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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 $\frac{2}{3}x^{3/2}$.

The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.

The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.

To set up the integral for the area moment of inertia about the y-axis ($I_y$) using a vertical strip, express the differential area $dA$ as the product of the strip's height (upper boundary minus lower boundary) and its width $dx$. Substitute this into the definition $I_y = \int x^2 dA$ and determine the integration limits based on the horizontal extent of the region bounded by functions of $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.

To set up the integral for the area moment of inertia about the y-axis ($I_y$) using a vertical strip, express the differential area $dA$ as the product of the strip's height (upper boundary minus lower boundary) and its width $dx$. Substitute this into the definition $I_y = \int x^2 dA$ and determine the integration limits based on the horizontal extent of the region bounded by functions of $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.