Find avg value over given interval
f(x) = x^2 + x - 2 ; [0,4]
f(x) = x^2 + x - 2 ; [0,4]
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f_average = 1/(b - a) ∫ f(x) dx [a , b]
so in this case:
f_average = 1/(4 - 0) ∫ x^2 + x - 2 dx [0 , 4]
= 1/4*[x^3/3 + x^2/2 - 2x] from 0 to 4
= 1/4* [64/3 - 0]
= 16/3
so in this case:
f_average = 1/(4 - 0) ∫ x^2 + x - 2 dx [0 , 4]
= 1/4*[x^3/3 + x^2/2 - 2x] from 0 to 4
= 1/4* [64/3 - 0]
= 16/3
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And you are always welcome Michael.
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The average (mean) value of f(x) over the interval [a,b] is (1/(b-a))integral from a to b of f(x) dx.