大一微积分题目求详解哦
\u5927\u4e00\u5fae\u79ef\u5206\u9898\u76ee\u6c42\u8be6\u89e3\u5df2\u77e5f(x)\u5728\u533a\u95f4[a,b]\u5185\u53ef\u79ef\uff0c\u5219f(x)\u5728\u533a\u95f4[a,b]\u5185\u7684\u5e73\u5747\u503c\u4e3af(x)\u5728[a,b]\u5185\u7684\u5b9a\u79ef\u5206\u9664\u4ee5\u533a\u95f4\u957f\u5ea6\u3002
\u5177\u4f53\u89c1\u56fe\uff1a
f(-x) = -f(x)
g(x) =√(1-x^2)
g(-x) = g(x)
∫(-1->1)[√(1-x^2) + xcosx] dx
=∫(-1->1)√(1-x^2) dx
=2∫(0->1)√(1-x^2) dx
let
x=siny
dx=cosy dy
x=0, y=0
x=1, y=π/2
∫(0->1)√(1-x^2) dx
=∫(0->π/2) (cosy)^2 dy
=(1/2) ∫(0->π/2) (1+cos2y) dy
=(1/2) [y+ (1/2)sin2y]|(0->π/2)
=π/4
∫(-1->1)[√(1-x^2) + xcosx] dx
=2∫(0->1)√(1-x^2) dx
=π/2
ans: B
绛旓細璁緁(x)=x-味sinx f鈥(x)=1-味cosx 鍥犱负0<味<1 |cosx|鈮1 鎵浠鈥(x)>0 鍗砯(x)鏄湪R鍗曡皟閫掑 涓攛鈫掆垶 f(x)鈫掆垶 鎵浠ュ浜庝换鎰弝0鈭圧 閮藉瓨鍦▁浣垮緱y0=x-味sinx鎴愮珛 涓旇В鏄敮涓鐨 鏈涢噰绾 璋㈣阿
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绛旓細绛旓細1銆佸銆2銆佷笉瀵癸紝鐩稿樊涓涓父鏁般3銆侀敊锛屽嚱鏁板煎ぇ锛屽畬鍏ㄤ笉琛ㄧず鍑芥暟鐨勭┖闂村彉鍖栫巼澶э紝鎵浠ュ鍑芥暟鐨勫间笉鑳界敱姝ゅ喅瀹氾紱4銆侀敊锛屽鍑芥暟澶э紝琛ㄧず绌洪棿鍙樺寲鐜囧ぇ锛屼笉琛ㄧず鍘熷嚱鏁扮殑鍑芥暟鍊煎ぇ銆倅 = lncosx + 鏍瑰彿(3x+2)dy = [1/cosx][-sinx] + [1/2鏍瑰彿(3x+2)]*3 dx = {-tanx + 3/[2鏍瑰彿...
绛旓細an)^2<|an|锛屽洜姝の(an)^2鏀舵暃 (an)^2/[1+(an)^2]<(an)^2锛屽洜姝ゅ師绾ф暟鏀舵暃锛屽張鍥犱负杩欐槸涓椤圭骇鏁帮紝姝ｉ」绾ф暟鐨勬敹鏁涗竴瀹氭槸缁濆鏀舵暃銆3銆佺敱绉垎涓煎畾鐞嗭紝瀛樺湪尉鈭(0,1)锛屼娇鈭 [0-->1] x^n*f(x)dx=尉^nf(尉)鐢眆(x)杩炵画寰梖(尉)鏈夌晫锛屛綹n-->0锛屽洜姝ゆ湰棰樼粨鏋滀负0 ...
绛旓細鎶婃暟鍒楀垎鎴恱^2椤广亁椤瑰拰甯告暟椤 x^2椤圭郴鏁=(n-1)/n锛岃秼鍚戜簬1 x椤圭郴鏁=2a/n^2+4a/n^2+...+2(n-1)a/n^2 =2a[1+2+...+(n-1)]/n^2 =a(n-1)/n锛岃秼鍚戜簬a 甯告暟椤=a^2/n^3+4a^2/n^3+...+(n-1)^2*a^2/n^3 =(a^2/n^3)*[1+4+...+(n-1)^2]=(a...
绛旓細3 0 1 1/e 6x 2xe^x^2 1.5x^2-2x fxcosxdx=fxdsinx=xsinx-fsinxdx=xsinx+cosx 1/3 16 鏃犵┓ 1/e (x+5)^3/3 fxlnxdx=flnxd(x^2/2)=x^2lnx/2-fx^2/2dlnx==x^2lnx/2-fx/2==x^2lnx/2-x^2/4
绛旓細xy'+y=lnx/x (xy)'=lnx/x 涓よ竟绉垎锛歺y=(lnx)^2/2+C 浠=1锛1/2=C 鎵浠y=(lnx)^2/2+1/2 y=((lnx)^2+1)/(2x)
绛旓細璁绉垎涓篴 f(x)=1/(1+x鏂)+ax绔嬫柟 涓よ竟绉垎锛屽緱 a=鈭(0,1)1/(1+x鏂)dx+a鈭(0,1)x绔嬫柟dx a=arctanx|(0,1)+a脳1/4 x^4|(0,1)a=蟺/4+a/4 4a=蟺+a 3a=蟺 a=蟺/3
