If the average value of the function f on the interval [1, 5] is 3, and the average value on the interval [5, 7] is 2, how do you find the average value on the interval [1, 7]?

Question

19254 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-23T18:09:38

# 8/3.#
This Solution is Erroneous.

Recall that, the #cancel("Average Value")# Average Rate of Change of a Function #f# on the Interval

#[a,b]# is, #(f(b)-f(a))/(b-a).#

Let us denote it by, #AV(f)[a,b]," so, "AV(f)[a,b]=(f(b)-f(a))/(b-a).#

Then, by what has been given,

#AV(f)[1,5]=3, AV(f)[5,7]=2," and, we need "AV(f)[1,7].#

#AV(f)[1,5]=3 rArr (f(5)-f(1))/(5-1)=3,#

# :. f(5)-f(1)=12............................................................(1).#

#"Similarly, "AV(f)[5,7]=2,#

# rArr f(7)-f(5)=4...........................................................(2).#

#:. AV(f)[1,7]=(f(7)-f(1))/(7-1)=(f(7)-f(1))/6,#

#=(f(7)-f(5)+f(5)-f(1))/6,#

#=(4+12)/6,.............[because, (1), &, (2)],#

# rArr AV(f)[1,7]=8/3.#

Answer 2 · 2017-07-23T19:12:46

# 8/3.#

The Average Value of a Function #f," on an Interval "[a,b]# is

#1/(b-a)int_a^bf(x)dx.#

Hence, by what is given, we have,

#1/(5-1)int_1^5f(x)dx=3 rArr int_1^5f(x)dx=12.............(1).#

Similarly, #int_5^7f(x)dx=2(7-5)=4................................(2).#

Therefore, the Reqd. Average value of

#f" over "[1,7]=[1,5]uu[5,7],# is,

#1/(7-1)int_1^7f(x)dx,#

#=1/6[int_1^5f(x)dx+int_5^7f(x)dx],#

#=1/6[12+4],.....................................[because, (1), &, (2)],#

#=16/6,#

#=8/3.#