Question

calcule a integral dupla dada por

Answer 1

$\large\boxed{\begin{array}{l}\sf 1\leqslant y\leqslant 3\\\sf 1-y\leqslant x\leqslant y-1\\\displaystyle\sf\int\int_R (x+y)d_A=\int_1^3\int_{1-y}^{y-1}(x+y)dxdy\\\displaystyle\sf\int_1^3\bigg[\dfrac{x^2}{2}+xy\bigg]_{1-y}^{y-1}dy\\\\\displaystyle\sf\int_1^3\bigg[\dfrac{(y-1)^2}{2}+(y-1)y-\bigg\{\dfrac{(1-y)^2}{2}+(1-y)y\bigg\}\bigg]dy\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int_1^3\bigg[\dfrac{y^2-2y+1}{2}+y^2-y-\dfrac{1-2y+y^2}{2}-y+y^2\bigg]dy\\\\\displaystyle\sf\int_1^3\bigg[\dfrac{y^2-2y+1}{2}+y^2-y+\dfrac{-1+2y-y^2}{2}-y+y^2\bigg]dy\\\\\displaystyle\sf\int_1^3\bigg[\dfrac{\diagup\!\!\!y^2-2y+\diagup\!\!\!1+2y^2-2y-\diagup\!\!\!1+\diagdown\!\!\!\!\!\!2y-\diagup\!\!\!y^2-\diagdown\!\!\!\!\!2y+2y^2}{2}\bigg]dy\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int_1^3\bigg[\dfrac{4y^2-4y}{2}\bigg]dy\\\\\displaystyle\sf\int_1^3(2y^2-2y)dy=\bigg[\dfrac{2}{3}y^3-y^2\bigg]_1^3\\\\\sf \dfrac{2}{3}\cdot3^3-3^2-\bigg(\dfrac{2}{3}\cdot1^3-1^2\bigg)\\\\\sf\dfrac{54}{3}-9-\dfrac{2}{3}+1=\dfrac{54-27-2+3}{3}=\dfrac{28}{3}~u\bullet a\end{array}}$