1.∫∫dxdy=∫-ln5→ln5dy ∫0→e^ydx
= ∫-ln5→ln5 e^y dy
=5-1/5=24/5
2.∫∫dxdy=∫0→√2 dy ∫y/2→y dx+∫0→√2 dy ∫y/2→2/y dx
=∫y/2→y y/2 dy +∫y/2→2 (2/y-y/2)dy
=y^2/4|0→√2 +[2lny-y^2/4]√2→2
=2+2ln2-ln2+1-1/2
=5/2+ln2
3.题有问题
1.∫∫dxdy=∫-ln5→ln5dy ∫0→e^ydx
= ∫-ln5→ln5 e^y dy
=5-1/5=24/5
2.∫∫dxdy=∫0→√2 dy ∫y/2→y dx+∫0→√2 dy ∫y/2→2/y dx
=∫y/2→y y/2 dy +∫y/2→2 (2/y-y/2)dy
=y^2/4|0→√2 +[2lny-y^2/4]√2→2
=2+2ln2-ln2+1-1/2
=5/2+ln2
3.题有问题