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。谢谢谢谢谢谢 求解微积分题目。在线等答案。采纳悬赏财富值。
\u6c42\u5e2e\u5fd9\u89e3\u7b54\u4e00\u9053\u5fae\u79ef\u5206\u9898\u76ee~\u6025\uff01\uff01\uff01\u5728\u7ebf\u7b49\u7b54\u6848\u3002\u65b0\u624b\uff0c\u8d22\u5bcc\u503c\u4e0d\u9519\u60ac\u8d4f\u4e0d\u9519\uff0c\u8bf7\u591a\u62c5\u5f85C
\u222bxsin\uff08x²\uff09dx=½\u222bsinx²dx²=-½cosx²
\u5e0c\u671b\u91c7\u7eb3\u554a\uff01
Zy'=-x+4y-3=0
解得x=1,y=1
代入得z=-1
极大值还是极小值，可以用公式，简便方法还是当x=0时，z=1-3y，可以小于-1，所以应该是极大值-1。
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^...
绛旓細瑙ｇ瓟锛氭湰绉垎棰锛屽彧瑕佸皢鏍瑰紡鍐呭啓鎴(x+1)²+1锛屽氨鎴愪簡涓鑸殑鏍囧噯浠ｆ崲鐨勭被鍨嬶紝浠anu = 1+x锛屽啀杩愮敤鏈夌悊鍒嗗紡绉垎娉曪紝灏卞彲浠ョН鍑烘潵浜嗐傝缁嗚В绛旓紝璇峰弬瑙佷笅鍥俱傜偣鍑绘斁澶с佽崸灞忔斁澶у啀鏀惧ぇ锛
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)
绛旓細璁 f(x)=鈭玔1,x] ln(1+t)/t dt 浠=1/t =鈭玔1,1/x] uln(1+1/u) d1/u =鈭玔1,1/x] -[ln(1+u)-lnu] / udu =鈭玔1,1/x] -ln(1+u) / udu+ 鈭玔1,1/x] lnu / udu =-f(1/x)+鈭玔1,1/x] lnu / udu =-f(1/x)+鈭玔1,1/x] lnu dlnu...
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...
绛旓細鏁協'(0)=lim(鈻硏->0)[鈻硑/鈻硏] (鐢辨瀬闄愬畾涔)=lim(鈻硏->0)[鈭(0,鈻硏)(e^t²-1)dt]/(鈻硏)³=lim(鈻硏->0){[e^(鈻硏)²-1]/[3(鈻硏)²] (0/0鍨嬶紝搴旂敤缃楁瘮杈炬硶鍒)=lim(鈻硏->0){[2(鈻硏)e^(鈻硏)²]/[6(鈻硏)] (0/0鍨嬶紝搴旂敤...
绛旓細鍒嗕韩涓绉嶈В娉曪紝搴旂敤娆ф媺鍏紡姹傝В銆(I)锛岃ak=cos[(k-1)蟺/60]锛宐k=sin[(k-1)蟺/60]锛宬=1.2锛屸︼紝30銆傘愪护伪=蟺/60銆戔埓鈭(ak+ibk)=鈭慹^[i(k-1)伪]锛屾槸棣栭」涓1銆佸叕姣攓=e^(i伪)鐨勭瓑姣旀暟鍒椼傗埓鈭慹^[i(k-1)伪]=[1-e^(i30伪)]/[1-e^(i伪)]=[(1+sin伪-cos伪)+...
