What is the average value of a function # y = x^2 (x^3 + 1)^(1/2) # on the interval #[0, 2]#?

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Answer 1 · 2016-06-02T20:58:23

#26/9#

The average value of the differentiable function #f(x)# on the interval #[a,b]# can be found through

#1/(b-a)int_a^bf(x)dx#

For #f(x)=x^2(x^3+1)^(1/2)# on #[0,2]#, this yields an average value of

#1/(2-0)int_0^2x^2(x^3+1)^(1/2)dx#

Use substitution: #u=x^3+1# and #du=3x^2dx#.

#=1/6int_0^2 3x^2(x^3+1)^(1/2)dx#

Note that the bounds will change--plug the current bounds into #u=x^3+1#.

#=1/6int_1^9u^(1/2)du#

#=1/6[2/3u^(3/2)]_1^9=1/6[2/3(9)^(3/2)-2/3(1)^(3/2)]=1/6[2/3(27)-2/3(1)]#

#=1/6[54/3-2/3]=26/9#