0 ≤ x ≤ 4
x/2 ≤ y ≤ 2
или
0≤ y ≤ 2
0≤ x ≤ 2y
Выбираем второй вариант. Поэтому
[m]∫∫_{D}y^2\cdot \frac{e^{xy}}{8}dxdy =∫_{0} ^{2}\frac{1}{8}y^2( ∫ ^{2y}_{0}e^{xy}dx)dy=[/m]
так как[m] (xy)`_{x}=y[/m] ⇒
[m] =∫_{0} ^{2}\frac{1}{8}y^2( ∫ ^{2y}_{0}\frac{1}{y}e^{xy}ydx))dy=[/m]
[m] =∫_{0} ^{2} \frac{1}{8}y ( e^{xy})|^{x=2y}_{x=0}dy=[/m]
[m] =∫_{0} ^{2} \frac{1}{8}y ( e^{2y^2}-e^{0})dy=[/m]
[m] =∫_{0} ^{2} \frac{1}{8}y ( e^{2y^2})dy-∫_{0} ^{2} \frac{1}{8}ydy=[/m]
[m] =∫_{0} ^{2} \frac{1}{2} ( e^{y^2})4ydy-∫_{0} ^{2} \frac{1}{8}ydy=[/m]
[m]=(\frac{1}{2} e^{2y^2})|_{0} ^{2}-(\frac{1}{8}\frac{y^2}{2})|_{0} ^{2} =[/m]