What is the arclength of f ( x ) = x 3 − e x on x ∈ [ − 1 , 0 ]?

Question

What is the arclength of   f  (  x  )  =      x          3        −      e          x       on   x  ∈  [  −  1  ,  0  ]?

Cristopher Barrera · Accepted Answer

Use the Arc Length theorem: Let f(x) be a continuous on [a, b], then the length of the curve

y = f (x), a \leq x \leq b

, is

L = \int_{a}^{b} \sqrt{1 + [\frac{d f (x)}{d x}]^{2}} d x

Now

f (x) = x^{3} - e^{x}; x \in [- 1, 0]

find

f^{'} (x) = \frac{d f (x)}{d x} = 3 x^{2} - e^{x}

[\frac{d f (x)}{d x}]^{2} = [3 x^{2} - e^{x}]^{2}

L = \int_{- 1}^{0} \sqrt{1 + [3 x^{2} - e^{x}]^{2}} d x

There is no closed form antiderivative so integrate using an integral calculator or estimate numerically:

L \approx 1.430

In general the Arc Length integral is evaluated numerically, there very few Arc Length Integral with a closed form antiderivative.

What is the arclength of f ( x ) = x <mrow class="MJX-TeXAtom-ORD"> 3