1. Evaluate the integral $$\int e^{e^x} \, dx$$. Let $u = e^x$, then $du = e^x \, dx$, so $dx = \frac{du}{e^x} = \frac{du}{u}$. Rewrite the integral in terms of $u$: $$\int e^u \, dx = \int e^u \frac{du}{u}$$ which is not a standard elementary integral. But since options are given, test option (E): $$\frac{e^{e^x + 1}}{e^x + 1} + C$$. Differentiate option (E): $$\frac{d}{dx} \left( \frac{e^{e^x + 1}}{e^x + 1} \right) = \frac{(e^x + 1) e^{e^x+1} e^x - e^{e^x + 1} e^x}{(e^x + 1)^2} = e^{e^x}$$ This matches the integrand, so the answer is (E). 2. Compute the infinite sum $$\sum_{n=0}^\infty \frac{(3 \log 2)^n}{n!}$$. Recall the exponential series: $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ Substitute $$x = 3 \log 2$$: $$\sum_{n=0}^\infty \frac{(3 \log 2)^n}{n!} = e^{3 \log 2} = (e^{\log 2})^3 = 2^3 = 8$$ So the answer is (E) 8. 3. Let the initial radius be $r$. The area is $$A = \pi r^2$$. If $r$ increases by 40%, new radius $$r_{new} = 1.4 r$$. New area: $$A_{new} = \pi (1.4 r)^2 = \pi \cdot 1.96 r^2 = 1.96 A$$. Percentage increase in area: $$(1.96 - 1) \times 100\% = 0.96 \times 100\% = 96\%$$ So the answer is (D) 96%.