Đáp án:
\(1)\quad I= \displaystyle\iint\limits_D(x^2y - 2)dxdy = - \dfrac{11}{6}\)
\(2)\quad I= \displaystyle\iint\limits_D(xy +4)dxdy = 72\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad I= \displaystyle\iint\limits_D(x^2y - 2)dxdy\qquad D:\begin{cases}y = x\\y = 2-x\\x=0\end{cases}\\
\text{Phương trình hoành độ giao điểm}\\
\quad x = 2 - x \Leftrightarrow x = 1\\
\text{Miền D:}\ D = \{(x,y):\ 0\leqslant x \leqslant 1;\ x \leqslant y \leqslant 2 - x\}\\
\text{Ta được:}\\
\quad I= \displaystyle\int\limits_0^1dx\displaystyle\int\limits_x^{2-x}(x^2y - 2)dy\\
\Leftrightarrow I = \displaystyle\int\limits_0^1\left[\left(\dfrac{x^2y^2}{2} - 2y\right)\Bigg|_x^{2-x}\right]dx\\
\Leftrightarrow I = \displaystyle\int\limits_0^1(-2x^3 + 2x^2 + 4x - 4)dx\\
\Leftrightarrow I = \left(-\dfrac{x^4}{2} + \dfrac{2x^3}{3} + 2x^2 - 4x\right)\Bigg|_0^1\\
\Leftrightarrow I = -\dfrac{11}{6}\\
2)\quad I = \displaystyle\iint\limits_D(xy + 4)dxdy\qquad D:\begin{cases}y = x+2\\y = 2-x\\x=3\end{cases}\\
\text{Phương trình hoành độ giao điểm}\\
\quad x +2= 2 - x \Leftrightarrow x = 0\\
\text{Miền D:}\ D = \{(x,y):\ 0\leqslant x \leqslant 3;\ 2-x\leqslant y \leqslant x+2\}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^3dx\displaystyle\int\limits_{2-x}^{x+2}(xy + 4)dy\\
\Leftrightarrow I = \displaystyle\int\limits_0^3\left[\left(\dfrac{xy^2}{2} + 4y\right)\Bigg|_{2-x}^{x+2} \right]dx\\
\Leftrightarrow I = \displaystyle\int\limits_0^3(4x^2 + 8x)dx\
\Leftrightarrow I = \left(\dfrac{4x^3}{3} + 4x^2\right)\Bigg|_0^3\\
\Leftrightarrow I = 72
\end{array}\)
