How do you give the value of C guaranteed by the mean value theorem for integrals for the function #f(x)=1/sqrt(x+2)# on the interval [-1,23]?

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Answer 1 · 2015-04-06T01:16:53

The First Mean Value Theorem says
#int_a^b f(x) dx = f(c)*(b-a)# for some value #c epsilon [a,b]#

In may be more convenient to write this as
#f(c) = (int_a^b f(x) dx)/(b-a)#

If #F'(x) = f(x) = 1/sqrt(x+2) = (x+2)-1/2#

then
#int_a^b f(x) dx#

#= 2(x+2)^(1/2)]_a^b#

#= 10-2 =8#

#(b-a) = 23- (-1) = 24#

#(int_a^b f(x)dx)/(b-a) = 1/3 =f(c)#

#f(c) = 1/sqrt(c+2) = 1/3#

#sqrt(c+2) = 3#

#c=7#