Finding the Area Between Curves Expressed as Functions of $x$ - AP Calculus AB Study Guide

The Core Idea: Finding the Area Between Curves Expressed as Functions of $x$

This topic extends the concept of finding the area under a single curve to the more general problem of finding the area of a region enclosed between two different curves. The fundamental idea is to think of this area as being composed of an infinite number of infinitesimally thin vertical rectangles. The height of each representative rectangle is the vertical distance between the upper curve and the lower curve, and its width is an infinitesimal change in $x$, denoted as $dx$.

By integrating the expression for the height of these rectangles, $(topfunction)-(bottomfunction)$, over a specified interval from $x=a$ to $x=b$, we are summing the areas of all these infinitesimally thin rectangles to find the total area of the region. This process requires identifying which function has a greater value (the "top" function) over the interval and determining the correct limits of integration, which are often the points where the curves intersect.

Key Formulas

The area of a region between two curves can be calculated using a definite integral. The specific formula depends on the relationship between the functions.

• Area with a Consistent Top Function: If $f(x)\ge g(x)$ for all $x$ in the interval $[a,b]$, the area $A$ of the region between the graphs of $f$ and $g$ from $x=a$ to $x=b$ is given by:

  $$A={\int }_a^b(f(x)-g(x))dx$$

  Here, $(f(x)-g(x))$ represents the height of a representative rectangle at a given $x$, and $dx$ is its infinitesimal width.

• Area with Crossing Curves (Absolute Value Form): If the functions $f(x)$ and $g(x)$ cross within the interval $[a,b]$, the total area between them can be found by integrating the absolute value of their difference. This automatically accounts for which function is on top.

  $$A={\int }_a^b|f(x)-g(x)|dx$$

• Area with Crossing Curves (Split Integral Form): Alternatively, if the curves cross at one or more points, the area must be calculated by splitting the integral into separate pieces for each subinterval where the top and bottom functions are consistent. For example, if $f$ and $g$ intersect at $x=c$ where $a<c<b$:

  $$A={\int }_a^c(\text{top}-\text{bottom})dx+{\int }_c^b(\text{top}-\text{bottom})dx$$

  You must determine which function is on top for each integral separately.

Understanding the "Top Minus Bottom" Principle

The core of finding the area between curves is the "top minus bottom" principle. The integrand, $(f(x)-g(x))$, represents the height of a vertical rectangle at a specific $x$-value. This is because the y-coordinate on the top curve is $f(x)$ and the y-coordinate on the bottom curve is $g(x)$. The vertical distance between them is simply the difference of these y-values.

A critical step in this process is determining the limits of integration, $a$ and $b$.

• Given Limits: Sometimes the problem will explicitly state the interval, such as "Find the area between the curves from $x=1$ to $x=4$."

• Intersection Points as Limits: More commonly, the problem will ask for the area of the region "enclosed" or "bounded" by the curves. In this case, the limits of integration are the x-coordinates of the points where the curves intersect. To find these, you must set the functions equal to each other ($f(x)=g(x)$) and solve for $x$.

If the curves intersect between the limits of integration, the function that is on "top" will change. This requires you to break the problem into multiple integrals, one for each subinterval, to ensure you are always integrating a positive height ($top-bottom$).

Core Concepts & Rules

• The area between two curves, $y=f(x)$ and $y=g(x)$, on an interval $[a,b]$ is found by integrating the difference between the upper function and the lower function.

• The integrand must always represent a positive distance, so it should be structured as (top function) - (bottom function)`. * The limits of integration, $a and $b$, are the x-values that define the left and right boundaries of the region.

• If the limits of integration are not provided, they are typically the x-coordinates of the intersection points of the two curves.

• To find intersection points, set the two functions equal to each other and solve for $x$.

• If the curves cross within the desired interval, you must split the area calculation into multiple definite integrals at each point of intersection.

• The integral ${\int }_a^b|f(x)-g(x)|dx$ provides a single-integral expression for the total area, which is particularly useful for calculator-based problems.

Step-by-Step Example 1: Basic Application

Problem: Find the area of the region enclosed by the graphs of $f(x)=x+1$ and $g(x)=x^2-1$.

Step 1: Find the points of intersection.

Set the two functions equal to each other to find the limits of integration.

$$x+1=x^2-1$$

$$0=x^2-x-2$$

$$0=(x-2)(x+1)$$

The curves intersect at $x=-1$ and $x=2$. These are our limits of integration, $a=-1$ and $b=2$.

Step 2: Determine the top and bottom functions.

Choose a test value of $x$ within the interval $[-1,2]$, such as $x=0$.

• $f(0)=0+1=1$

• $g(0)=0^2-1=-1$

Since $f(0)>g(0)$, the function $f(x)=x+1$ is the top function and $g(x)=x^2-1$ is the bottom function on the interval $[-1,2]$.

Step 3: Set up the definite integral.

The area $A$ is the integral of (top function - bottom function) from $x=-1$ to $x=2$.

$$A={\int }_{-1}^2((x+1)-(x^2-1))dx$$

$$A={\int }_{-1}^2(-x^2+x+2)dx$$

Step 4: Evaluate the integral.

Find the antiderivative and apply the Fundamental Theorem of Calculus.

$$A={\left[-\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+2x\right]}_{-1}^2$$

$$A=\left(-\frac{1}{3}(2{)}^3+\frac{1}{2}(2{)}^2+2(2)\right)-\left(-\frac{1}{3}(-1{)}^3+\frac{1}{2}(-1{)}^2+2(-1)\right)$$

$$A=\left(-\frac{8}{3}+\frac{4}{2}+4\right)-\left(\frac{1}{3}+\frac{1}{2}-2\right)$$

$$A=\left(-\frac{8}{3}+2+4\right)-\left(\frac{2}{6}+\frac{3}{6}-\frac{12}{6}\right)$$

$$A=\left(6-\frac{8}{3}\right)-\left(-\frac{7}{6}\right)$$

$$A=\left(\frac{18}{3}-\frac{8}{3}\right)+\frac{7}{6}$$

$$A=\frac{10}{3}+\frac{7}{6}=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$$

The area of the region is $4.5$ square units.

Step-by-Step Example 2: Exam-Style Application

Problem: Find the area of the region bounded by the graphs of $y=\sin (x)$ and $y=\cos (x)$ between $x=0$ and $x=\frac{\pi }{2}$.

Step 1: Find points of intersection within the interval.

The problem gives the interval $[0,\frac{\pi }{2}]$. We must check if the curves cross within this interval.

$$\sin (x)=\cos (x)$$

This occurs when $x=\frac{\pi }{4}$. Since $\frac{\pi }{4}$ is in the interval $[0,\frac{\pi }{2}]$, we must split the area calculation into two parts.

Step 2: Determine the top and bottom functions on each subinterval.

• Subinterval 1: $[0,\frac{\pi }{4}]$

  Let's test $x=\frac{\pi }{6}$. $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$ and $\sin (\frac{\pi }{6})=\frac{1}{2}$. Since $\frac{\sqrt{3}}{2}>\frac{1}{2}$, $\cos (x)$ is the top function on this interval.

• Subinterval 2: $[\frac{\pi }{4},\frac{\pi }{2}]$

  Let's test $x=\frac{\pi }{3}$. $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$ and $\cos (\frac{\pi }{3})=\frac{1}{2}$. Since $\frac{\sqrt{3}}{2}>\frac{1}{2}$, $\sin (x)$ is the top function on this interval.

Step 3: Set up the two definite integrals.

The total area $A$ is the sum of the areas from the two subintervals.

$$A={\int }_0^{\pi /4}(\cos (x)-\sin (x))dx+{\int }_{\pi /4}^{\pi /2}(\sin (x)-\cos (x))dx$$

Step 4: Evaluate the integrals.

• First Integral:

  $${\int }_0^{\pi /4}(\cos (x)-\sin (x))dx=[\sin (x)-(-\cos (x)){]}_0^{\pi /4}=[\sin (x)+\cos (x){]}_0^{\pi /4}$$

  $$=\left(\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)\right)-(\sin (0)+\cos (0))=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-(0+1)=\sqrt{2}-1$$

• Second Integral:

  $${\int }_{\pi /4}^{\pi /2}(\sin (x)-\cos (x))dx=[-\cos (x)-\sin (x){]}_{\pi /4}^{\pi /2}$$

  $$=\left(-\cos \left(\frac{\pi }{2}\right)-\sin \left(\frac{\pi }{2}\right)\right)-\left(-\cos \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{4}\right)\right)$$

  $$=(0-1)-\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)=-1-(-\sqrt{2})=\sqrt{2}-1$$

Step 5: Sum the results.

$$A=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2$$

Using Your Calculator

For many area problems, especially on the calculator-active section of the AP exam, a graphing calculator is an essential tool.

Problem: Find the area of the region enclosed by $f(x)=e^x$ and g(x) = 4 - x^2`. **Step 1: Graph the Functions** * Press `Y=`. * Enter `Y1 = e^(X)`. * Enter `Y2 = 4 - X^2`. * Press `GRAPH`. Adjust the `WINDOW` if necessary to see the enclosed region and intersection points clearly. **Step 2: Find the Intersection Points (Limits of Integration)** * Press `2nd` then `TRACE` to access the $CALC menu.

• Select $5:intersect$.

• The calculator will ask for "First curve?", "Second curve?", and "Guess?". Move the cursor near an intersection point and press ENTER for each prompt.

• The calculator finds the first intersection at approximately $x\approx -1.9646$. To save this value for accuracy, go to the home screen (2nd + MODE), press X,T,\theta,n$, then $STO\to$, then $ALPHA$, then $A$. Press ENTER`.

• Repeat the process for the other intersection point, which is at approximately $x\approx 1.0580$. Store this value as B`. **Step 3: Calculate the Integral** * From the home screen, press `MATH` and select `9: fnInt(`. * **Method A (Top - Bottom):** From the graph, we see that `g(x) = 4 - x^2` is the top function. The syntax is `fnInt(top - bottom, X, lower, upper)`. * Enter: `fnInt(Y2 - Y1, X, A, B)` * (You can access `Y1` and `Y2` by pressing `VARS` → `Y-VARS` → $1: Function...`) * **Method B (Absolute Value):** This method avoids having to determine the top function. The syntax is `fnInt(abs(Y1 - Y2), X, lower, upper)`. * (Access $abs( from the MATH → NUM menu).

  • Enter: fnInt(abs(Y1 - Y2), X, A, B)

Both methods will return the answer: approximately 10.305.

AP Exam Quick Hit

Common Question Types

• Calculator-Active Area FRQ: You will be given two non-trivial functions (e.g., involving exponentials, logarithms, or trig functions) and asked to find the area of the region they enclose. This tests your ability to use a calculator to find intersection points and evaluate a definite integral.

• Non-Calculator Area Problem: You will be given two simpler functions (e.g., polynomials, basic trig) and asked to find the enclosed area. This tests your ability to find intersections algebraically and compute an antiderivative by hand.

• "Set Up, Do Not Evaluate": You may be asked to write an integral expression for the area of a complex region, possibly one that requires being split into multiple integrals, without actually solving it. This tests your conceptual understanding of the setup.

Common Mistakes

• Incorrect Order of Subtraction: Integrating $(bottom-top)$ instead of (top - bottom)`. This will result in the negative of the correct area. The area of a region must be positive. - **Forgetting to Split the Integral:** If the curves cross, failing to find the intersection point and splitting the integral will lead to an incorrect answer. The single integral will calculate the *net* signed area between the curves, not the *total* geometric area. - **Using y-values as Limits:** The limits of integration for an integral with respect to $x (i.e., with a $dx$) must always be x-values.

• Algebraic Errors: Simple mistakes in solving $f(x)=g(x)$ for the intersection points or errors in finding the antiderivative are common on non-calculator sections.

• Premature Rounding: When using a calculator, do not round the intersection points before storing them and using them in the integral calculation. Store the full decimal values in your calculator to maintain accuracy.