Đáp án:
$3)\quad I = \displaystyle\iint\limits_D(x-y)dxdy=-\dfrac{16}{3}$
$4)\quad I = \displaystyle\int\limits_0^3dy\displaystyle\int\limits_0^2(2x^2 - y)dx=-\dfrac{5}{2}$
Giải thích các bước giải:
\(\begin{array}{l}
3)\quad I = \displaystyle\iint\limits_D(x-y)dxdy\quad \text{với}\quad D:\begin{cases}x = 0\\y = x\\x+y=4\end{cases}\\
\text{Phương trình hoành độ giao điểm:}\\
\quad x = 4 - x \Leftrightarrow x = 2\\
\text{Miền D được biểu diễn:}\\
D = \{(x,y):0\leqslant x \leqslant 2;\ x \leqslant y \leqslant 4 - x\}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^2dx\displaystyle\int\limits_x^{4-x}(x-y)dy\\
\Leftrightarrow I = \displaystyle\int\limits_0^2\left[\left(xy - \dfrac{y^2}{2}\right)\Bigg|_x^{4-x}\right]\\
\Leftrightarrow I = \displaystyle\int\limits_0^2(-2x^2 + 8x - 8)dx\\
\Leftrightarrow I = \left(-\dfrac{2x^3}{3} + 4x^2 - 8x\right)\Bigg|_0^2\\
\Leftrightarrow I = -\dfrac{16}{3}\\
4)\quad I = \displaystyle\int\limits_0^3dy\displaystyle\int\limits_0^2(2x^2 - y)dx\\
\Leftrightarrow I = \displaystyle\int\limits_0^3\left[\left(\dfrac{2x^3}{3} - xy\right)\Bigg|_0^1\right]dy\\
\Leftrightarrow I = \displaystyle\int\limits_0^3\left(\dfrac{2}{3} - y\right)dy\\
\Leftrightarrow I = \left(\dfrac{2y}{3} - \dfrac{y^2}{2}\right)\Bigg|_0^3\\
\Leftrightarrow I = -\dfrac{5}{2}
\end{array}\)
