What is the average value of #y=x^2sqrt(x^3+1#) on the interval[0,2]? Calculus Derivatives Average Rate of Change Over an Interval 1 Answer GiĆ³ Jun 17, 2015 I found the Average value: #A_v=26/9# Explanation: Try this: Answer link Related questions How do you find the average rate of change of a function from graph? How do you find the average rate of change of a function between two points? How do you find the average rate of change of #f(x) = sec(x)# from #x=0# to #x=pi/4#? How do you find the average rate of change of #f(x) = tan(x)# from #x=0# to #x=pi/4#? How do you find the rate of change of y with respect to x? How do you find the average rate of change of #y=x^3+1# from #x=1# to #x=3#? What is the relationship between the Average rate of change of a fuction and derivatives? What is the difference between Average rate of change and instantaneous rate of change? What does the Average rate of change of a linear function represent? What is the relationship between the Average rate of change of a function and a secant line? See all questions in Average Rate of Change Over an Interval Impact of this question 9297 views around the world You can reuse this answer Creative Commons License