Home > Calculus > The Average Value of a Function

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

Answer 1

Scarlett Allison

The average value is #3#.

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Paisley Johnson

To find the average value of a function ( f(x) ) on the interval ([a,b]), you use the formula:

[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx ]

For the function ( f(x) = x^2 ) on the interval ([0,3]), the average value is:

[ \text{Average value} = \frac{1}{3-0} \int_{0}^{3} x^2 , dx ]

[ = \frac{1}{3} \left[ \frac{x^3}{3} \right]_{0}^{3} ]

[ = \frac{1}{3} \left( \frac{27}{3} - 0 \right) ]

[ = \frac{1}{3} \cdot 9 ]

[ = 3 ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.