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#
