x+x) dx
= ln|x| + x²/2
= [ln2 + 2²/2] - [ln1 + 1²/2]
= 3/2 + ln2

∫[0,π/4] (sinx+cosx) dx
= -cosx + sinx
= [-cos(π/4) + sin(π/4)] - [-cos(0) + sin(0)]
= [-√2/2 + √2/2] - [-1 + 0]
= 1

∫[1,2] (√x + 1/x²) dx
= ∫[1,2] [x^(1/2) + x^(-2)] dx
= (2/3)x^(3/2) - 1/x
= [(2/3)(2)^(3/2) - 1/2] - [(2/3)(1)^(3/2) - 1]
= (4√2)/3 - 1/6
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lim(X→0)【[∫上限X下线0(sin2)tdt]\/xcosx】要过程
= lim[x→0] sin²x \/ (cosx - xsinx)= (0)² \/ (1 - 0)= 0 ∫[4,9] √x(1+√x) dx = ∫[4,9] (√x+x) dx = (2\/3)x^(3\/2) + x²\/2 = [(2\/3)(9)^(3\/2) + 9²\/2] - [(2\/3)(4)^(3\/2) + 4²\/2]= 271\/6 ...

求极限lim(x→0)∫(x,0)(cost^2dx)\/x
分子为关于x的变上限积分.当x趋于0,其也趋于0.故为0\/0型,用洛必达法则 若积分从x到0(反之加负号),则 =lim[cosx^2(-1)]\/1,x趋于0 =-1

求极限 lim x→0 ∫ x 0 sintdt xtanx .
∫x0sintdt=-cost.x0=1-cosx,故limx→0∫x0sintdtxtanx=limx→01-cosxxtanx=limx→0cosx(1-cosx)xsinx=limx→01-cosxxsinx=limx→0sinxsinx+xcosx=limx→0cosxcosx+cosx-xsinx=12.

lim(x→0)∫cost^2dt\/x上限x下限0
分子分母同导;lim→0(∫[0,x]cost^2dt)\/x =lim→0(cosx^2)\/1 =cos0 =1

limx→0 上限为x下限为0 2tcostdt\/1-cosx 的极限
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∫上限π\/2下限0 sin2x cosx dx
∫ sin2x cosx dx =∫ 2*sinx (cosx)^2 dx =-2∫ (cosx)^2 d cos x =-2\/3*(cosx)^3| =2\/3

limx->0{【∫(下限为0,上限为x)sintdt\/】\/【∫(下限为0,上限为x)tdt...
最后再求极限,先求两个的积分 分子的积分很简单 = 1-cosx = 2[sin(x\/2)]^2 分母的积分也很简单 = 0.5x^2 再求极限,此时sin(x\/2)可以用x\/2替代(当x趋向0时,这两个是等价无穷小量),即 [2*(x\/2)^2]\/0.5x^2 = 1

lim在x趋近于0的时候∫上限为x下限为0(1-cost)dt这个整体比上1\/2x²...
用洛必达法则:x→0时,原式→(1-cosx)\/(4x)→sinx\/4 →0.注:原式的分母是2x^2,不是(1\/2x^2).

求lim(x→0)[∫(上限x^2,下限0)tantdt]\/(sint)^4
洛必达 原式=lim(x->0) tan(x^2)\/[4cosx(sinx)^3]=lim(x->0) 2(x^3)\/(4cosxx^3)=(1\/2)lim(x->0) 1\/cosx =1\/2

limx→0 ∫上限x下限0(1-cost)dt\/sin^2x?
分享解法如下。由积分中值定理有,分子=x(1-cosξ),0<ξ<x。显然,x→0时,ξ→0。∴ 1-cosξ~ξ²\/2。∴原式=lim(x→0,ξ→0)(xξ²\/2)\/sin²x=lim(x→0)(x³\/2)\/sin²x=0。

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