How do you find the average value of the function for #f(x)=1+x, -1<=x<=1#?

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How do you find the average value of the function for #f(x)=1+x, -1<=x<=1#?

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Answer 1 · 2018-02-11T14:10:20

#1#

Explanation:

#"average of f(x) in interval [a,b] is : "#
#(1/(b-a)) int_a^b f(x)#
#"So here we have"#
#a = -1#
#b = 1#
#f(x) = 1+x#
#=> avg = (1/2) int_{-1}^1 (1+x)(dx)#
#= (1/2) [x + x^2/2 + C]_{-1}^1#
#= (1/2) [1 + 1^2/2 + cancel(C) - (-1) - (-1)^2/2 - cancel(C)]#
#= (1/2) [1 + cancel(1/2) + 1 - cancel(1/2)]#
#= (1/2) [1 + 1]#
#= 1#

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How do you find the average value of the function for #f(x)=1+x, -1<=x<=1#?

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