What is the surface area of the solid created by revolving $f (x) = \frac{e^{x + 1}}{x + 1}$ $f(x)=e^(x+1)/(x+1)$ over $x \in [0, 1]$ $x in [0,1]$ around the x-axis?

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x + 1}}{x + 1}$ $f(x)=e^(x+1)/(x+1)$ over $x \in [0, 1]$ $x in [0,1]$ around the x-axis?

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Answer 1 · 2016-11-28T22:15:19

The area of the solid of revolution around the x-axis of a curve is calculated as:

$π \int_{a}^{b} f^{2} (x) d x$ $pi int_a^b f^2(x)dx$

Explanation:

$V = π \int_{0}^{1} {(\frac{e^{x + 1}}{x + 1})}^{2} d x = π \int_{0}^{1} \frac{e^{2 (x + 1)}}{{(x + 1)}^{2}} d x$ $V= pi int_0^1((e^(x+1))/(x+1))^2dx = pi int_0^1e^(2(x+1))/(x+1)^2dx$

Substitute $t = 2 (x + 1); d x = \frac{d t}{2}$ $t=2(x+1); dx=dt/2$

$V = 2 π \int_{2}^{4} \frac{e^{t}}{t^{2}} d t$ $V=2pi int_2^4e^t/t^2dt$

We can calculate this integral by parts:

$\int \frac{e^{t}}{t^{2}} d t = - \int e^{t} d (t^{- 1}) = - \frac{e^{t}}{t} + \int \frac{e^{t}}{t}$ $int e^t/t^2dt = -int e^t d(t^(-1)) = -e^t/t + int e^t/t$

$\int \frac{e^{t}}{t} d t$ $int e^t/tdt$ is not easily solved, you can find it on manuals:

$\int \frac{e^{t}}{t} d t = log | x | + \infty \sum 0 \frac{x^{n}}{n \cdot n!}$ $int e^t/tdt =log|x| + sum_0^oox^n/(n*n!)$ ,

so

$V = 2 π (- \frac{e^{4}}{4} + \frac{e^{2}}{2} + log 4 - log 2 + \infty \sum 0 \frac{4^{n}}{n \cdot n!} - \infty \sum 0 \frac{2^{n}}{n \cdot n!})$ $V=2pi(-e^4/4+e^2/2+log 4 -log 2 +sum_0^oo4^n/(n*n!)-sum_0^oo2^n/(n*n!))$

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x + 1}}{x + 1}$ $f(x)=e^(x+1)/(x+1)$ over $x \in [0, 1]$ $x in [0,1]$ around the x-axis?

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What is the surface area of the solid created by revolving f(x)=ex+1x+1f(x)=e^(x+1)/(x+1) over x∈[0,1]x in [0,1] around the x-axis?

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Explanation:

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What is the surface area of the solid created by revolving $f (x) = \frac{e^{x + 1}}{x + 1}$ $f(x)=e^(x+1)/(x+1)$ over $x \in [0, 1]$ $x in [0,1]$ around the x-axis?