Question

$\iint \left(\frac{x}{y}\right)$over the triangle with vertices $\left(0,0\right)\mathbf{,}\left(1,1\right)\mathbf{,}\left(1,2\right)$

Short answer

The required solution is$\frac{In2}{2}$.

Step-by-step solution

 Step 1: Definition of double integral

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

Drawing the area bounded by the curve

The area is bounded by the triangle with vertices $\left(0,0\right),\left(1,1\right),\left(1,2\right)$..

Calculation of the area under the curve

Integrating over the variable y

$$\begin{gathered} {\iint }_A\frac{x}{y}dxdy={\int }_{x=0}^1\left[{\int }_x^{2x}\frac{dy}{y}\right]xdx \\ ={\int }_0^1\left(I{\overline{)ny}}_x^{2x}\right)xdx \\ ={\int }_0^1\left(In\left(2x\right)-Inx\right)xdx \end{gathered}$$

Further calculation

Integrating over the variable x

$$\begin{gathered} {\int }_0^1\left(In\left(2x\right)-Inx\right)xdx={\int }_0^1\left(In2+Inx-Inx\right)xdx \\ ={\overline{)\frac{In2}{2}{x}^2}}_0^1 \\ =\frac{In2}{2} \end{gathered}$$

Therefore, the value is$\frac{In2}{2}$