AP Calculus BC 8.6 Finding the Area Between Curves That Intersect at More Than Two Points Study Notes - New Syllabus
AP Calculus BC 8.6 Finding the Area Between Curves That Intersect at More Than Two Points Study Notes- New syllabus
AP Calculus BC 8.6 Finding the Area Between Curves That Intersect at More Than Two Points Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Finding the Area Between Curves That Intersect at More Than Two Points
Finding the Area Between Curves That Intersect at More Than Two Points
Finding the Area Between Curves That Intersect at More Than Two Points
When two curves intersect at more than two points, the region between them may be split into multiple parts. In such cases, we cannot simply use a single integral; instead, we must divide the area into separate sections where the same function is consistently the upper (or rightmost) function.
The “top” or “rightmost” curve can change between intervals, so the integral must be split at each intersection point where this switch happens.
General Approach:
- Solve \( f(x) = g(x) \) to determine the \( x \)-coordinates where the curves meet.
- For each interval between successive intersection points, identify which curve is above by testing a sample point.
- The area for each interval is given by \( A_i = \displaystyle \int_{x_{i}}^{x_{i+1}} \big[ \text{(top function)} – \text{(bottom function)} \big] \, dx \)
- Since area is always positive, take the sum of all \( A_i \) values.
Total Area Formula:
\( A_{\text{total}} = \sum_{i=1}^{n} \displaystyle \int_{x_i}^{x_{i+1}} \big| f(x) – g(x) \big| \, dx \)
Steps to Solve:
- Sketch both curves and mark all intersection points.
- Determine the order of the curves (which is on top/right) between each pair of intersection points.
- Set up separate integrals for each interval, using the correct top/right minus bottom/left function.
- Add the absolute values of these integrals to find the total enclosed area.
Example:
Find the total area between \( y = x^2 \) and \( y = |x| \).
▶️ Answer/Explanation
Find intersections.
For \( y = x^2 \) and \( y = |x| \):
- Case 1: \( x \geq 0 \) → \( x^2 = x \) → \( x(x – 1) = 0 \) → \( x = 0, 1 \)
- Case 2: \( x < 0 \) → \( x^2 = -x \) → \( x(x + 1) = 0 \) → \( x = 0, -1 \)
Intersection points: \( (-1, 1), (0, 0), (1, 1) \).
Determine top function in each section.
- For \( -1 \leq x \leq 0 \), top is \( y = -x \), bottom is \( y = x^2 \).
- For \( 0 \leq x \leq 1 \), top is \( y = x \), bottom is \( y = x^2 \).
Set up the total area.
\( A = \displaystyle \int_{-1}^{0} \left[ -x – x^2 \right] dx + \displaystyle \int_{0}^{1} \left[ x – x^2 \right] dx \)
Evaluate.
First integral: \( \left[ -\dfrac{x^2}{2} – \dfrac{x^3}{3} \right]_{-1}^{0} = \left( 0 – 0 \right) – \left( -\dfrac{1}{2} + \dfrac{1}{3} \right) = \dfrac{1}{2} – \dfrac{1}{3} = \dfrac{1}{6} \)
Second integral: \( \left[ \dfrac{x^2}{2} – \dfrac{x^3}{3} \right]_{0}^{1} = \left( \dfrac{1}{2} – \dfrac{1}{3} \right) – (0) = \dfrac{1}{6} \)
Total area: \( \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3} \)
Example:
Find the total area between \( y = \sin x \) and \( y = \cos x \) on \( [0, 2\pi] \).
▶️ Answer/Explanation
Find intersections.
\( \sin x = \cos x \) → \( \tan x = 1 \) → \( x = \dfrac{\pi}{4}, \dfrac{5\pi}{4} \) within \( [0, 2\pi] \).
Determine top function in each section.
- On \( [0, \dfrac{\pi}{4}] \), \( \cos x > \sin x \).
- On \( [\dfrac{\pi}{4}, \dfrac{5\pi}{4}] \), \( \sin x > \cos x \).
- On \( [\dfrac{5\pi}{4}, 2\pi] \), \( \cos x > \sin x \).
Set up the total area.
\( A = \displaystyle \int_{0}^{\pi/4} \left[ \cos x – \sin x \right] dx + \displaystyle \int_{\pi/4}^{5\pi/4} \left[ \sin x – \cos x \right] dx + \displaystyle \int_{5\pi/4}^{2\pi} \left[ \cos x – \sin x \right] dx
Evaluate each integral.
\( \int \left[ \cos x – \sin x \right] dx = \sin x + \cos x \)
First: \( \left[ \sin x + \cos x \right]_{0}^{\pi/4} = \left( \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \right) – (0 + 1) = \sqrt{2} – 1 \)
Second: \( \left[ -\cos x – \sin x \right]_{\pi/4}^{5\pi/4} = \left( -\left( -\dfrac{\sqrt{2}}{2} \right) – \dfrac{\sqrt{2}}{2} \right) – \left( -\dfrac{\sqrt{2}}{2} – \dfrac{\sqrt{2}}{2} \right) = \sqrt{2} \)
Third: \( \left[ \sin x + \cos x \right]_{5\pi/4}^{2\pi} = \left( 0 + 1 \right) – \left( -\dfrac{\sqrt{2}}{2} – \dfrac{\sqrt{2}}{2} \right) = 1 + \sqrt{2} \)
Add them.
Total area = \( (\sqrt{2} – 1) + \sqrt{2} + (1 + \sqrt{2}) = 3\sqrt{2} \)