Question

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] ?

Answer 1

`f(x)=2e^(2x)+C` or `f(x)=-2e^(2x)+C`

Explanation:

the arc length from `x=0` to `x=3` is `int_0^3sqrt(1+(f'(x))^2)dx`

if the arc length is also equivalent to `int_0^3sqrt(1+16e^(4x))dx`, you can see that `(f'(x))^2=16e^(4x)`

solving for `f(x)`:
`f'(x)=+-sqrt(16e^(4x))`
`f'(x)=+-4e^(2x)`

now there are `2` cases:
`1)` if `f'(x)=4e^(2x)`, then `f(x)=int(4e^(2x))dx=2e^(2x)+C`
`2)` if `f'(x)=-4e^(2x)`, then `f(x)=int(-4e^(2x))dx=-2e^(2x)+C`

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