Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

What is the average value of the function $f(x) = x^2$ on the interval $[0,3]$ ?

1 Answer

Mar 23, 2016

The average value is $3$ .

Explanation:

The average value of a function $f$ on an interval $[a,b]$ is

$1/(b-a) int_a^b f(x) dx$

So the value we seek is

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

$= 1/3 x^3/3]_0^3$

$= (3)^3/9 - (0)^3/9 = 3$

Related questions

The profit (in dollars) from the sale of x lawn mowers is ...
What is the average value of the function $f(x)=(x-1)^2$ on the interval from x=1 to x=5?
What is the average value of the function $u(x) = 10xsin(x^2)$ on the interval $[0,sqrt pi]$ ?
What is the average value of the function $f(x)=cos(x/2)$ on the interval $[-4,0]$ ?
What is the average value of the function $f(t)=te^(-t^2 )$ on the interval $[0,5]$ ?
What is the average value of the function $f(x) = x - (x^2)$ on the interval $[0,2]$ ?
What is the average value of the function $f(t) = t (sqrt (1 + t^2) )$ on the interval $[0,5]$ ?
What is the average value of the function $f(x) = sec x tan x$ on the interval $[0,pi/4]$ ?
What is the average value of the function $f(x) = 2x^3(1+x^2)^4$ on the interval $[0,2]$ ?
What is the average value of the function $f(x)=18x+8$ on the interval $[0,10]$ ?

See all questions in The Average Value of a Function

Impact of this question

9432 views around the world

You can reuse this answer
Creative Commons License