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

Mar 28, 2017

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

$L = {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

#### Explanation:

Given $f \left(x\right) = {e}^{{x}^{2} - x}$

Find the Arc Length in the interval $\left[0 , 15\right]$

Compute $f ' \left(x\right)$

$f ' \left(x\right) = \left(2 x - 1\right) {e}^{{x}^{2} - x}$

${\left(f ' \left(x\right)\right)}^{2} = \left(4 {x}^{2} - 4 x + 1\right) {e}^{2 {x}^{2} - 2 x}$

I used WolframAlpha to calculate this:

$L = {\int}_{0}^{15} \sqrt{1 + \left(4 {x}^{2} - 4 x + 1\right) {e}^{2 {x}^{2} - 2 x}} \mathrm{dx} \approx 1.59 \times {10}^{91}$