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

Question

3186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T07:56:55

The average value is #=26/9#

If #f(x)# is a continuous function on the interval #[a,b]#, then the average value is

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

Here, #g(x)# is continuous on the interval #[0,2]#

Therefore, the average value is

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

First, compute the indefinite integral

#I=intx^2sqrt(1+x^3)dx#

Let #u=1+x^3#, #=>#, #du=3x^2dx#

Therefore,

#I=1/3intu^(1/2)du=1/3u^(3/2)/(3/2)#

#=2/9(1+x^3)^(3/2)+C#

And finally

#barx=1/2*2/9[(1+x^3)^(3/2)]_0^2#

#=1/9*((27)-(1))#

#=26/9#