计算lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
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计算lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
计算lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
计算lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
lim(r->0)[1/πr²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
由积分中值定理,在D内存在点(a,b),使:
∫∫e^(x²-y²)cos(x+y)dxdy=πr²e^(a²-b²)cos(a+b)
所以:lim(r->0)[1/πr²]∫∫e^(x²-y²)cos(x+y)dxdy
=lim(r->0)[1/πr²]πr²e^(a²-b²)cos(a+b)
=lim(r->0)e^(a²-b²)cos(a+b)
=1
由积分中值定理,有
(a,b)∈D,使得
lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy
=lim(r->0)[1/πr²]*πr²*e^(a²-b²)cos(a+b)
=lim(r->0)e^(a²-b²)cos(a+b)
=e^0*cos0
=1
因为r->0,得
a->0,b->0
即
a²-b²->0
a+b->0
由积分中值定理,有
(a,b)∈D,使得
lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy
=lim(r->0)[1/πr²]*πr²*e^(a²-b²)cos(a+b)
=lim(r->0)e^(a²-b²)cos(a+b)
=e^0*c...
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由积分中值定理,有
(a,b)∈D,使得
lim(r->0)[1/∏r²]∫∫e^(x²-y²)cos(x+y)dxdy
=lim(r->0)[1/πr²]*πr²*e^(a²-b²)cos(a+b)
=lim(r->0)e^(a²-b²)cos(a+b)
=e^0*cos0
=1
因为r->0,得
a->0,b->0
即
a²-b²->0
a+b->0lim(r->0)[1/πr²]∫∫e^(x²-y²)cos(x+y)dxdy,其中D为x²+y²≤r²
由积分中值定理,在D内存在点(a,b),使:
∫∫e^(x²-y²)cos(x+y)dxdy=πr²e^(a²-b²)cos(a+b)
所以:lim(r->0)[1/πr²]∫∫e^(x²-y²)cos(x+y)dxdy
=lim(r->0)[1/πr²]πr²e^(a²-b²)cos(a+b)
=lim(r->0)e^(a²-b²)cos(a+b)
=1
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