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Area Between Two Curves - AP Calculus AB Rundown
Introduction
Finding the area between two curves is a common application of definite integrals in AP Calculus AB. The core idea is to integrate the difference between the two functions over a specified interval. This rundown will cover the essential concepts, formulas, and techniques you need to master this topic.
The Fundamental Idea
The area between two curves, $ f(x) $ and $ g(x) $ , over an interval $ [a, b] $ is found by integrating the absolute difference between the functions. It is important to find which function is on top to ensure that the result is positive.
Formula
The area, $ A $ , between the curves $ f(x) $ and $ g(x) $ from $ x = a $ to $ x = b $ is given by:
$$ A = \int_{a}^{b} |f(x) - g(x)| , dx
$$
Where $ f(x) $ and $ g(x) $ are continuous functions on the interval $ [a, b] $ . If $ f(x) \geq g(x) $ on $ [a, b] $ , then the formula simplifies to:
$$ A = \int_{a}^{b} [f(x) - g(x)] , dx
$$
Key Point: $ f(x) $ is the “top” function and $ g(x) $ is the “bottom” function within the interval $ [a, b] $ .
Steps to Calculate Area Between Curves
- Identify the Functions: Determine the equations of the two curves, $ f(x) $ and $ g(x) $ .
- Find the Intersection Points: Determine the interval of integration $ [a, b] $ by finding the $ x $ -coordinates of the points where the two curves intersect. This is done by setting $ f(x) = g(x) $ and solving for $ x $ . These solutions are your limits of integration, $ a $ and $ b $ . Finding Intersection Points
- Determine Which Function is on Top: Determine which function has a greater value within the interval $ [a, b] $ . Choose a test value $ c $ such that $ a < c < b $ . Evaluate $ f(c) $ and $ g(c) $ . The larger value is the “top” function. Determining the Top Function
- Set up the Integral: Set up the definite integral with the appropriate limits of integration and the difference between the top and bottom functions.
- Evaluate the Integral: Evaluate the definite integral to find the area. This might require u-substitution or other integration techniques.
Examples
Example 1
Find the area between the curves $ f(x) = x^2 $ and $ g(x) = x $ from $ x = 0 $ to $ x = 1 $ .
- Functions: $ f(x) = x^2 $ and $ g(x) = x $
- Interval: $ [0, 1] $ (Given)
- Top Function: On the interval $ [0, 1] $ , $ g(x) = x $ is greater than $ f(x) = x^2 $ . For example, at $ x = 0.5 $ , $ f(0.5) = 0.25 $ and $ g(0.5) = 0.5 $ .
- Integral:
$$ A = \int_{0}^{1} (x - x^2) , dx
$$ 5. Evaluation:
$$ A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \left( \frac{1}{2} - \frac{1}{3} \right) - \left( 0 - 0 \right) = \frac{1}{6}
$$
Therefore, the area between the curves is $ \frac{1}{6} $ .
Example 2
Find the area between the curves $ f(x) = x^2 - 4 $ and $ g(x) = x - 2 $ .
- Functions: $ f(x) = x^2 - 4 $ and $ g(x) = x - 2 $
- Intersection Points: $$ x^2 - 4 = x - 2 \ x^2 - x - 2 = 0 \ (x - 2)(x + 1) = 0 \ x = 2, x = -1 $$ Interval: $ [-1, 2] $
- Top Function: On the interval $ [-1, 2] $ , $ g(x) = x - 2 $ is greater than $ f(x) = x^2 - 4 $ . For example, at $ x = 0 $ , $ f(0) = -4 $ and $ g(0) = -2 $ .
- Integral:
$$ A = \int_{-1}^{2} [(x - 2) - (x^2 - 4)] , dx = \int_{-1}^{2} (-x^2 + x + 2) , dx
$$ 5. Evaluation:
$$ A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} = \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = -\frac{8}{3} + 2 + 4 - \frac{1}{3} - \frac{1}{2} + 2 = 8 - 3 - \frac{1}{2} = 5 - \frac{1}{2} = \frac{9}{2}
$$
Therefore, the area between the curves is $ \frac{9}{2} $ .
Integrating with Respect to $ y $
Sometimes, it’s easier to integrate with respect to $ y $ if the functions are easily expressible in terms of $ y $ , i.e., $ x = f(y) $ and $ x = g(y) $ . In this case, the area is:
$$ A = \int_{c}^{d} |f(y) - g(y)| , dy
$$
Where $ c $ and $ d $ are the $ y $ -coordinates of the intersection points, and $ f(y) $ and $ g(y) $ are the functions of $ y $ . $ f(y) $ is the “right” function and $ g(y) $ is the “left” function.
To find the intersection points of two curves