# 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]?

Jul 23, 2017

$\frac{8}{3.}$
This Solution is Erroneous.

#### Explanation:

Recall that, the $\cancel{\text{Average Value}}$ Average Rate of Change of a Function $f$ on the Interval

$\left[a , b\right]$ is, $\frac{f \left(b\right) - f \left(a\right)}{b - a} .$

Let us denote it by, $A V \left(f\right) \left[a , b\right] , \text{ so, } A V \left(f\right) \left[a , b\right] = \frac{f \left(b\right) - f \left(a\right)}{b - a} .$

Then, by what has been given,

$A V \left(f\right) \left[1 , 5\right] = 3 , A V \left(f\right) \left[5 , 7\right] = 2 , \text{ and, we need } A V \left(f\right) \left[1 , 7\right] .$

$A V \left(f\right) \left[1 , 5\right] = 3 \Rightarrow \frac{f \left(5\right) - f \left(1\right)}{5 - 1} = 3 ,$

$\therefore f \left(5\right) - f \left(1\right) = 12. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right) .$

$\text{Similarly, } A V \left(f\right) \left[5 , 7\right] = 2 ,$

$\Rightarrow f \left(7\right) - f \left(5\right) = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right) .$

$\therefore A V \left(f\right) \left[1 , 7\right] = \frac{f \left(7\right) - f \left(1\right)}{7 - 1} = \frac{f \left(7\right) - f \left(1\right)}{6} ,$

$= \frac{f \left(7\right) - f \left(5\right) + f \left(5\right) - f \left(1\right)}{6} ,$

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

$\Rightarrow A V \left(f\right) \left[1 , 7\right] = \frac{8}{3.}$

Jul 23, 2017

$\frac{8}{3.}$

#### Explanation:

The Average Value of a Function $f , \text{ on an Interval } \left[a , b\right]$ is

$\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx} .$

Hence, by what is given, we have,

$\frac{1}{5 - 1} {\int}_{1}^{5} f \left(x\right) \mathrm{dx} = 3 \Rightarrow {\int}_{1}^{5} f \left(x\right) \mathrm{dx} = 12. \ldots \ldots \ldots \ldots \left(1\right) .$

Similarly, ${\int}_{5}^{7} f \left(x\right) \mathrm{dx} = 2 \left(7 - 5\right) = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right) .$

Therefore, the Reqd. Average value of

$f \text{ over } \left[1 , 7\right] = \left[1 , 5\right] \cup \left[5 , 7\right] ,$ is,

$\frac{1}{7 - 1} {\int}_{1}^{7} f \left(x\right) \mathrm{dx} ,$

$= \frac{1}{6} \left[{\int}_{1}^{5} f \left(x\right) \mathrm{dx} + {\int}_{5}^{7} f \left(x\right) \mathrm{dx}\right] ,$

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

$= \frac{16}{6} ,$

$= \frac{8}{3.}$