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

1 Answer
mason m
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)#

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