Question

How do you find the arc length of the curve `f(x)=x^(3/2)` over the interval [0,1]?

Answer 1

`1/27(13sqrt13-8)`

Explanation:

The arc length of the curve `f(x)` on the interval `[a,b]` is defined as:

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

Here, we see that `f(x)=x^(3/2)` and `f^'(x)=3/2x^(1/2)`. Thus:

`s=int_0^1sqrt(1+(3/2x^(1/2))^2)dx`

`=int_0^1sqrt(1+9/4x)dx`

`=int_0^1sqrt((4+9x)/4)dx`

`=1/2int_0^1sqrt(4+9x)dx`

Substitute with `u=4+9x` such that `du=9dx`.

`=1/18int_0^1 9sqrt(4+9x)dx`

Substitute `u` and change the bounds from `x` to `u`: `x=0->u=4+9(0)=4;x=1->u=4+9(1)=13`.

`=1/18int_4^13sqrtudu`

`=1/18int_4^13u^(1/2)du`

`=1/18[u^(3/2)/(3/2)]_4^13`

`=1/18(2/3)[u^(3/2)]_4^13`

`=1/27[u^(3/2)]_4^13`

`=1/27(13^(3/2)-4^(3/2))`

`=1/27(13^1*13^(1/2)-(2^2)^(3/2))`

`=1/27(13sqrt13-2^3)`

`=1/27(13sqrt13-8)`