Answer
$$\int^{\ln9}_0e^{\theta}(e^\theta-1)^{1/2}d\theta=\frac{32\sqrt2}{3}$$
Work Step by Step
$$A=\int^{\ln9}_0e^{\theta}(e^\theta-1)^{1/2}d\theta$$
We set $u=e^\theta-1$, which means $$du=e^\theta d\theta$$
For $\theta=\ln9$, we have $$u=e^{\ln9}-1=9-1=8$$
For $\theta=0$, we have $$u=e^0-1=1-1=0$$
Therefore, $$A=\int^8_0u^{1/2}du$$ $$A=\frac{2u^{3/2}}{3}\Big]^8_0$$ $$A=\frac{2}{3}(8^{3/2}-0^{3/2})$$ $$A=\frac{2\times16\sqrt2}{3}$$ $$A=\frac{32\sqrt2}{3}$$
