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

1 Answer
Sep 6, 2016

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)