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

1 Answer
Yonas Yohannes
Mar 22, 2016

L ~~ 1.430L≈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 + [(df(x))/(dx)]^2)dxL=∫ba√1+[df(x)dx]2dx
Now f(x) = x^3-e^x; x in [-1,0] f(x)=x3−ex;x∈[−1,0]
find f'(x) =(df(x))/(dx)=3x^2-e^x
[(df(x))/dx]^2 = [3x^2-e^x]^2
L=int_-1^0 sqrt(1+[3x^2-e^x]^2)dx There is no closed form antiderivative so integrate using an integral calculator or estimate numerically:
L ~~ 1.430

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

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