Question

f(x,y)=xye^(x^2+y^2)+1在区域D:a<=x<=b,c<=y<=d上的二重积分

Answer 1

∫∫f(x,y)dxdy
=∫(a,b)dx ∫(c,d) xye^(x^2+y^2)+1dy
=∫(a,b)dx [d-c+x/2*∫(c,d) e^(x^2+y^2) d(x^2+y^2)]
=∫(a,b)dx [d-c+x/2*e^(x^2+y^2) |(c,d)]
=∫(a,b) [d-c+x/2*[e^(x^2+d^2)-e^(x^2+c^2)] dx
=(d-c)(b-a)+∫(a,b) x/2*e^(x^2+d^2)dx-∫(a,b) x/2*e^(x^2+c^2)dx
=(d-c)(b-a)+1/4*[e^(x^2+d^2)-e^(x^2+c^2)]|(a,b)
=(d-c)(b-a)+1/4*[e^(b^2+d^2-e^(b^2+c^2)-e^(a^2+d^2)+e^(a^2+c^2)]