RMS & Mean Value? Help

For the function 𝑓(𝑥) = 𝑥^2 + 3, find the mean value over the interval [3, 5] and
the R.M.S. value over the interval [0, 2]

1 Answer
SCooke
Apr 27, 2018

Mean #= 19.335# for [3,5]
RMS# = 4.49# for [0,2]

Explanation:

The mean value of a function over an interval [a,b] is:

#f(x) = 1/(b-a)int_a^b(f(x)dx#

The RMS is: #sqrt(1/(b-a)int_a^b(f(x)^2dx)#

#f(x) = x^2 + 3#

Mean over [3,5]
#=1/(5-3)int_3^5(x^2 +3)dx#
#=1/(5-3)[(x^3)/3 +3x]_3^5#
#=1/2((5^3)/3 +3(5) - (3^3)/3 +3(3))#
#=1/2(56.67 - 18) = 19.335#

RMS over [0,2]

#sqrt(1/(2-0)int_0^2(x^2 +3)^2dx)#
Find the value under the radical first, then take the square root.
#1/(2-0)int_0^2(x^2 +3)^2dx#
#1/2int_0^2(x^4 + 6x^2 + 9)dx#

#=1/2[(x^5)/5 + 6x^3/3 + 9x]_0^2#

#=1/2[(2^5)/5 + 6(2)^3/3 + 9(2)]#
#= 1/2[6.4 + 16 + 18] = 20.2#

RMS = #sqrt(20.2) = 4.49#

https://ask.fxplus.ac.uk/tools/HELM/pages/workbooks_1_50_jan2008/Workbook14/14_2_meanvalue_n_rms_functn.pdf

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