Question

Tính tích phân 2 lớp:
∫∫(1+3x+2y)dxdy

Answer 1

`\\\int _(\:)^(\:)\int _(\:)^(\:)\:1+3x+2y\:dxdy=Cy+(3x^2)/(2)y+xy+xy^2+C`

Explanation:

`\\\int _(\:)^(\:)\int _(\:)^(\:)\:1+3x+2y\:dxdy`

`\int _(\:)^(\:)\left(1+3x+2y\right)\:dx\:=x+2yx\:+(3x^2)/(2)+C`

`=\int \left(x+2yx+(3x^2)/(2)+C\right)dy`

`=Cy+(3x^2)/(2)y+xy+xy^2+C`

Answer 2

We know the formula

`\boxed{\displaystyle\int x^ndx=(x^(n+1))/(n+1)+c}`

• c is constant
• Here c=1

`\\ \displaystyle\longmapsto \int (1+3x+2y)dx`

`\\ \displaystyle\longmapsto (3x^(1+1))/(1+1)+(2y^(1+1))/(1+1)+1`

`\\ \sf\longmapsto (3x^2)/(2)+(2y^2)/(2)+1`

`\\ \sf\longmapsto (3x^2)/(2)+y^2+1`

`\\ \sf\longmapsto (3x^2+2y^2+2)/(2)`