关于2重积分！求教
\u6c422\u91cd\u79ef\u5206\u95ee\u9898\u6211\u4eec\u6765\u4ea4\u6d41\u4e00\u4e0b\u5427\uff01\u6211\u7684\u89e3\u6cd5\u5982\u4e0b
4\u6761\u76f4\u7ebf\u6784\u6210\u7684\u83f1\u5f62~~
x+y\u22642 -x+y\u22642 x-y\u22642 -x-y\u22642
\u7c7b\u4f3c\u7684\u7edd\u5bf9\u503c\u65b9\u7a0b\u4e00\u6837\u7684\u601d\u8def\uff1a\u5206\u522b\u8ba8\u8bba\u7edd\u5bf9\u503c\u5185\u7684\u90e8\u5206\u5927\u4e8e0\u548c\u5c0f\u4e8e0\u7684\u60c5\u51b5\uff0c\u53bb\u7edd\u5bf9\u503c
=(1/3)∫<0,1>y³e^(-y²)dy
=(1/6)∫<0,1>y²e^(-y²)d(y²)
=(1/6)(-1/e+∫<0,1>e^(-y²)d(y²)) (应用分部积分法)
=(1/6)(-1/e+(1-1/e))
=(1/6)(1-2/e)。
绛旓細鏄剧劧瀵箈绉垎锛屽氨鎶妝浣滀负甯告暟 鈭玿y dx=y鈭玿dx=y *x²/2 鍐嶄唬鍏鐨勪笂涓嬮檺2鍜寉 鍗冲緱鍒皔 *(4-y²)/2=2y -y³/2 鍐嶅y绉垎寰楀埌y² -y^4 /8 浠ｅ叆涓婁笅闄2鍜1 绛変簬(4-2) -(1-1/8)=9/8
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