Question

这三题怎么做求大神解释

Answer 1

约定:∫[a,b]表示[a,b]定积
(1)原式=4∫[0,1]dy∫[0,1](x^2+y^2)dx
∫[0,1](x^2+y^2)dx=(x^3/3+y^2x)|[0,1]=y^2+1/3
原式=4∫[0,1](y^2+1/3)dy
=4(y^3/3+y/3)|[0,1]
=8/3
(3)原式=∫[0,1]dx∫[0,1-x](y+2x)dy
∫[0,1-x](y+2x)dy=(y^2/2+2xy)|[0,1-x]=(-3/2)x^2+x+(1/2)
原式=∫[0,1]((-3/2)x^2+x+(1/2))dx
=(-1/2)x^3+x^2/2+(x/2)|[0,1]
=1/2
(4)原式=∫[0,1]dx∫[x^2,x](x^2y)dy
∫[x^2,x](x^2y)dy=(1/2)(x^2y^2)|[x^2,x]=(1/2)(x^4-x^6)
原式=∫[0,1]((1/2)(x^4-x^6))dx
=(1/10)x^5-(1/14)x^7|[0,1]
=1/35

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