Exploring Accumulations of Change - AP Calculus BC Study Guide

The Core Idea: Exploring Accumulations of Change

The fundamental concept of this topic is that the definite integral is a tool for calculating accumulation. If we know the rate at which a quantity is changing, the definite integral allows us to find the total net change in that quantity over a specific interval. For example, if we know the velocity (rate of change of position) of an object, the definite integral can tell us the object's net displacement (net change in position).

When we cannot find this accumulation exactly, we can approximate it. The primary method for this approximation is the Riemann sum, which involves dividing the interval into smaller pieces, calculating the change over each small piece using simple geometric shapes (like rectangles), and then summing up these individual changes. The definite integral can also be determined exactly if the rate function forms common geometric shapes, where we can use standard area formulas to find the total accumulation.

Key Formulas & Rules

The following methods are used to approximate the value of a definite integral, ${\int }_a^{b}f(x)dx$, by dividing the interval $[a,b]$ into $n$ subintervals of width $\Delta x=\frac{b-a}{n}$. For subintervals of unequal width, $\Delta x_i$ represents the width of the $i$-th subinterval.

Riemann Sums

A Riemann sum approximates the definite integral by summing the areas of rectangles. The height of each rectangle is determined by the function's value at a specific point within each subinterval.

• Left Riemann Sum: Uses the left endpoint of each subinterval to determine the height of the rectangle.

  $${\int }_a^{b}f(x)dx\approx \sum _{i=0}^{n-1}f(x_i)\Delta x$$

• Right Riemann Sum: Uses the right endpoint of each subinterval to determine the height of the rectangle.

  $${\int }_a^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_i)\Delta x$$

• Midpoint Riemann Sum: Uses the midpoint of each subinterval to determine the height of the rectangle.

  $${\int }_a^{b}f(x)dx\approx \sum _{i=1}^{n}f\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x$$

Trapezoidal Sum

The trapezoidal sum approximates the definite integral by summing the areas of trapezoids formed in each subinterval. The area of a trapezoid is $\frac{1}{2}({\text{base}}_1+{\text{base}}_2)\times \text{height}$. In this context, the "bases" are the parallel function values $f(x_{i-1})$ and $f(x_i)$, and the "height" is the width of the subinterval $\Delta x$.

• Trapezoidal Sum Formula:

  $${\int }_a^{b}f(x)dx\approx \frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots +2f(x_{n-1})+f(x_n)]$$

  For subintervals of unequal width, the sum is calculated one trapezoid at a time:

  $$\sum _{i=1}^n\frac{1}{2}(f(x_{i-1})+f(x_i))\Delta x_i$$

Understanding the Connection Between Area and Net Change

The definite integral ${\int }_a^{b}f(x)dx$ represents the accumulated net change of a quantity whose rate of change is given by $f(x)$. Geometrically, this same integral represents the net area between the graph of $y=f(x)$ and the x-axis from $x=a$ to $x=b$.

This dual interpretation is critical. The "net change" and "net area" are two sides of the same coin.

• When $f(x)$ is positive, the rate is positive, and the quantity is increasing. Geometrically, this corresponds to an area above the x-axis, which contributes positively to the integral's value.

• When $f(x)$ is negative, the rate is negative, and the quantity is decreasing. Geometrically, this corresponds to an area below the x-axis, which contributes negatively to the integral's value.

Therefore, the definite integral is not the total area but the net area: (Area above the x-axis) - (Area below the x-axis). This value is precisely equal to the net change in the original quantity over the interval.

Core Concepts & Rules

• Integral as Net Change: The definite integral of a rate of change, ${\int }_a^b{f}^{\prime }(x)dx$, gives the net change in the original function, $f(x)$, over the interval $[a,b]$.

• Approximation with Sums: A definite integral can be approximated by summing the areas of geometric shapes. A Riemann sum uses rectangles, while a trapezoidal sum uses trapezoids.

• Approximation Methods: The primary methods for approximation are the Left Riemann Sum, Right Riemann Sum, Midpoint Riemann Sum, and the Trapezoidal Sum. The choice of method determines the point in each subinterval used to define the height of the shape.

• Exact Value from Geometry: If the graph of the function $f(x)$ over the interval $[a,b]$ is composed of common geometric figures (e.g., lines, semicircles, rectangles), the exact value of ${\int }_a^{b}f(x)dx$ can be found by calculating the areas of these figures.

• Area and Sign: Areas of regions above the x-axis are added to the total value of the integral, while areas of regions below the x-axis are subtracted.

Step-by-Step Example 1: Approximating from a Table

A pump removes water from a tank at a rate modeled by the differentiable function $R(t)$, where $R(t)$ is measured in liters per minute and $t$ is measured in minutes. Selected values of $R(t)$ are shown in the table below. Use a trapezoidal sum with the four subintervals indicated by the table to approximate ${\int }_0^{12}R(t)dt$.

Step 1: Understand the Goal

The expression ${\int }_0^{12}R(t)dt$ represents the total net change in the volume of water removed from the tank, in liters, from $t=0$ to $t=12$ minutes. We need to approximate this value using a trapezoidal sum. Notice the subintervals have unequal widths.

Step 2: Set up the Trapezoidal Sum

We will calculate the area of four trapezoids, one for each subinterval: $[0,3]$, $[3,5]$, $[5,9]$, and $[9,12]$. The formula for the area of one trapezoid is $\frac{1}{2}(h_1+h_2)\times w$, where $h_1$ and $h_2$ are the function values (the rates) and $w$ is the width of the interval ($\Delta t$).

• Trapezoid 1 ($t=0$ to $t=3$):

  Width $\Delta t_1=3-0=3$. Heights are $R(0)=15$ and $R(3)=12$.

  Area = $\frac{1}{2}(15+12)\times 3$

• Trapezoid 2 ($t=3$ to $t=5$):

  Width $\Delta t_2=5-3=2$. Heights are $R(3)=12$ and $R(5)=8$.

  Area = $\frac{1}{2}(12+8)\times 2$

• Trapezoid 3 ($t=5$ to $t=9$):

  Width $\Delta t_3=9-5=4$. Heights are $R(5)=8$ and $R(9)=5$.

  Area = $\frac{1}{2}(8+5)\times 4$

• Trapezoid 4 ($t=9$ to $t=12$):

  Width $\Delta t_4=12-9=3$. Heights are $R(9)=5$ and $R(12)=3$.

  Area = $\frac{1}{2}(5+3)\times 3$

Step 3: Calculate the Sum

Now, sum the areas of the four trapezoids.

$${\int }_0^{12}R(t)dt\approx \frac{1}{2}(27)(3)+\frac{1}{2}(20)(2)+\frac{1}{2}(13)(4)+\frac{1}{2}(8)(3)$$

$$=(13.5)(3)+(10)(2)+(6.5)(4)+(4)(3)$$

$$=40.5+20+26+12=98.5$$

Step 4: Interpret the Result

The approximate value of the integral is 98.5. Based on the context, this means approximately 98.5 liters of water were removed from the tank during the first 12 minutes.

Step-by-Step Example 2: Evaluating from a Graph

The graph of the function $f(x)$ is shown below. It consists of a line segment from $(-4,0)$ to $(0,4)$ and a semicircle of radius 4 centered at $(4,0)$. Find the exact value of ${\int }_{-4}^{8}f(x)dx$.