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

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

The definite integral ${\int }_a^{b}f(x)dx$ provides a way to calculate the net area between the curve of a function $f(x)$ and the $x$-axis over an interval $[a,b]$. This topic extends that fundamental concept to a more general application: finding the area of a region enclosed between two distinct curves.

The central principle is to conceptualize this area as an infinite sum of the areas 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 $dx$. By integrating this height, which is the difference between the two functions, over a specified interval, we can find the exact area of the region bounded by them. This method relies on expressing the bounding curves as functions of $x$ and integrating with respect to $x$.

Key Formulas

The area of a region bounded by curves is calculated using a definite integral. The specific formula depends on the orientation of the curves.

Area Between Two Curves (Functions of $x$)

If a region is bounded above by the curve $y=f(x)$ and below by the curve $y=g(x)$ over the interval $[a,b]$, where $f(x)\ge g(x)$ for all $x$ in $[a,b]$, the area $A$ of the region is given by:

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

• $f(x)$: The "upper" or "top" function. Its graph forms the upper boundary of the region.

• $g(x)$: The "lower" or "bottom" function. Its graph forms the lower boundary of the region.

• $[a,b]$: The interval on the $x$-axis over which the area is being calculated. These bounds are often the $x$-coordinates of the points where the curves intersect, or they may be explicitly given.

• $(f(x)-g(x))$: This expression represents the height of a representative vertical rectangle at any given value of $x$ in the interval.

Understanding Intersecting Curves

A critical nuance arises when the two curves, $y=f(x)$ and $y=g(x)$, intersect or cross one another within the interval of interest. When this occurs, the function that serves as the "upper" boundary changes, and so does the "lower" boundary.

According to the Essential Knowledge (CHA-5.A.2, CHA-5.A.3), a single integral is insufficient to calculate the total area in such cases. The region must be divided into sub-regions at each point of intersection. The total area is then found by calculating the area of each sub-region separately and summing the results.

For example, if $f(x)$ and $g(x)$ intersect at $x=c$ within the interval $[a,b]$, and:

• $f(x)\ge g(x)$ on $[a,c]$

• $g(x)\ge f(x)$ on $[c,b]$

Then the total area is the sum of two integrals:

$$A_{\text{total}}={\int }_a^c[f(x)-g(x)]dx+{\int }_c^b[g(x)-f(x)]dx$$

Notice that the order of subtraction within the integrand is reversed for the second interval to ensure the height of the representative rectangle, $(upper-lower)$, is always a positive quantity.

Core Concepts & Rules

• Area as a Definite Integral: The area of a region bounded by curves expressed as functions of $x$ is found by integrating with respect to $x$.

• The Integrand: The integrand must be the expression $(upperfunction)-(lowerfunction)$. This ensures the result is a positive value representing area.

• Determining Bounds: The limits of integration, $a$ and $b$, are the leftmost and rightmost $x$-values of the region. They are either provided in the problem statement or must be found by setting the functions equal to each other and solving for the $x$-coordinates of their intersection points.

• Handling Intersections: If the curves cross, the region must be split into multiple sub-regions. A separate integral is required for each sub-region, and the total area is the sum of the results of these integrals.

• Calculator Use: Technology can be used to find intersection points (the bounds of integration) and to evaluate the resulting definite integrals numerically.

Step-by-Step Example 1: Basic Application

Problem: Find the area of the region enclosed by the graphs of $y=5-x^2$ and $y=x-1$.

Step 1: Find the Bounds of Integration

The bounds are the $x$-coordinates of the intersection points. Set the two functions equal to each other and solve for $x$.

$$5-x^2=x-1$$

$$0=x^2+x-6$$

$$0=(x+3)(x-2)$$

The points of intersection occur at $x=-3$ and $x=2$. Therefore, our interval of integration is $[-3,2]$.

Step 2: Identify the Upper and Lower Functions

Choose a test value of $x$ within the interval $[-3,2]$, such as $x=0$.

• For $y=5-x^2$: $y(0)=5-0^2=5$.

• For $y=x-1$: $y(0)=0-1=-1$.

Since $5>-1$, the function $y=5-x^2$ is the upper function, and $y=x-1$ is the lower function on this interval.

Step 3: Set Up the Definite Integral

Using the formula $A={\int }_a^b[\text{upper}(x)-\text{lower}(x)]dx$:

$$A={\int }_{-3}^2[(5-x^2)-(x-1)]dx$$

Step 4: Simplify and Integrate

First, simplify the integrand. Be careful with the parentheses.

$$A={\int }_{-3}^2(5-x^2-x+1)dx$$

$$A={\int }_{-3}^2(-x^2-x+6)dx$$

Now, find the antiderivative:

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

Step 5: Evaluate Using the Fundamental Theorem of Calculus

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

$$A=\left(-\frac{8}{3}-\frac{4}{2}+12\right)-\left(-\frac{-27}{3}-\frac{9}{2}-18\right)$$

$$A=\left(-\frac{8}{3}-2+12\right)-\left(9-\frac{9}{2}-18\right)$$

$$A=\left(10-\frac{8}{3}\right)-\left(-9-\frac{9}{2}\right)$$

$$A=\left(\frac{30-8}{3}\right)-\left(\frac{-18-9}{2}\right)$$

$$A=\frac{22}{3}-\left(-\frac{27}{2}\right)=\frac{22}{3}+\frac{27}{2}$$

$$A=\frac{44}{6}+\frac{81}{6}=\frac{125}{6}$$

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

Problem: Find the total area of the region(s) bounded by the graphs of $f(x)=x^3-3x$ and $g(x)=x$.

Step 1: Find All Intersection Points

Set the functions equal to find all points where the regions begin or end.

$$x^3-3x=x$$

$$x^3-4x=0$$

$$x(x^2-4)=0$$

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

The intersections occur at $x=-2$, $x=0$, and $x=2$. This indicates that the area must be calculated over two separate intervals: $[-2,0]$ and $[0,2]$.

Step 2: Identify Upper and Lower Functions on Each Sub-Interval

• Interval 1: $[-2,0]$

  • Choose a test point, e.g., $x=-1$.

  • $f(-1)=(-1{)}^3-3(-1)=-1+3=2$.

  • $g(-1)=-1$.

  • On $[-2,0]$, $f(x)\ge g(x)$. The upper function is $x^3-3x$.

• Interval 2: $[0,2]$

  • Choose a test point, e.g., $x=1$.

  • $f(1)=(1{)}^3-3(1)=1-3=-2$.

  • $g(1)=1$.

  • On $[0,2]$, $g(x)\ge f(x)$. The upper function is $x$.

Step 3: Set Up the Sum of Two Integrals

The total area is the sum of the areas of the two regions.

$$A_{\text{total}}={\int }_{-2}^0[(x^3-3x)-(x)]dx+{\int }_0^2[(x)-(x^3-3x)]dx$$

Step 4: Simplify and Integrate

• First Integral:

  $$A_1={\int }_{-2}^0(x^3-4x)dx={\left[\frac{x^4}{4}-2x^2\right]}_{-2}^0$$

  $$A_1=(0)-\left(\frac{(-2{)}^4}{4}-2(-2{)}^2\right)=-\left(\frac{16}{4}-2(4)\right)=-(4-8)=4$$

• Second Integral:

  $$A_2={\int }_0^2(-x^3+4x)dx={\left[-\frac{x^4}{4}+2x^2\right]}_0^2$$

  $$A_2=\left(-\frac{(2{)}^4}{4}+2(2{)}^2\right)-(0)=\left(-\frac{16}{4}+2(4)\right)=-4+8=4$$

Step 5: Calculate the Total Area

$$A_{\text{total}}=A_1+A_2=4+4=8$$

Using Your Calculator

Technology can be used to find the area of a region, which is especially useful when intersection points are not simple integers or when the functions are difficult to integrate by hand.

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

Step 1: Graph the Functions

• Enter $Y_1=cos(X)$ and $Y_2=X^2-2$ into your calculator's graphing utility.

• Visually inspect the graph. You will see two intersection points. Note that $Y_1$ is the upper function and $Y_2$ is the lower function for the enclosed region.

Step 2: Find the Bounds of Integration

• Use your calculator's intersection-finding tool (e.g., 2nd -> TRACE [CALC] -> 5: intersect).

• Find the left intersection point. Let's call it $a$. Store this value in your calculator's memory (e.g., $STO->A$). You should find $a\approx -1.4546$.

• Find the right intersection point. Let's call it $b$. Store this value (e.g., $STO->B$). You should find $b\approx 1.4546$.

Step 3: Set Up and Evaluate the Integral

• The area is given by the integral ${\int }_a^b(\cos (x)-(x^2-2))dx$.

• Use your calculator's numerical integration function (e.g., MATH -> 9: fnInt( on a TI-84).

• Enter the expression as follows, using the stored variables for the bounds to maintain precision:

  fnInt(Y1 - Y2, X, A, B)

• The calculator will evaluate the integral and return the area. The result is approximately $5.816$.

AP Exam Quick Hit

Common Question Types

• Calculator-Active Area Problem: You will be given two complex functions (e.g., involving logarithms, exponentials, or complicated polynomials) and asked to find the area of the region between them. The solution requires using a calculator to find the intersection points and evaluate the definite integral.

  • Example: "Let $R$ be the region enclosed by the graphs of $f(x)=e^{x/2}-1$ and $g(x)=\sin (x^2)$. Find the area of $R$."

• Area with Multiple Intersections (Non-Calculator): You will be given two simpler functions (polynomials, basic trig) that intersect multiple times. You must find all intersection points algebraically and set up a sum of two or more integrals to find the total area.

  • Example: "Find the area of the region bounded by $y=\sin (x)$ and $y=\sin (2x)$ on the interval $[0,\pi ]$."

• Area with Given Bounds: You will be asked to find the area between two curves on a specified interval $[a,b]$, which may or may not correspond to the natural intersection points. You must check for intersections within the given interval.

  • Example: "Find the area of the region between $y=x^2$ and $y=x+2$ on the interval $[0,3]$."

Common Mistakes

• Incorrectly Identifying Upper/Lower Functions: Integrating $lower-upper$ will result in the negative of the correct area. Always test a point in the interval or analyze the graphs to confirm which function has the greater value.

• Forgetting to Split the Integral: If the curves cross, integrating $f(x)-g(x)$ over the entire interval will incorrectly calculate a net area, where some parts cancel out, rather than the total area. The integral must be split at each intersection point.

• Parentheses Errors in Subtraction: A very common algebraic mistake is failing to distribute the negative sign when subtracting the lower function. For example, writing $\int (5-x^2-x-1)dx$ instead of the correct $\int (5-x^2-(x-1))dx$.

• Errors in Finding Bounds: Simple algebraic mistakes when solving for intersection points will lead to incorrect limits of integration and a completely wrong final answer, even if the integral setup and evaluation are otherwise perfect.

• Prematurely Rounding: On calculator-active questions, do not round the intersection points before using them as the limits of integration. Store the full decimal values in your calculator's memory and use the stored values for the final calculation. Round only the final answer to the required number of decimal places.