Volume of Solids of Revolution

$f(x) = \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{e^x})^{n - j} =$ $\lim_{n \rightarrow \infty} \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^n (\dfrac{1}{e^x})^{j} =$ $\sum_{j = 0}^{\infty} (\dfrac{1}{e^{x + 1}})^j = \dfrac{e^{x + 1}}{e^{x + 1} - 1}$ on $x > -1$ .

Let $g(x) = \dfrac{1}{ax + b}$ .

$g(0) = f(0) = \dfrac{e}{e - 1} \implies b = \dfrac{e - 1}{e}$ and $g(1) = f(1) = \dfrac{e^2}{e^2 - 1} \implies a = \dfrac{e - 1}{e^2} \implies g(x) = \dfrac{e^2}{(e - 1)(x + e)}$ .

Let $I_{1} = \pi\int_{0}^{1} (f(x))^2 dx = \pi\int_{0}^{1} (\dfrac{e^{x + 1}}{e^{x + 1} - 1})^2 dx$ .

Let $u = e^{x + 1} \implies dx = \dfrac{1}{u} du \implies$

$I_{1} = \pi\int_{e}^{e^2} \dfrac{u}{(u - 1)^2} du$

Let $w = u - 1 \implies dw = du \implies I_{1} = \pi\int_{e - 1}^{e^2 - 1} \dfrac{w + 1}{w^2} dw = \pi\int_{e - 1}^{e^2 - 1} \dfrac{1}{w} + w^{-2} dw = \pi(\ln(w) - \dfrac{1}{w}|_{e - 1}^{e^2 - 1}) =$

$\pi(\ln(e + 1) + \dfrac{e}{e^2 - 1})$

$I_{2} = \pi\int_{0}^{1} (g(x))^2 dx = \dfrac{e^4\pi}{(e - 1)^2}\int_{0}^{1} (x + e)^{-2} dx =$

$-\dfrac{e^4\pi}{(e - 1)^2}(\dfrac{1}{x + e})|_{0}^{1} =$ $\dfrac{\pi e^4}{(e^2 - 1)(e - 1)e}$

$\implies$ the volume $V = \pi\int_{0}^{1} (g^2(x) - f^2(x)) dx = \pi(\dfrac{e(e^2 - e + 1)}{(e^2 - 1)(e - 1)} - \ln(e + 1)) \approx \boxed{0.2854552}$