# What is the average value of a function y=50x+5 on the interval [3,9]?

Apr 30, 2017

$\frac{1}{9 - 3} {\int}_{3}^{9} \left(50 x + 5\right) \mathrm{dx} = 305$

#### Explanation:

The average value of a function $f \left(x\right)$ on the interval $\left[a , b\right]$ is given by:

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

So here, the average value is:

$\frac{1}{9 - 3} {\int}_{3}^{9} \left(50 x + 5\right) \mathrm{dx}$

$= \frac{5}{6} {\int}_{3}^{9} \left(10 x + 1\right) \mathrm{dx}$

Integrating term by term:

$= \frac{5}{6} {\left[5 {x}^{2} + x\right]}_{3}^{9}$

$= \frac{5}{6} \left[\left(5 {\left(9\right)}^{2} + 9\right) - \left(5 {\left(3\right)}^{2} + 3\right)\right]$

$= \frac{5}{6} \left(414 - 48\right)$

$= 305$

Apr 30, 2017

$a v g = 305$

#### Explanation:

Use the average function $a v g = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

$a v g = \frac{1}{9 - 3} {\int}_{3}^{9} \left(50 x + 5\right) \mathrm{dx} = \frac{1}{6} {\int}_{3}^{9} \left(50 x + 5\right) \mathrm{dx}$

$a v g = \frac{1}{6} {\left[\frac{50 {x}^{2}}{2} + 5 x\right]}_{3}^{9} = \frac{1}{6} {\left[25 {x}^{2} + 5 x\right]}_{3}^{9}$

$a v g = \frac{1}{6} \left[\left(25 \cdot {9}^{2} + 5 \cdot 9\right) - \left(25 \cdot {3}^{2} + 5 \cdot 3\right)\right]$

$a v g = \frac{1}{6} \left(2070 - 240\right)$

$a v g = 305$