Introduction

Welcome back! In today’s topic, we’re only going to cover one thing, namely how to find the average value of a function over an interval. The main method we’ll be using involves using definite integrals, which are integrals of functions but taken on a specific interval. Finding average value has all sorts of use cases, like trying to see which item on a menu is ordered the most over the course of a day.

The formula for calculating the average value of a function on the interval $[a,b]$ is $\frac{1}{b-a}{\int }_a^{b}f(x)dx$. To understand the underlying principle for this formula, all we have to do is remember how we calculate averages. As you may know, the average of a set of values is calculated by dividing the sum of the values by the number of values. Our formula essentially does the same thing, with the integral adding up all of the values, and $\frac{1}{b-a}$ giving us the number of values. Since today’s topic is relatively simple, let’s jump straight into practice.

Practice

Answers

Problem 1

Our interval in this case is $[2,7]$, so all we have to do is plug in the values into the formula.

Plugging in:

Average value $=\frac{1}{7-2}{\int }_2^{7}5t^2-7dt$

Average value $=\frac{1}{5}{\left[\frac{5}{3}{t}^3-7t\right]}_2^7$

Average value $=\frac{1}{5}\left(\left[\frac{5}{3}(7{)}^3-7(7)\right]-\left[\frac{5}{3}(2{)}^3-7(2)\right]\right)$

Average value $=\frac{1}{5}\left(\left[\frac{5}{3}(343)-49\right]-\left[\frac{5}{3}(8)-14\right]\right)$

Average value $=\frac{1}{5}\left(\left[\frac{1715}{3}-\frac{147}{3}\right]-\left[\frac{40}{3}-\frac{42}{3}\right]\right)$

Average value $=\frac{1}{5}\left(\left[\frac{1568}{3}\right]-\left[\frac{-2}{3}\right]\right)$

Average value $=\frac{1}{5}\left(\frac{1570}{3}\right)$

Average value $=\frac{1570}{15}$

Our answer is approximately $104.6$ repeating, but since there’s no such thing as $0.6$ of a ticket, we’ll round down to $104$. Therefore, the average number of tickets sold per hour from 2:00 PM to 7:00 PM is $104$.

Problem 2

Average value $=\frac{1}{\pi -0}{\int }_0^{\pi }\sin (x)\cos (x)dx$

Here, we can use u-substitution by letting $u=\sin (x)$ and $du=\cos (x)dx$. Since we are using a definite integral however, our bounds change to $u(0)$ and $u(\pi )$, which are both equal to $0$.

Average value $=\frac{1}{\pi }{\int }_0^{0}udu$

Now, we integrate as normal.

Average value $=\frac{1}{\pi }{\left[\frac{u^2}{2}\right]}_0^0$

Here, we substitute $\sin (x)$ back in for $u$, and change the bounds back as well.

Average value $=\frac{1}{\pi }{\left[\frac{{\sin }^2(x)}{2}\right]}_0^{\pi }$

Average value $=\frac{1}{\pi }\left[\frac{{\sin }^2(\pi )}{2}-\frac{{\sin }^2(0)}{2}\right]$

Average value $=\frac{1}{\pi }\left[\frac{0}{2}-\frac{0}{2}\right]$

Average value $=\frac{1}{\pi }(0)$

Average value $=0$

Therefore, the average value of the function $f(x)=\sin (x)\cos (x)$ on the interval $[0,\pi ]$ is $0$.

Problem 3

Average value $=\frac{1}{5-(-2)}{\int }_{-2}^5{x}^4-3x^3+7x^2-12x+4dx$

Average value $=\frac{1}{7}{\left[\frac{x^5}{5}-\frac{3x^4}{4}+\frac{7x^3}{3}-6x^2+4x\right]}_{-2}^5$

Average value $=\frac{1}{7}([\frac{(5{)}^5}{5}-\frac{3(5{)}^4}{4}+\frac{7(5{)}^3}{3}-6(5{)}^2+4(5)]-[\frac{(-2{)}^5}{5}-\frac{3(-2{)}^4}{4}+\frac{7(-2{)}^3}{3}-6(-2{)}^2+4(-2)])$

Average value $=\frac{1}{7}\left(\left[625-\frac{3(625)}{4}+\frac{7(125)}{3}-6(25)+20\right]-\left[\frac{-32}{5}-\frac{3(16)}{4}+\frac{7(-8)}{3}-6(4)-8\right]\right)$

Average value $=\frac{1}{7}\left(\left[\frac{2500}{4}-\frac{1875}{4}+\frac{875}{3}-150+20\right]-\left[\frac{-32}{5}-\frac{48}{4}+\frac{-56}{3}-24-8\right]\right)$

Average value $=\frac{1}{7}\left(\left[\frac{625}{4}+\frac{875}{3}-130\right]-\left[\frac{-32}{5}-12+\frac{-56}{3}-32\right]\right)$

Average value $=\frac{1}{7}\left(\left[\frac{4463}{12}\right]-\left[\frac{-1036}{15}\right]\right)$

Average value $=\frac{1}{7}\left(\frac{26459}{60}\right)$

Average value $=\frac{26459}{420}$

Therefore, the average value of the function $f(x)=x^4-3x^3+7x^2-12x+4$ on the interval $[-2,5]$ is $\frac{26459}{420}$ or approximately $62.998$.

Problem 4

Average value $=\frac{1}{5-0}{\int }_0^{5}4x^3+12x^2-4dx$

Average value $=\frac{1}{5}{\left[x^4+4x^3-4x\right]}_0^5$

Average value $=\frac{1}{5}\left(\left[(5{)}^4+4(5{)}^3-4(5)\right]-\left[(0{)}^4+4(0{)}^3-4(0)\right]\right)$

Average value $=\frac{1}{5}\left(\left[625+500-20\right]-\left[0+0-0\right]\right)$

Average value $=\frac{1}{5}(1105)$

Average value $=221$

Therefore, the average number of toys produced per hour from 8:00 AM to 1:00 PM is $221$.

Problem 5

Average value $=\frac{1}{4-1}{\int }_1^4\frac{1}{2\sqrt{x}}dx$

Average value $=\frac{1}{3}{\left[\sqrt{x}\right]}_1^4$

Average value $=\frac{1}{3}\left(\sqrt{4}-\sqrt{1}\right)$

Average value $=\frac{1}{3}(2-1)$

Average value $=\frac{1}{3}(1)$

Average value $=\frac{1}{3}$

Therefore, the average value of the function $f(x)=\frac{1}{2\sqrt{x}}$ on the interval $[1,4]$ is $\frac{1}{3}$.