Question

$\mathbf{\int }\mathbf{\int }\mathbf{2}\mathit{x}\mathit{y}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$over the triangle with vertices role="math" localid="1658828431182" $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{(}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$.

Short answer

The required solution is$\frac{7}{4}$.

Step-by-step solution

Definition of double integral

The double integral of $\mathit{f}\left(x,y\right)$over the area Ain the$\left(x,y\right)$plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathit{f}\left(x,y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$.

Drawing the area bounded by the curve

The triangle with vertices (0,0),(2,1) and (3,0).

Integration over the bounded curve

Now the total integral region can be divided into two parts and then adding them can be found in the final area.

$$\begin{gathered} \int {\int }_{A}ydydx=\mathrm{I} \\ ={\mathrm{I}}_1+{\mathrm{I}}_2 \end{gathered}$$

The area of the left triangle

Calculation of the value, ${\mathrm{I}}_1$:

$$\begin{gathered} {\mathrm{I}}_1={\int }_{\mathrm{X}=0}^2\left[{\int }_{\mathrm{Y}=0}^{\mathrm{X}/2}2\mathrm{ydy}\right]\mathrm{xdx} \\ ={\int }_0^2\left({\overline{){\mathrm{y}}^2}}_0^{\mathrm{x}/2}\right)\mathrm{xdx} \\ =\frac{1}{4}{\int }_0^2{\mathrm{x}}^3\mathrm{dx} \\ ={\overline{)\frac{1}{16}{\mathrm{x}}^4}}_0^2 \\ =1 \end{gathered}$$

The area of the right side triangle

Calculation of the value,${\mathrm{I}}_2$:

$$\begin{gathered} {\mathrm{I}}_2={\int }_{\mathrm{X}=2}^3\left[{\int }_{\mathrm{Y}=0}^{3-\mathrm{X}}2\mathrm{ydy}\right]\mathrm{xdx} \\ ={\int }_2^3{\left(3-\mathrm{x}\right)}^2\mathrm{xdx} \\ ={\overline{)\left(\frac{9}{2}{\mathrm{x}}^2-2{\mathrm{x}}^3+\frac{1}{4}{\mathrm{x}}^4\right)}}_2^3 \\ =\frac{3}{4} \end{gathered}$$

Total area bounded by the triangle

Hence:

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2 \\ =1+\frac{3}{4} \\ =\frac{7}{4} \end{gathered}$$

Therefore, the value is$\frac{7}{4}$.