分部积分法求∫(x∧2)e∧(x∧2)dx 求解∫e^(x^2)dx，谢谢。
\u6c42\u79ef\u5206\u222b\u3010\uff08x^2\uff09e^\uff08x^2\uff09\u3011dx\u7ed3\u679c\u4e3a\uff1aB/2 = \u221a\u03c0 /2
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u8bbe\u539f\u79ef\u5206\u7b49\u4e8eA
\u2235 B\uff1d \u222b e^(-x^2)dx \u79ef\u5206\u533a\u95f4\u4e3a\u8d1f\u65e0\u7a77\u5230\u6b63\u65e0\u7a77
\u2235 B\uff1d \u222b e^(-y^2)dy \u79ef\u5206\u533a\u95f4\u4e3a\u8d1f\u65e0\u7a77\u5230\u6b63\u65e0\u7a77
\u53c8\uff0c\u88ab\u79ef\u51fd\u6570e^\uff08-x^2\uff09\u5728\u6b63\u8d1f\u65e0\u7a77\u4e0a\u5076\u51fd\u6570
\u2234A=B/2
\u2234B^2= (\u222b e^(-x^2)dx)*(\u222b e^(-y^2)dy) = \u222b \u222b e^(-(x^2+y^2))dx dy
\u5c06\u4e0a\u8ff0\u79ef\u5206\u5316\u5230\u6781\u5750\u6807\u4e2d
\u2234 x^2+y^2=r^2
\u222b \u222b e^(-(x^2+y^2))dx dy = \u222b \u222b r e^(-r^2)dr d\u03b8 r\u4ece0\u5230\u6b63\u65e0\u7a77\uff0c\u03b8\u4ece0\u52302\u03c0
= \u222b 1/2 d\u03b8 \u03b8\u4ece0\u52302\u03c0= \u03c0
\u2234B\uff1d\u221a\u03c0
\u2234B/2 = \u221a\u03c0 /2
\u6269\u5c55\u8d44\u6599\u6c42\u51fd\u6570\u79ef\u5206\u7684\u65b9\u6cd5\uff1a
\u8bbef\uff08x\uff09\u662f\u51fd\u6570f\uff08x\uff09\u7684\u4e00\u4e2a\u539f\u51fd\u6570\uff0c\u6211\u4eec\u628a\u51fd\u6570f\uff08x\uff09\u7684\u6240\u6709\u539f\u51fd\u6570F\uff08x\uff09+C\uff08C\u4e3a\u4efb\u610f\u5e38\u6570\uff09\u53eb\u505a\u51fd\u6570f\uff08x\uff09\u7684\u4e0d\u5b9a\u79ef\u5206\uff0c\u8bb0\u4f5c\uff0c\u5373\u222bf\uff08x\uff09dx=F\uff08x\uff09+C\u3002
\u5176\u4e2d\u222b\u53eb\u505a\u79ef\u5206\u53f7\uff0cf\uff08x\uff09\u53eb\u505a\u88ab\u79ef\u51fd\u6570\uff0cx\u53eb\u505a\u79ef\u5206\u53d8\u91cf\uff0cf\uff08x\uff09dx\u53eb\u505a\u88ab\u79ef\u5f0f\uff0cC\u53eb\u505a\u79ef\u5206\u5e38\u6570\uff0c\u6c42\u5df2\u77e5\u51fd\u6570\u4e0d\u5b9a\u79ef\u5206\u7684\u8fc7\u7a0b\u53eb\u505a\u5bf9\u8fd9\u4e2a\u51fd\u6570\u8fdb\u884c\u79ef\u5206\u3002
\u82e5f\uff08x\uff09\u5728[a,b]\u4e0a\u6052\u4e3a\u6b63\uff0c\u53ef\u4ee5\u5c06\u5b9a\u79ef\u5206\u7406\u89e3\u4e3a\u5728Oxy\u5750\u6807\u5e73\u9762\u4e0a\uff0c\u7531\u66f2\u7ebf\uff08x\uff0cf\uff08x\uff09\uff09\u3001\u76f4\u7ebfx=a\u3001x=b\u4ee5\u53cax\u8f74\u56f4\u6210\u7684\u9762\u79ef\u503c\uff08\u4e00\u79cd\u786e\u5b9a\u7684\u5b9e\u6570\u503c\uff09\u3002
\u79ef\u5206\u7684\u4e00\u4e2a\u4e25\u683c\u7684\u6570\u5b66\u5b9a\u4e49\u7531\u6ce2\u6069\u54c8\u5fb7\u00b7\u9ece\u66fc\u7ed9\u51fa\u3002\u9ece\u66fc\u7684\u5b9a\u4e49\u8fd0\u7528\u4e86\u6781\u9650\u7684\u6982\u5ff5\uff0c\u628a\u66f2\u8fb9\u68af\u5f62\u8bbe\u60f3\u4e3a\u4e00\u7cfb\u5217\u77e9\u5f62\u7ec4\u5408\u7684\u6781\u9650\u3002
\u8def\u5f84\u79ef\u5206\u662f\u591a\u5143\u51fd\u6570\u7684\u79ef\u5206\uff0c\u79ef\u5206\u7684\u533a\u95f4\u4e0d\u518d\u662f\u4e00\u6761\u7ebf\u6bb5\uff08\u533a\u95f4[a,b]\uff09\uff0c\u800c\u662f\u4e00\u6761\u5e73\u9762\u4e0a\u6216\u7a7a\u95f4\u4e2d\u7684\u66f2\u7ebf\u6bb5\uff1b\u5728\u9762\u79ef\u79ef\u5206\u4e2d\uff0c\u66f2\u7ebf\u88ab\u4e09\u7ef4\u7a7a\u95f4\u4e2d\u7684\u4e00\u4e2a\u66f2\u9762\u4ee3\u66ff\u3002\u5bf9\u5fae\u5206\u5f62\u5f0f\u7684\u79ef\u5206\u662f\u5fae\u5206\u51e0\u4f55\u4e2d\u7684\u57fa\u672c\u6982\u5ff5\u3002
\u5982\u679c\u4e00\u4e2a\u51fd\u6570\u7684\u79ef\u5206\u5b58\u5728\uff0c\u5e76\u4e14\u6709\u9650\uff0c\u5c31\u8bf4\u8fd9\u4e2a\u51fd\u6570\u662f\u53ef\u79ef\u7684\u3002\u4e00\u822c\u6765\u8bf4\uff0c\u88ab\u79ef\u51fd\u6570\u4e0d\u4e00\u5b9a\u53ea\u6709\u4e00\u4e2a\u53d8\u91cf\uff0c\u79ef\u5206\u57df\u4e5f\u53ef\u4ee5\u662f\u4e0d\u540c\u7ef4\u5ea6\u7684\u7a7a\u95f4\uff0c\u751a\u81f3\u662f\u6ca1\u6709\u76f4\u89c2\u51e0\u4f55\u610f\u4e49\u7684\u62bd\u8c61\u7a7a\u95f4\u3002
先用分部积分法：∫x^2e(x^2)dx=(1/2)∫xd(e^x^2)=(1/2)xe^(x^2)-(1/2)∫e^x^2dx，这里求∫e^x^2dx，设t=x^2，dx=1/[2t^(1/2)]
原式=∫e^tdt/t^(1/2)
用泰勒展开式e^t=1+t+t^2/2!+t^3/3!+..+t^n/n!
=∫[1/t^(1/2)+t^(1/2)+t^(3/2)/2!+t^(5/2)/3!+..+t^(n-1/2)/n!]dt 逐项积分：
=2t^(1/2)+(2/3)t^(3/2)+(2/5)t^(5/2)/2!+(2/7)t^(7/2)/3!+..+(n+1/2)*t^(n+1/2)/n!+C
所以∫x^2e^(x^2)dx
=(1/2)xe^(x^2)-(1/4)[2*x+(2/3)x^3+(2/5)x^5/2!+(2/7)x^7/3!+..+(n+1/2)x^(2n+1)/n!] +C
解：
∫x²·e^(-x)dx
=∫x²
d[-e^(-x)]
=x²·[-e^(-x)]-∫2x·[-e^(-x)]dx
=-x²·e^(-x)+∫2x·e^(-x)dx
=-x²·e^(-x)-2x·e^(-x)+∫2e^(-x)dx
=-x²·e^(-x)-2x·e^(-x)-2e^(-x)+c
如果满意记得采纳哦！
你的好评是我前进的动力。
(*^__^*)
嘻嘻……
我在沙漠中喝着可口可乐，唱着卡拉ok，骑着狮子赶着蚂蚁，手中拿着键盘为你答题！！！
百鸟朝凤先用分部积分法：∫x^2e(x^2)dx=(1/2)∫xd(e^x^2)=(1/2)xe^(x^2)-(1/2)∫e^x^2dx，这里求∫e^x^2dx，设t=x^2，dx=1/[2t^(1/2)
绛旓細鍏堢敤鍒嗛儴绉垎娉锛氣埆x^2e(x^2)dx=(1/2)鈭玿d(e^x^2)=(1/2)xe^(x^2)-(1/2)鈭玡^x^2dx锛岃繖閲姹傗埆e^x^2dx锛岃t=x^2锛宒x=1/[2t^(1/2)]鍘熷紡=鈭玡^tdt/t^(1/2)鐢ㄦ嘲鍕掑睍寮寮廵^t=1+t+t^2/2!+t^3/3!+..+t^n/n!=鈭玔1/t^(1/2)+t^(1/2)+t^(3/2)/2!...
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細鑻ユ湁甯姪锛岃閲囩撼銆
绛旓細鈭(0,+鈭)[(x^2)e^(-2x)]dx 鐢鍒嗛儴绉垎娉 = -1/2 * 鈭玔(x^2)]de^(-2x) 鍏堢渷鐣 (0,+鈭)= -1/2 * [(x^2)]e^(-2x) - 鈭(e^(-2x) dx^2 ]= -1/2 * [(x^2)]e^(-2x) - 鈭2x e^(-2x) dx ]= -1/2 * [(x^2)]e^(-2x...
绛旓細璁＄畻杩囩▼濡備笅锛鈭玿^2e^(-x)dx =鈭玿^2e^(-x)(-1)d(-x)=-鈭玿^2de^(-x)=-x^2e^(-x)+鈭玡^(-x)dx^2 =-x^2e^(-x)+鈭玡^(-x)2xdx =-x^2e^(-x)-2鈭玿de^(-x)=-x^2e^(-x)-2xe^(-x)+2鈭玡^(-x)(-1)d(-x)=-x^2e^(-x)-2xe^(-x)-2鈭玠e^(-x)=...
绛旓細鈭玿²e^xdx=x²e^x-2xe^x+2e^x+C銆侰涓哄父鏁般傗埆x²e^xdx =鈭玿²d(e^x)=x²e^x-鈭玡^xd(x²)=x²e^x-鈭2xd(e^x)=x²e^x-2xe^x+鈭2d(e^x)=x²e^x-2xe^x+2e^x+C ...
绛旓細鏂规硶濡備笅鍥炬墍绀猴紝璇蜂綔鍙傝冿紝绁濆涔犳剦蹇細
绛旓細鍒嗛儴绉垎娉銆傚師寮 = 鈭1/2*x^2d(e^(2x)) = 1/2*x^2*e^(2x) - 鈭玿*e^(2x)dx = 1/2*x^2*e^(2x) - 鈭1/2*x d(e^(2x))= 1/2*x^2*e^(2x) - 1/2*x*e^(2x) + 1/2*鈭玡^(2x)dx = 1/2*x^2*e^(2x) - 1/2*x*e^(2x) + 1/4*e^(2x) + C...
绛旓細瑙佸浘
绛旓細鈭玿²e^(2x)dx =-1/2鈭玿de^(-2x)=-1/2*xe^(-2x)+1/2鈭玡^(-2x)dx =-1/2*xe^(-2x)-1/4*e^(-2x)+C2 =1/2鈭玿²e^x²dx²=1/2鈭玿²de^x²=1/2x²e^(x²)-1/2鈭玡^x²dx²=1/2x²e^(x²)...
绛旓細dx)*(鈭 e^(-y^2)dy) = 鈭 鈭 e^(-(x^2+y^2))dx dy 灏嗕笂杩绉垎鍖栧埌鏋佸潗鏍囦腑 鈭 x^2+y^2=r^2 鈭 鈭 e^(-(x^2+y^2))dx dy = 鈭 鈭 r e^(-r^2)dr d胃 r浠0鍒版鏃犵┓锛屛镐粠0鍒2蟺 = 鈭 1/2 d胃 胃浠0鍒2蟺= 蟺 鈭碆锛濃垰蟺 鈭碆/2 = 鈭毾 /2 ...
