Introduction

Hello again! In today’s topic, we’ll cover how we can use integrals to calculate the total amount of change over specific time intervals. In this specific scenario, we can use definite integrals over specific intervals to find out the net change over a given interval. Examples of this include the total number of people who enter the line for a rollercoaster over 4 hours or how many people watch a video in the span of a week.

To understand why this works, we have to remember what an integral represents and how it works in the case of rate of change functions. Integrals are essentially summations of values of a function over periods of time. By taking the integral of a rate of change function, we essentially are able to add up an infinite amount of instantaneous rates of change, resulting in the net change.

Essentially, the definite integral of a function represents an accumulation of a rate of change, and the definite integral of said rate of change over an interval gives us the net change of that quantity over that interval. 

Accumulation of Change

Accumulation of change is when you add up all of the change over a specific interval. This can be done by taking the integral of the rate of change function over the interval, which gives the net change over the interval. Now although this may seem simple… it is. Let’s go over an example.

Example

Explanation:

As mentioned earlier, taking the definite integral of a rate of change function over a given period will give you the net change over that period. Therefore, all we need to do for this problem is take the definite integral of the function over the given period.

First, let’s integrate $f(x)=4x^3-6x^2+8x-10$ over the interval $[2,6]$. Doing so gives us $[x^4-2x^3+4x^2-10x{]}_2^6$. 

Plugging in:

Plugging in our values gives us $[(6{)}^4-2(6{)}^3+4(6{)}^2-10(6)]-[(2{)}^4-2(2{)}^3+4(2{)}^2-10(2)]$

Simplifying:

$[1296-2(216)+4(36)-60]-[16-2(8)+4(4)-20]$

$[1296-532+144-60]-[16-16+16-20]$

$[848]-[-4]$

$[852]$

Answer:

Therefore, the total number of people who entered the amusement park from $\text{2:00}$ P.M. to $\text{6:00}$ P.M. is $852$ people. 

Now although our last answer happened to nicely fit the context of the problem, that might not always be the case. Say you’re given a problem that asks for the number of apple pies made in 2 hours. If your result ends up being a decimal, you’d have to round down to the nearest integer since there’s no such thing as $0.673$ of a pie in real life.

As for the rest of the topic, that’s pretty much it! Most of the topics in this unit are on the shorter side, but there are a lot of them, meaning the difficulty of this unit is about the same as the other ones. Anyways, let’s wrap up this topic with some practice problems.

Practice 

This time around, the practice section will be mostly multiple choice to give you a small break in case you’re trying to cram this unit before the test.

Find the total distance traveled by a particle on the interval $[1,e^2]$ if its velocity is given by $\frac{2x}{x^2}$

Answer: 

First, we can simplify $\frac{2x}{x^2}$ by factoring out the $x$, giving us $\frac{2}{x}$. Integrating this over the interval $[1,e^2]$ gives us $[2\ln (x){]}_1^{e^2}$.

From here, all we have to do is plug in our values and solve.

$[2\ln (e^2)-2\ln (1)]$

$[2(2)-2(0)]$

$[4-0]$

$4$

Therefore, the total distance traveled by a particle on the interval $[1,e^2]$ whose velocity is given by $\frac{2x}{x^2}$ is $4$.