數學雙重積分問題！

Question

∫(上限是1;下限是0) ∫(上限是1;下限是0) 下面三個都一樣！
1.∫∫(2x^2+3y)*10^6 dxdy
2.∫∫1000+2x*10^3 dxdy
3.∫∫(xy+2x^2)*10^6 dxdy
請問一下這三題要怎麼積分呢？
可以教我一下嗎？過程越詳細越好！謝謝！

Answer 1

1. ∫01∫01 (2x²+3y)*106 dxdy= 106∫01∫01 (2x²+3y) dxdy= 106∫01 [2(1/3)x³+3y(x))]x=01 dy= 106∫01 [(2/3)x³+3xy]x=01 dy= 106∫01 ((2/3)(1)³+3(1)y)-((2/3)(0)³+3(0)y)) dy= 106∫01 (2/3)+3y dy= 106 [(2/3)y+3(1/2)y²]01 = 106 [(2/3)y+(3/2)y²]01 =106 (((2/3)(1)+(3/2)(1)²)- ((2/3)(0)+(3/2)(0)²))=106 ((2/3)+(3/2))=106 ((4/6)+(9/6))=106 (13/6)=13000000/6=6500000/32. ∫01∫01 1000+2x*10³ dxdy=∫01∫01 1000+2x*1000 dxdy= 1000∫01∫01 1+2x dxdy= 1000∫01 [x+x²] 01 dy= 1000∫01 ((1+1²)-(0+0²)) dy= 1000∫01 2 dy= 2000[y]01 = 2000(1-0)= 20003.∫01∫01 (xy+2x²)*106 dxdy= 106∫01∫01 (xy+2x²) dxdy= 106∫01∫01 (xy+2x²) dxdy= 106∫01 [(1/2)x²y+2(1/3)x³]01 dy= 106∫01 [(1/2)x²y+(2/3)x³]x=01 dy= 106∫01 ((1/2)(1²)y+(2/3)(1³))-((1/2)(0²)y+(2/3)(0³)) dy= 106∫01 ((1/2)y+(2/3)) dy= 106 [(1/2)(1/2)y²+(2/3)y]01 = 106 [(1/4)y²+(2/3)y]01 = 106 ((1/4)(1²)+(2/3)(1))-((1/4)(0²)+(2/3)(0)) = 106 ((1/4)+(2/3))=106 ((3/12)+(8/12))=106 (11/12)= 11000000/12= 2750000/3