Question

tính tích phân 2 lớp:
∫∫(1+3x+2y)dxdy

Answer 1

We know the formula

$\boxed{\displaystyle\int x^ndx=\dfrac{x^{n+1}}{n+1}+c}$

• c is constant
• Here c=1

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

$\\ \displaystyle\longmapsto \dfrac{3x^{1+1}}{1+1}+\dfrac{2y^{1+1}}{1+1}+1$

$\\ \sf\longmapsto \dfrac{3x^2}{2}+\dfrac{2y^2}{2}+1$

$\\ \sf\longmapsto \dfrac{3x^2}{2}+y^2+1$

$\\ \sf\longmapsto \dfrac{3x^2+2y^2+2}{2}$

Answer 2

Answer:

$\\\int _{\:}^{\:}\int _{\:}^{\:}\:1+3x+2y\:dxdy=Cy+\frac{3x^2}{2}y+xy+xy^2+C$

Step-by-step explanation:

$\\\int _{\:}^{\:}\int _{\:}^{\:}\:1+3x+2y\:dxdy$

$\int _{\:}^{\:}\left(1+3x+2y\right)\:dx\:=x+2yx\:+\frac{3x^2}{2}+C$

$=\int \left(x+2yx+\frac{3x^2}{2}+C\right)dy$

$=Cy+\frac{3x^2}{2}y+xy+xy^2+C$