Answer:
Average length of the given function
A(x) [tex]= \frac{4}{3}[/tex]
Step-by-step explanation:
Step(i):-
Given function f(x) = x³ on [0,2]
Given interval by partitioning the interval into four subintervals of equal length
The average length of four subintervals of equal length
[tex]Average length = \frac{1}{b-a} \int\limits^b_a {f(x)} \, dx[/tex]
Step(ii):-
[tex]Average length = \frac{1}{2-0} \int\limits^2_0 {x^3} \, dx[/tex]
Now integrating
[tex]\int\limits {x^{n} } \, dx = \frac{x^{n+1} }{n+1}[/tex]
[tex]Average length = \frac{1}{2-0}( \frac{x^3}{3} )_{0} ^{2}[/tex]
Final answer
Average length of the given function
[tex]= \frac{1}{2-0}( \frac{2^3}{3} ) = \frac{4}{3}[/tex]
