# For which function does the integral integral 0^3 sqrt(1+16e^4x)dx give the length of the curve on the interval [0,3] ?

Jun 6, 2018

$f \left(x\right) = 2 {e}^{2 x} + C$ or $f \left(x\right) = - 2 {e}^{2 x} + C$

#### Explanation:

the arc length from $x = 0$ to $x = 3$ is ${\int}_{0}^{3} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

if the arc length is also equivalent to ${\int}_{0}^{3} \sqrt{1 + 16 {e}^{4 x}} \mathrm{dx}$, you can see that ${\left(f ' \left(x\right)\right)}^{2} = 16 {e}^{4 x}$

solving for $f \left(x\right)$:
$f ' \left(x\right) = \pm \sqrt{16 {e}^{4 x}}$
$f ' \left(x\right) = \pm 4 {e}^{2 x}$

now there are $2$ cases:
1) if $f ' \left(x\right) = 4 {e}^{2 x}$, then $f \left(x\right) = \int \left(4 {e}^{2 x}\right) \mathrm{dx} = 2 {e}^{2 x} + C$
2) if $f ' \left(x\right) = - 4 {e}^{2 x}$, then $f \left(x\right) = \int \left(- 4 {e}^{2 x}\right) \mathrm{dx} = - 2 {e}^{2 x} + C$

there's no more information so there are $2$ possible answers.