一道有关微积分题目求大佬解答 微积分的一道判断题，求解答
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2019-09-04 \u56de\u7b54\u8005: \u6bdb\u5e05\u9879\u5dcd\u7136 1\u4e2a\u56de\u7b54
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2014-10-13 \u56de\u7b54\u8005: \u77e5\u9053\u7f51\u53cb 1\u4e2a\u56de\u7b54 3
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\u7b54\uff1a\u8bbe f(x)=\u222b[1,x] ln(1+t)/t dt \u4ee4u=1/t =\u222b[1,1/x] uln(1+1/u) d1/u =\u222b[1,1/x] -[ln(1+u)-lnu] / udu =\u222b[1,1/x] -ln(1+u) / udu+ \u222b[1,1/x] lnu / udu =-f(1/x)+\u222b[1,1/x] lnu / udu =-f(1/x)+\u222b[1,1/x] lnu dlnu =-f(1/x)+(lnu)^2/2 | [1,1/x] =-f(1...
2013-09-07 \u56de\u7b54\u8005: \u77e5\u9053\u7f51\u53cb 1\u4e2a\u56de\u7b54
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\u7b54\uff1a\u222bx^3.\u221a(4-x^2) dx =-(1/3)\u222bx^2. d(4-x^2)^(3/2) =-(1/3)x^2. (4-x^2)^(3/2) +(2/3)\u222bx (4-x^2)^(3/2) dx =-(1/3)x^2. (4-x^2)^(3/2) -(2/15)\u222b d(4-x^2)^(5/2) =-(1/3)x^2. (4-x^2)^(3/2) -(2/15)(4-x^2)^(5/2) +C
2019-12-02 \u56de\u7b54\u8005: tllau38 2\u4e2a\u56de\u7b54 1
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x->0
(1+x)^(1/x)
=e^[ln(1+x)/x]
ln(1+x) = x -(1/2)x^2 +(1/3)x^3 +o(x^3)
=e^[ 1 - (1/2)x + (1/3)x^2 +o(x^2)]
=e. e^[ - (1/2)x + (1/3)x^2 +o(x^2)]
利用泰勒公式把 e^[ - (1/2)x + (1/3)x^2 +o(x^2)] 拆开
=e. { 1 +[ - (1/2)x + (1/3)x^2 +o(x^2)] +(1/2)[ - (1/2)x + (1/3)x^2 +o(x^2)]^2 +o(x^2) }
=e. { 1 +[ - (1/2)x + (1/3)x^2 +o(x^2)] +(1/2)[ (1/4)x^2 +o(x^2)] +o(x^2) }
=e. [1 -(1/2)x + (11/24)x^2 +o(x^2) ]
e^[(1+x)^(1/x)]
=e^【e. [1 -(1/2)x + (11/24)x^2 +o(x^2) ]】
=e^e. e^[ -(e/2)x + (11e/24)x^2 +o(x^2) ]
利用泰勒公式把 e^[ -(e/2)x + (11e/24)x^2 +o(x^2) ] 拆开
=e^e. { 1+[-(e/2)x + (11e/24)x^2+o(x^2)]+(1/2)[-(e/2)x+(11e/24)x^2 +o(x^2)]^2 }
=e^e. { 1+[-(e/2)x + (11e/24)x^2+o(x^2)]+(1/2)[(e^2/4)x^2+o(x^2)] }
=e^e. { 1 -(e/2)x + [(11e/24) + (e^2/8) ]x^2+o(x^2) }
同样地
(1+x)^(e/x)
=e^[e.ln(1+x)/x]
=e^[ e -(e/2)x + (e/3)x^2 +o(x^2)]
=e^e. e^[-(e/2)x + (e/3)x^2 +o(x^2)]
=e^e.{ 1 +[-(e/2)x + (e/3)x^2 +o(x^2)] +(1/2)[-(e/2)x + (e/3)x^2 +o(x^2)]^2 +o(x^2) }
=e^e.{ 1 +[-(e/2)x + (e/3)x^2 +o(x^2)] +(1/2)[(e^2/4)x^2 +o(x^2)] +o(x^2) }
=e^e.{ 1 -(e/2)x + [(e/3)+(e^2/8)]x^2 +o(x^2) }
e^[(1+x)^(1/x)] -(1+x)^(e/x)
=e^e. { 1 -(e/2)x + [(11e/24) + (e^2/8) ]x^2+o(x^2) }
-e^e.{ 1 -(e/2)x + [(e/3)+(e^2/8)]x^2 +o(x^2) }
=e^e. [(e/8)x^2 +o(x^2) ]
lim(x->0) { e^[(1+x)^(1/x)] -(1+x)^(e/x) }/x^2
=lim(x->0) e^e. [(e/8)x^2 ]/x^2
=(e/8) e^e
=(1/8) e^(e+1)
绛旓細1} 1+鈭憑2 鈮 n} (1/n-1(n-1))t^(n-1) dt = 1+鈭憑2 鈮 n} (1/n-1(n-1))路鈭珄0,1} t^(n-1) dt (鐢辩骇鏁板湪(-1,1)鍐呴棴涓鑷存敹鏁, 鍙愰」绉垎)= 1+鈭憑2 鈮 n} (1/n-1(n-1))路1/n = 1-鈭憑2 鈮 n} 1/(n²(n-1)),鍗虫墍姹傝瘉....
绛旓細(1)鈭(0->3) 鈭(x+1)dx =(2/3)(x+1)^(3/2)|(0->3)=(2/3)(8-1)=14/3 ans : C (2)f(x) =x(cosx)^3/(x^2+1)f(-x) =-f(x)=> 鈭(-5->5) x(cosx)^3/(x^2+1) dx =0 ans :D (3)y=鈭(3->x^2) tf(t^2) dt y'=x^2.f(x^4) .(x^2)...
绛旓細瑙佸浘鐗囷紝姣忛杩囩▼閮藉緢璇︾粏銆傜偣鍑诲彲浠ョ湅澶у浘銆傝繕鏈変粈涔堜笉鏄庣櫧鐨勫湴鏂瑰彲浠ョ粰鎴戠暀瑷銆
绛旓細1锛塪A/dt=鈭玡^(-3t)cos2tdt =1/2*鈭玡^(-3t)d(sin2t)=1/2*[e^(-3t)sin2t+3*鈭玡^(-3t)sin2tdt]=1/2*e^(-3t)sin2t-3/4*鈭玡^(-3t)d(cos2t)=1/2*e^(-3t)sin2t-3/4*[e^(-3t)cos2t+3鈭玡^(-3t)cos2tdt]=1/2*e^(-3t)sin2t-3/4*e^(-3t)cos2t-9...
绛旓細x->0 (1+x)^(1/x)=e^[ln(1+x)/x]ln(1+x) = x -(1/2)x^2 +(1/3)x^3 +o(x^3)=e^[ 1 - (1/2)x + (1/3)x^2 +o(x^2)]=e. e^[ - (1/2)x + (1/3)x^2 +o(x^2)]鍒╃敤娉板嫆鍏紡鎶 e^[ - (1/2)x + (1/3)x^2 +o(x^2)] 鎷嗗紑 =e. {...
绛旓細瑙ｇ瓟濡傚浘
绛旓細鍩烘湰涓婂彲浠ワ紝浣嗚绠楅涓鑸墠闈㈣鍐欌滆В鈥濓紝璇佹槑棰樿鍐欌滆瘉鏄庘濈8璇佹槑棰樿繖鏍峰仛锛氾紙1锛夎y=lnx-2(x-1)/(x+1)y=lnx-2+4/(x+1)姹傚寰楋細y'=1/x-4/(x+1)^2 褰搚'=0鏃讹紝1/x-4/(x+1)^2=0,x^2+2x+1=4x x=1 褰搙>1鏃讹紝y'>0锛屽崟璋冨 鑰屽綋x=1鏃讹紝y(1)=0锛屾晠x>1...
绛旓細x<n+1>/xn=(n+10)/(3n-2),鈭磝n/x<n-1>=(n+9)/(3n-5),鈥︹3/x2=12/4,x2/x1=11,绱箻寰梮n/x1=[11*12*鈥︹*(n+9)]/[1*4*鈥︹*锛3n-5)],鐢(n+9)/(3n-5)<1/2寰2n+18<3n-5,n>23,鈭磝n/x1<11*12*鈥︹*33/[1*4*鈥︹*64*2^(n-23)]鈫0锛坣鈫掆垶),...
绛旓細1.浠=sint 鍘熷紡=鈭(0鈫捪/2)sin^2(t)*cost*costdt =鈭(0鈫捪/2)(1-cos(2t))/2*(1+cos(2t))/2dt =1/4鈭(0鈫捪/2)(1-cos^2(2t))dt =1/4鈭(0鈫捪/2)(1-(1+cos(4t))/2)dt =1/4鈭(0鈫捪/2)(1-cos(4t))/2dt =1/8鈭(0鈫捪/2)dt-1/8鈭(0鈫捪/2)...
绛旓細鈭2x/鈭(x^2+4x) dx =2鈭玠鈭(x^2+4x) - 4鈭 {1/鈭(x^2+4x)} dx = 2鈭(x^2+4x) - 4鈭 {1/鈭(x^2+4x)} dx x^2+4x =(x+2)^2-4 let x+2 = 2seca dx= 2secatanada 鈭 {1/鈭(x^2+4x)} dx =鈭 secada = ln|seca+tana|+C'= ln| (x+2)/2 ...
