# What is the arc length of f(x)= e^(3x) +x^2e^x  on x in [1,2] ?

Aug 19, 2017

The arc length $s$ of the curve $f$ on $x \in \left[a , b\right]$ is given by

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

Here, $f \left(x\right) = {e}^{3 x} + {x}^{2} {e}^{x}$ so $f ' \left(x\right) = 3 {e}^{2 x} + 2 x {e}^{x} + {x}^{2} {e}^{x}$. The arc length is then:

$s = {\int}_{1}^{2} \sqrt{1 + {\left(3 {e}^{2 x} + 2 x {e}^{x} + {x}^{2} {e}^{x}\right)}^{2}} \mathrm{dx}$

Put this into an online graphing service like WolframAlpha or into any other calculator to find an approximation for the integral:

$s \approx \textcolor{b l u e}{97.658636}$