What is the surface area of the solid created by revolving $f (x) = x e^{- x} - x e^{x}, x \in [1, 3]$ $f(x) = xe^-x-xe^(x) , x in [1,3]$ around the x axis?

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What is the surface area of the solid created by revolving $f (x) = x e^{- x} - x e^{x}, x \in [1, 3]$ $f(x) = xe^-x-xe^(x) , x in [1,3]$ around the x axis?

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Answer 1 · 2016-01-13T01:59:26

Determine the sign, then integrate by parts. Area is:

$A = 39.6345$ $A=39.6345$

Explanation:

You have to know whether $f (x)$ $f(x)$ is negative or positive in $[1, 3]$ $[1,3]$ . Therefore:

$x e^{- x} - x e^{x}$ $xe^-x-xe^x$

$x (e^{- x} - e^{x})$ $x(e^-x-e^x)$

To determine a sign, the second factor will be positive when:

$e^{- x} - e^{x} > 0$ $e^-x-e^x>0$

$\frac{1}{e^{x}} - e^{x} > 0$ $1/e^x-e^x>0$

$e^{x} \cdot \frac{1}{e^{x}} - e^{x} \cdot e^{x} > e^{x} \cdot 0$ $e^x*1/e^x-e^x*e^x>e^x*0$

Since $e^{x} > 0$ $e^x>0$ for any $x \in (- \infty, + \infty)$ $x in(-oo,+oo)$ the unequality doesn't change:

$1 - e^{x + x} > 0$ $1-e^(x+x)>0$

$1 - e^{2 x} > 0$ $1-e^(2x)>0$

$e^{2 x} < 1$ $e^(2x)<1$

${ln e}^{2 x} < ln 1$ $lne^(2x) < ln1$

$2 x < 0$ $2x<0$

$x < 0$ $x<0$

So the function is only positive when x is negative and vice versa. Since there is also an $x$ $x$ factor in $f (x)$ $f(x)$

$f (x) = x (e^{- x} - e^{x})$ $f(x)=x(e^-x-e^x)$

When one factor is positive, the other is negative, so f(x) is always negative. Therefore, the Area:

$A = - \int_{1}^{3} f (x) d x$ $A=-int_1^3f(x)dx$

$A = - \int_{1}^{3} (x e^{- x} - x e^{x}) d x$ $A=-int_1^3(xe^-x-xe^x)dx$

$A = - \int_{1}^{3} x e^{- x} d x + \int_{1}^{3} x e^{x} d x$ $A=-int_1^3xe^-xdx+int_1^3xe^xdx$

$A=-int_1^3x*(-(e^-x)')dx+int_1^3x(e^x)'dx$

$A=int_1^3x*(e^-x)'dx+int_1^3x(e^x)'dx$

$A=[xe^-x]_1^3-int_1^3(x)'e^-xdx+[x(e^x)]_1^3-int_1^3(x)'e^xdx$

$A=[xe^-x]_1^3-int_1^3e^-xdx+[x(e^x)]_1^3-int_1^3e^xdx$

$A=[xe^-x]_1^3-[-e^-x]_1^3+[x(e^x)]_1^3-[e^x]_1^3$

$A=(3e^-3-1*e^-1)+(e^-3-e^-1)+(3e^3-1*e^1)-(e^3-e^1)$

$A=3/e^3-1/e+1/e^3-1/e+3e^3-e-e^3+e$

$A=4/e^3 -2/e+2e^3$

Using calculator:

$A=39.6345$

Answer 2 · 2016-01-13T03:11:56

Area = 11,336.8 square units

Explanation:

the given $f(x)=xe^-x -xe^x$

for simplicity let $f(x)=y$

and $y=xe^-x -xe^x$

the first derivative $y'$ is needed in the computation of the surface area.

Area $= 2pi int_1^3 y$ $ds$

where $ds$ $=sqrt(1+ (y')^2)$ $dx$

Area $= 2pi int_1^3 y$ $sqrt(1+ (y')^2)$ $dx$

Determine the first derivative $y'$ :

differentiate $y=x(e^-x - e^x)$ using the derivative of product formula

$y' = 1*(e^-x-e^x)+x*(e^-x *(-1)-e^x)$

$y'=e^-x - e^x -x*e^-x -x*e^x$

after simplification and factoring, the result is

the first derivative $y'=e^-x*(1-x)-e^x*(1+x)$

Compute now the Area:

Area = $2 pi int_1^3 y$ $ds$

Area $= 2pi int_1^3 y$ $sqrt(1+ (y')^2)$ $dx$

Area
$= 2pi int_1^3 x(e^-x - e^x)$ $sqrt(1+ (e^-x*(1-x)-e^x*(1+x))^2$ $dx$

For complicated integrals like this, we may use Simpson's Rule:

so that

Area
$= 2pi int_1^3 x(e^-x - e^x)$ $sqrt(1+ (e^-x*(1-x)-e^x*(1+x))^2$ $dx$

Area = -11,336.804

this involves the direction of revolution so that there can be negative surface area or positive surface area. Let us just consider the positive value Area = 11336.804 square units

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What is the surface area of the solid created by revolving $f (x) = x e^{- x} - x e^{x}, x \in [1, 3]$ $f(x) = xe^-x-xe^(x) , x in [1,3]$ around the x axis?

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What is the surface area of the solid created by revolving f(x) = xe^-x-xe^(x) , x in [1,3]f(x)=xe−x−xex,x∈[1,3]f(x) = xe^-x-xe^(x) , x in [1,3] around the x axis?

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Explanation:

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What is the surface area of the solid created by revolving $f (x) = x e^{- x} - x e^{x}, x \in [1, 3]$ $f(x) = xe^-x-xe^(x) , x in [1,3]$ around the x axis?