What is the arclength of #f(x)=e^(x^2-x) # in the interval #[0,15]#?

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Answer 1 · 2017-03-28T21:10:53

The general formula for Arc Length of a function of x is:

#L = int_a^bsqrt(1+(f'(x))^2)dx#

Given #f(x)=e^(x^2-x)#

Find the Arc Length in the interval #[0,15]#

Compute #f'(x)#

#f'(x) = (2x-1)e^(x^2-x)#

#(f'(x))^2=(4x^2-4x+1)e^(2x^2-2x)#

I used WolframAlpha to calculate this:

#L = int_0^15sqrt(1+(4x^2-4x+1)e^(2x^2-2x))dx~~1.59xx10^91#