f(x)在[-1,1]连续,证明 ∫∫f(x+y)dxdy=∫[-1,1]f(t)dt,D:|x|+|y|≤1.
1个回答
以y轴分为界线,将区域为分两部分,左边为D1,右边为D2
当积分区域为D1时：
∫∫f(x+y)dxdy
=∫[-1---->0] dx ∫[-1-x---->1+x] f(x+y)dy
对内层积分换元
令x+y=t,则dy=dt,t：-1--->2x+1
=∫[-1---->0] dx ∫[-1---->2x+1] f(t)dt
交换积分次序：
=∫[-1---->1] dt ∫[(t-1)/2---->0] f(t) dx
=(1/2)∫[-1---->1] (1-t)f(t) dt
当积分区域为D2时：
∫∫f(x+y)dxdy
=∫[0---->1] dx ∫[x-1---->1-x] f(x+y)dy
对内层积分换元
令x+y=t,则dy=dt,t：2x-1--->1
=∫[0---->1] dx ∫[2x-1---->1] f(t)dt
交换积分次序：
=∫[-1---->1] dt ∫[0---->(t+1)/2] f(t) dx
=(1/2)∫[-1---->1] (t+1)f(t) dt
综上：左边=(1/2)∫[-1---->1] (1-t)f(t) dt+(1/2)∫[-1---->1] (t+1)f(t) dt
=∫[-1---->1] f(t) dt=右边
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