Domain A is bounded by curve x² = 4y, ordinate x = 2 and x-axis. The value of{\iint }_A^{}ydxd\mathrm{y \; is: }

The correct option is 1/5.

Let us evaluate the double integral step-by-step.

First, we identify the region of integration, $A$.
The region is bounded by:
1. The parabola $x^2=4y$, which can be written as $y=\frac{x^2}{4}$.
2. The vertical line (ordinate) $x=2$.
3. The x-axis, which is the line $y=0$.

From these boundaries, we determine the limits of integration for the double integral:
For a given $x$ in the interval $[0,2]$, the variable $y$ varies from the x-axis ($y=0$) up to the curve ($y=\frac{x^2}{4}$).

Thus, we set up the double integral as follows:

$${\iint }_A^{}ydxdy={\int }_0^2\left[{\int }_0^{\frac{x^2}{4}}ydy\right]dx$$

Now, we perform the inner integration with respect to $y$:

$${\int }_0^{\frac{x^2}{4}}ydy={\left[\frac{y^2}{2}\right]}_0^{\frac{x^2}{4}}$$

Substituting the upper and lower limits:

$$=\frac{1}{2}{\left(\frac{x^2}{4}\right)}^2-0=\frac{1}{2}·\frac{x^4}{16}=\frac{x^4}{32}$$

Now, substitute this result back into the outer integral and integrate with respect to $x$ from $0$ to $2$:

$${\int }_0^2\frac{x^4}{32}dx=\frac{1}{32}{\left[\frac{x^5}{5}\right]}_0^2$$

Evaluating this at the limits:

$$=\frac{1}{32}(\frac{2^5}{5}-0)$$

Since $2^5=32$, we have:

$$=\frac{1}{32}·\frac{32}{5}=\frac{1}{5}$$

Thus, the value of the double integral is indeed 1/5.