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]?

2 Answers
Jul 23, 2017

8/3.
This Solution is Erroneous.

Explanation:

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.

Jul 23, 2017

8/3.

Explanation:

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.