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The antiderivative of $\sqrt{u}$ is found by rewriting the square root as a fractional power, $u^{1/2}$, and then applying the power rule for integration. The power rule states that $\int u^n du = \frac{u^{n+1}}{n+1}$.

Conditions: The integrand is $\sqrt{u}$, which is equivalent to $u^{1/2}$.; The power rule for integration is applicable.; $u \ge 0$ for the real-valued square root form.

The antiderivative of $\sqrt{u}$ is found by rewriting the square root as a fractional power, $u^{1/2}$, and then applying the power rule for integration. The power rule states that $\int u^n du = \frac{u^{n+1}}{n+1}$.

Conditions: The integrand is $\sqrt{u}$, which is equivalent to $u^{1/2}$.; The power rule for integration is applicable.; $u \ge 0$ for the real-valued square root form.

The antiderivative of $\sqrt{x}$ is $\frac{2}{3}x^{3/2}$. This is found by rewriting $\sqrt{x}$ as $x^{1/2}$ and applying the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$.

Conditions: The function is $\sqrt{x}$.; The power rule for integration is applied.; The exponent $n = 1/2$.

The antiderivative of $\sqrt{x}$ is $\frac{2}{3}x^{3/2}$. This is found by rewriting $\sqrt{x}$ as $x^{1/2}$ and applying the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$.

Conditions: The function is $\sqrt{x}$.; The power rule for integration is applied.; The exponent $n = 1/2$.

The final numerical value of the area moment of inertia $I_y$ for the specific shaded region about the centroidal y-axis is approximately $0.762 \text{ m}^4$. This value is obtained by evaluating the definite integral $\int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$, which simplifies to $\frac{2^4}{3} - \sqrt{2} \left(\frac{2}{7}\right) 2^{3.5}$.

Conditions: The region is bounded by $y = 2$ (top) and $y = \sqrt{2} x^{1/2}$ (bottom) from $x=0$ to $x=2$.; The axis is the centroidal y-axis.; Units are in meters.

The final numerical value of the area moment of inertia $I_y$ for the specific shaded region about the centroidal y-axis is approximately $0.762 \text{ m}^4$. This value is obtained by evaluating the definite integral $\int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$, which simplifies to $\frac{2^4}{3} - \sqrt{2} \left(\frac{2}{7}\right) 2^{3.5}$.

Conditions: The region is bounded by $y = 2$ (top) and $y = \sqrt{2} x^{1/2}$ (bottom) from $x=0$ to $x=2$.; The axis is the centroidal y-axis.; Units are in meters.

The final value is $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This is obtained by evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at the converted bounds $x=3$ and $x=2$ and subtracting the results.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

The final value is $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This is obtained by evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at the converted bounds $x=3$ and $x=2$ and subtracting the results.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.