What is the arclength of $f (x) = x^{3} - x e^{x}$ $f(x)=x^3-xe^x$ on $x \in [- 1, 0]$ $x in [-1,0]$ ?

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Answer 1 · 2016-01-17T09:41:28

Arc length $s = 1.54116$ $s=1.54116$ units

The formula to determine length of arc s:
$s=int_a^bsqrt(1+f' (x)^2) dx$
$a=-1$ and $b=0$ the limits

$f(x)=x^3-x e^x$

$f' (x) 3 x^2 -[x*e^x*1+1*e^x]$
$f' (x) = 3x^2 - x e^x - e^x$

$s=int_-1^0 sqrt(1+(3x^2 - x e^x - e^x)^2) dx$

There is no simple formula to evaluate the integral, so try using Simpson's Rule:

$s= 1.54116$ units

Just observe the graph from $x=-1$ to $x=0$
graph{y=x^3-x e^x [-2.5, 2.5, -1.25, 1.25]}

What is the arclength of f(x)=x^3-xe^xf(x)=x3−xexf(x)=x^3-xe^x on x in [-1,0]x∈[−1,0]x in [-1,0]?