Question

What is the arclength of `f(x)=x+xsqrt(x+3)` on `x in [-3,0]`?

Answer 1

`s~~6.54696`

Explanation:

The formula for arc length of a curve defined by `f(x)` is:

`s=int_a^bsqrt(1+f'(x)^2)`

First let's find the derivative `f'(x)`:

`f(x)=x+x(x+3)^(1/2)`

`f'(x)=1+(x+3)^(1/2)+1/2x(x+3)^(-1/2)` (by the product rule)

`f'(x)=1+sqrt(x+3)+x/(2sqrt(x+3))`

Now let's evaluate `f'(x)^2`

`f'(x)^2=[1+sqrt(x+3)+x/(2sqrt(x+3))]^2`

`=1+sqrt(x+3)+x/(2sqrt(x+3))+sqrt(x+3)+(x+3)+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x/(2sqrt(x+3))+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x^2/(4(x+3))`

`=1+2sqrt(x+3)+cancel(2)x/(cancel(2)sqrt(x+3))+(x+3)+cancel(2)(x/cancel(2))+x^2/(4(x+3))`

`=x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+4`

Now you would need to evaluate this (very nasty) integral:

`int_-3^0sqrt(x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+5)`

I punched it into Wolfram|Alpha and got the following result:

So the answer is

`s~~6.54696`