Question

$\int\limits^2_1$ $\int\limits^3_0$ +(2xy-x²+2y)dydx

Answer 1

$\int_{1}^{2}\int_{0}^{3}{(2xy-x^{2}+2y)}\,dy\,dx\\ \\ \\ =\int_{1}^{2}{\left[\int_{0}^{3}{(2xy-x^{2}+2y)}\,dy \right ]}\,dx\\ \\ \\ =\int_{1}^{2}{\left[\dfrac{2xy^{2}}{2}-x^{2}y+\dfrac{2y^{2}}{2} \right ]_{y=0}^{y=3}}\,dx\\ \\ \\ =\int_{1}^{2}{\left[xy^{2}-x^{2}y+y^{2} \right ]_{y=0}^{y=3}}\,dx\\ \\ \\ =\int_{1}^{2}{\left[(x(3)^{2}-x^{2}(3)+(3)^{2})-(x(0)^{2}-x^{2}(0)+(0)^{2}) \right ]}\,dx\\ \\ \\ =\int_{1}^{2}{\left[9x-3x^{2}+9-0 \right ]}\,dx\\ \\ \\ =\int_{1}^{2}{\left[9x-3x^{2}+9 \right ]}\,dx\\ \\ \\ =\left[\dfrac{9x^{2}}{2}-\dfrac{3x^{3}}{3}+9x \right ]_{x=1}^{x=2}\\ \\ \\ =\left[\dfrac{9x^{2}}{2}-x^{3}+9x \right ]_{x=1}^{x=2}\\ \\ \\ =\dfrac{9(2)^{2}}{2}-(2)^{3}+9(2)-\left(\dfrac{9(1)^{2}}{2}-(1)^{3}+9(1)\right)$


$=18-8+18-\left(\dfrac{9}{2}-1+9\right)\\ \\ \\ =28-\left(\dfrac{9-2+18}{2}\right)\\ \\ \\ =28-\dfrac{25}{2}\\ \\ \\ =\dfrac{56-25}{2}\\ \\ \\ =\dfrac{31}{2}$