What is the arc length of #y=1/3 (x^2 + 2)^(3/2) (0 <= x <= 1) #?

Question

#y=1/3 (x^2 + 2)^(3/2) (0 <= x <= 1) #

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Answer 1 · 2018-06-07T11:32:36

#4/3#

We get

#y'=xsqrt(x^2+1)#
so our integral is given by

#int_0^1sqrt(1+x^2+2)dx#
Note that

#1+x^2(x^2+2)=x^4+2x^2+1=(x^2+1)^2#
so we have

#[x^3/3+x]_0^1=4/3#