c) $$ \int_{0}^{1} \int_{\sqrt{y}}^{2-\sqrt{y}} x y d x d y=\frac{1}{5} $$

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Consideramos la integral

$$
I=\int_{0}^{1} \int_{\sqrt{y}}^{2-\sqrt{y}} x\, y\, dx\, dy.
$$

**Paso 1.** Integramos respecto a $$x$$:

$$
\int_{\sqrt{y}}^{2-\sqrt{y}} x\, dx=\left.\frac{x^2}{2}\right|_{x=\sqrt{y}}^{x=2-\sqrt{y}}=\frac{(2-\sqrt{y})^2}{2}-\frac{y}{2}.
$$

Observamos que

$$
(2-\sqrt{y})^2=4-4\sqrt{y}+y,
$$

por lo que

$$
(2-\sqrt{y})^2 - y=4-4\sqrt{y}.
$$

Entonces, la integral se transforma en

$$
I=\int_{0}^{1} y\ \frac{4-4\sqrt{y}}{2}\, dy = \int_{0}^{1} y\,(2-2\sqrt{y})\, dy.
$$

**Paso 2.** Factorizamos y separamos la integral:

$$
I=2\int_{0}^{1} y\,(1-\sqrt{y})\, dy = 2\left[\int_{0}^{1} y\, dy-\int_{0}^{1} y\sqrt{y}\, dy\right].
$$

Notamos que $$y\sqrt{y}=y^{3/2}$$.

**Paso 3.** Calculamos las integrales individuales:

$$
\int_{0}^{1} y\, dy=\left.\frac{y^2}{2}\right|_{0}^{1}=\frac{1}{2},
$$

$$
\int_{0}^{1} y^{3/2}\, dy=\left.\frac{y^{5/2}}{\frac{5}{2}}\right|_{0}^{1}=\frac{2}{5}.
$$

**Paso 4.** Sustituimos los resultados:

$$
I=2\left(\frac{1}{2}-\frac{2}{5}\right)=2\left(\frac{5}{10}-\frac{4}{10}\right)=2\left(\frac{1}{10}\right)=\frac{2}{10}=\frac{1}{5}.
$$

Por lo tanto, se verifica que

$$
\int_{0}^{1} \int_{\sqrt{y}}^{2-\sqrt{y}} x\, y\, dx\, dy=\frac{1}{5}.
$$
La integral $$ \int_{0}^{1} \int_{\sqrt{y}}^{2-\sqrt{y}} x\, y\, dx\, dy $$ es igual a $$ \frac{1}{5} $$.

c) \( \int_{0}^{1} \int_{\sqrt{y}}^{2-\sqrt{y}} x y d x d y=\frac{1}{5} \)

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