求帮忙解答一道微积分题目。在线等答案。。。。。。急 求帮忙解答一道微积分题目。在线等答案~~采纳时加悬赏财富值。...
\u6c42\u5e2e\u5fd9\u89e3\u7b54\u4e00\u9053\u5fae\u79ef\u5206\u9898\u76ee~\u8981\u8fc7\u7a0b\u3002\u9898\u4e0d\u96be\uff0c\u5927\u5bb6\u5e2e\u5e2e\u5fd9\u3002\u6025\uff0c\u5728\u7ebf\u7b49\u7b54\u6848\u65b9\u6cd5\u4e00\uff1a
\u2235f(x)=x³cosx\u4e3a\u5947\u51fd\u6570
\u2234\u222b(-\u03c0/2, \u03c0/2) x³cosx dx=0
\u65b9\u6cd5\u4e8c\uff1a
\u222b(-\u03c0/2, \u03c0/2) x³cosx dx
=\u222b(-\u03c0/2, \u03c0/2) x³ d(sinx)
=x³sinx |(-\u03c0/2, \u03c0/2)-\u222b(-\u03c0/2, \u03c0/2) 3x²sinx dx
=\u222b(-\u03c0/2, \u03c0/2) 3x² d(cosx)
=3[x²cosx |(-\u03c0/2, \u03c0/2)-\u222b(-\u03c0/2, \u03c0/2) 2xcosx dx]
=-6\u222b(-\u03c0/2, \u03c0/2) xd(sinx)
=-6[xsinx |(-\u03c0/2, \u03c0/2)-\u222b(-\u03c0/2, \u03c0/2) sinx dx]
=-6cosx |(-\u03c0/2, \u03c0/2)
=0
Zx'=2x-y-1=0
Zy'=-x+4y-3=0
\u89e3\u5f97x=1,y=1
\u4ee3\u5165\u5f97z=-1
\u6781\u5927\u503c\u8fd8\u662f\u6781\u5c0f\u503c\uff0c\u53ef\u4ee5\u7528\u516c\u5f0f\uff0c\u7b80\u4fbf\u65b9\u6cd5\u8fd8\u662f\u5f53x=0\u65f6\uff0cz=1-3y\uff0c\u53ef\u4ee5\u5c0f\u4e8e-1\uff0c\u6240\u4ee5\u5e94\u8be5\u662f\u6781\u5927\u503c-1\u3002
交换积分次序，则根据图像可知积分域为
0≤y≤1, 0≤x≤³√y
∴∫(0,1)dx∫(x³,1)f(x,y)dy=∫(0,1)dy∫(0,³√y)f(x,y)dx
这道题主要考察积分区域，在坐标系中找到积分区域交换积分次序就行了
x积分区间是[0 1]
y积分区间是[x^3 1]
画图观察 那么其实积分区间就是图像y=x^3与y=1和y轴围成的区域。
交换x y 的积分次序得 积分0~1 dy 积分 0~三次根号下y f(x,y)dx
绛旓細鏂规硶涓锛氣埖f(x)=x³cosx涓哄鍑芥暟 鈭粹埆(-蟺/2, 蟺/2) x³cosx dx=0 鏂规硶浜岋細鈭(-蟺/2, 蟺/2) x³cosx dx =鈭(-蟺/2, 蟺/2) x³ d(sinx)=x³sinx |(-蟺/2, 蟺/2)-鈭(-蟺/2, 蟺/2) 3x²sinx dx =鈭(-蟺/2, 蟺/2) 3x&#...
绛旓細绉垎鍩熶负0鈮鈮1, x³鈮鈮1 浜ゆ崲绉垎娆″簭锛屽垯鏍规嵁鍥惧儚鍙煡绉垎鍩熶负 0鈮鈮1, 0鈮鈮³鈭歽 鈭粹埆(0,1)dx鈭(x³,1)f(x,y)dy=鈭(0,1)dy鈭(0,³鈭歽)f(x,y)dx
绛旓細瑙ｅ緱x=1,y=1 浠ｅ叆寰梲=-1 鏋佸ぇ鍊艰繕鏄瀬灏忓硷紝鍙互鐢ㄥ叕寮忥紝绠渚挎柟娉曡繕鏄綋x=0鏃讹紝z=1-3y锛屽彲浠ュ皬浜-1锛屾墍浠ュ簲璇ユ槸鏋佸ぇ鍊-1銆
绛旓細1.瑙ｏ細璁綟(x锛寉)=x³+y³-xy-1=0锛屽垯dy/dx=-(∂F/∂x)/(∂F/∂y)=-(3x²-y)/(3y²-x)2.瑙ｏ細y=(x²+1)arctanx-x+ln2锛沝y=[2xarctanx+1-1]dx=2xarctanxdx 鍥涖姹傜Н鍒 1.鈭玔2^x+(2/x)-x²]dx=(2^...
绛旓細z'x=2x-y-1 z'y=-x+4y-3 z'x=0 z'y=0 y=1,x=1 z鏋佸=1-1+2-1-3+1 =-1
绛旓細绗5棰橈細lim[sin(xy)/y] x->1, y->0 f(x,y)=sin(xy), g(x,y)=y; x0=1, y0=0; dx=x-1, dy=y f'x(x,y)=ycos(xy), f'y(x,y)=xcos(xy)g'x(x,y)=0, g'y(x,y)=1 鈭磍im[sin(xy)/y] x->1, y->0 =lim{[ycos(xy)*dx+xcos(xy)*dy]/[0...
绛旓細z = ln[(x-y)/(x+y)] = ln|x-y| - ln|x+y| ∂z/∂x = 1/(x-y) - 1/(x+y) = 2y/(x^2-y^2)∂z/∂y = -1/(x-y) - 1/(x+y) = -2x/(x^2-y^2)dz = 2(ydx-xdy)/(x^2-y^2)
绛旓細閫塁 璐熸棤绌锋椂鐨勬瀬闄愪负5鍙互淇濊瘉瀛樺湪涓涓狹1>0锛屼娇寰楀綋x<-M1鏃讹紝|f(x)-5|<1锛屽嵆4<f(x)<6 姝ｆ棤绌锋椂鐨勬瀬闄愪负0鍙互淇濊瘉瀛樺湪涓涓狹2>0锛屼娇寰楀綋x>M2鏃讹紝|f(x)-0|<1锛屽嵆-1<f(x)<1 鍥犱负f(x)鍦≧涓婅繛缁紝鎵浠ュ湪[-M1,M2]涓婂彇鍒版渶澶ф渶灏忓笺傚洜涓篺(0)=10锛屾墍浠-M1,M2]涓婄殑鏈澶...
绛旓細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. { 1 +[ - (1/2)x + (1/3)x^2 +o(...
绛旓細缁濆鏀舵暃锛屾牴鍊煎鏁涙硶锛歭im[n->鏃犵┓](n^2/4^n)^(1/n)=1/4<1鏀舵暃
