# How do you find the average value of #f(x)=4-x^2# as x varies between #[-2,2]#?

##### 1 Answer

Nov 5, 2017

#### Explanation:

#"since f(x) is continuous on the closed interval"#

#[-2,2]" then the average value of f(x) from x = -2"#

#"to x = 2 is the integral"#

#•color(white)(x)1/(b-a)int_a^bf(x)dx#

#"here "[a,b]=[-2,2]#

#=1/4int_-2^2(4-x^2)dx#

#=1/4[4x-1/3x^3]_-2^2#

#=1/4[(8-8/3)-(-8+8/3)]#

#1/4(32/3)=8/3#