# What is the arclength of f(x)=x^3-e^x on x in [-1,0]?

Mar 22, 2016

$L \approx 1.430$

#### Explanation:

Use the Arc Length theorem: Let f(x) be a continuous on [a, b], then the length of the curve y = f(x), a ≤ x ≤ b, is
$L = {\int}_{a}^{b} \sqrt{1 + {\left[\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right]}^{2}} \mathrm{dx}$
Now f(x) = x^3-e^x; x in [-1,0]
find $f ' \left(x\right) = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 3 {x}^{2} - {e}^{x}$
${\left[\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right]}^{2} = {\left[3 {x}^{2} - {e}^{x}\right]}^{2}$
$L = {\int}_{-} {1}^{0} \sqrt{1 + {\left[3 {x}^{2} - {e}^{x}\right]}^{2}} \mathrm{dx}$ 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.