AP Calculus BC 8.4 Finding the Area Between Curves Expressed as Functions of x Study Notes - New Syllabus
AP Calculus BC 8.4 Finding the Area Between Curves Expressed as Functions of x Study Notes- New syllabus
AP Calculus BC 8.4 Finding the Area Between Curves Expressed as Functions of x 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 Expressed as Functions of x
Finding the Area Between Curves Expressed as Functions of x
Area Under the Curve and the x-axis
The definite integral \( \displaystyle \int_{a}^{b} f(x) \, dx \) measures the net signed area between the graph of \( y = f(x) \) and the x-axis from \( x = a \) to \( x = b \).
- If \( f(x) \geq 0 \) on \( [a, b] \), the definite integral gives the actual area between the curve and the x-axis.
- If \( f(x) \leq 0 \) on \( [a, b] \), the definite integral is negative; the actual area is the absolute value of the result.
- If \( f(x) \) changes sign in \( [a, b] \), split the interval at the points where \( f(x) = 0 \) and compute each part separately, taking absolute values.
Total Area Formula:
\(A_{\text{total}} = \displaystyle \int_{a}^{b} |f(x)| \, dx \)
Procedure:
- Find where \( f(x) = 0 \) to locate the points where the curve crosses the x-axis.
- Split the integration range into intervals where \( f(x) \) keeps the same sign.
- Integrate \( |f(x)| \) over each sub-interval and sum to get the total area.
Example:
Find the total area enclosed between \( y = x^2 – 4 \) and the x-axis.
▶️ Answer/Explanation
Find where
\( x^2 – 4 = 0 \) → \( x = -2, 2 \).
Between
\( -2 \) and \( 2 \), \( x^2 – 4 \leq 0 \), so the curve is below the x-axis.
Total area:
\( A_{\text{total}} = \displaystyle \int_{-2}^{2} |x^2 – 4| \, dx \)
Since \( x^2 – 4 \leq 0 \), \(|x^2 – 4| = -(x^2 – 4) = 4 – x^2\).
Evaluate:
\( A_{\text{total}} = \displaystyle \int_{-2}^{2} (4 – x^2) \, dx \)
\( = \left[ 4x – \dfrac{x^3}{3} \right]_{-2}^{2} = \left(8 – \dfrac{8}{3}\right) – \left(-8 + \dfrac{8}{3}\right) \)
\( = \dfrac{16}{1} – \dfrac{16}{3} = \dfrac{32}{3} \)
Final Answer: \( \displaystyle \dfrac{32}{3} \) units²
Finding the Area Between Curves Expressed as Functions of x
When we have two curves, \( y = f(x) \) and \( y = g(x) \), the area between them over an interval \([a, b]\) can be found by integrating the difference of the functions. The function on top minus the function on the bottom gives the height of a thin vertical strip, and integration sums all these strips across the interval.
Formula:
\( \text{Area} = \displaystyle \int_{a}^{b} \left[ f(x) – g(x) \right] \, dx \)
Here, \( f(x) \) is the upper function and \( g(x) \) is the lower function over the interval.
Steps to Solve:
- Sketch or visualize the curves to determine which is on top over the given interval.
- Find points of intersection by setting \( f(x) = g(x) \) to determine integration limits (if not given).
- Set up the integral of the top function minus the bottom function over the interval.
- Integrate and evaluate to find the total area.
Example:
Find the area between \( y = x^2 \) and \( y = x + 2 \) from \( x = -1 \) to \( x = 2 \).
▶️ Answer/Explanation
The top function is \( y = x + 2 \) and the bottom is \( y = x^2 \) over the given interval.
\( \text{Area} = \displaystyle \int_{-1}^{2} \left[ (x+2) – (x^2) \right] dx \)
\( = \int_{-1}^{2} (-x^2 + x + 2) \, dx \)
\( = \left[ -\dfrac{x^3}{3} + \dfrac{x^2}{2} + 2x \right]_{-1}^{2} \)
At \( x = 2 \): \(-\dfrac{8}{3} + 2 + 4 = -\dfrac{8}{3} + 6 = \dfrac{10}{3}\)
At \( x = -1 \): \(\dfrac{1}{3} + \dfrac{1}{2} – 2 = \dfrac{5}{6} – 2 = -\dfrac{7}{6}\)
Area \( = \dfrac{10}{3} – (-\dfrac{7}{6}) = \dfrac{20}{6} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2} \ \text{square units}\)
Example:
Find the area between \( y = \sqrt{x} \) and \( y = x – 2 \) where they intersect.
▶️ Answer/Explanation
First, solve \( \sqrt{x} = x – 2 \).
Let \( t = \sqrt{x} \), so \( t^2 = x \). Then \( t = t^2 – 2 \) gives \( t^2 – t – 2 = 0 \) ⇒ \( (t-2)(t+1) = 0 \).
So \( t = 2 \) (since \( t \ge 0 \)), meaning \( \sqrt{x} = 2 \) ⇒ \( x = 4 \).
Also check \( t = -1 \) is not valid for \(\sqrt{x}\).
Find other intersection: check \( x = 1 \) ⇒ \(\sqrt{1} = 1\), but \( 1 – 2 = -1 \), so not equal. Actually, set \(\sqrt{x} = x – 2\): Square both sides ⇒ \( x = x^2 – 4x + 4 \) ⇒ \( 0 = x^2 – 5x + 4 \) ⇒ \( (x-4)(x-1) = 0 \) ⇒ \( x = 1 \) and \( x = 4 \).
On [1,4], \( y = \sqrt{x} \) is above \( y = x-2 \).
Area:
\( \displaystyle \int_{1}^{4} \left[ \sqrt{x} – (x – 2) \right] dx \)
\( = \int_{1}^{4} \left[ x^{1/2} – x + 2 \right] dx \)
\( = \left[ \dfrac{2}{3}x^{3/2} – \dfrac{x^2}{2} + 2x \right]_{1}^{4} \)
At \( x = 4 \): \(\dfrac{2}{3}(8) – 8 + 8 = \dfrac{16}{3}\)
At \( x = 1 \): \(\dfrac{2}{3}(1) – \dfrac{1}{2} + 2 = \dfrac{2}{3} – \dfrac{1}{2} + 2 = \dfrac{4}{6} – \dfrac{3}{6} + \dfrac{12}{6} = \dfrac{13}{6}\)
Area = \( \dfrac{16}{3} – \dfrac{13}{6} = \dfrac{32}{6} – \dfrac{13}{6} = \dfrac{19}{6} \ \text{square units}\)
Example:
Find the total area between \( y = x^2 – 4 \) and \( y = 0 \) over \( x \in [-3, 3] \).
▶️ Answer/Explanation
The curves intersect where \( x^2 – 4 = 0 \) ⇒ \( x = -2, 2 \).
On [-3, -2] and [2, 3], \( x^2 – 4 \) is positive (above x-axis). On [-2, 2], \( x^2 – 4 \) is negative, so area is computed as absolute value.
Total area:
\( \text{Area} = \int_{-3}^{-2} (x^2 – 4) dx – \int_{-2}^{2} (x^2 – 4) dx + \int_{2}^{3} (x^2 – 4) dx \)
By symmetry, compute one side and double it.
Area = \( 2\left[ \int_{2}^{3} (x^2 – 4) dx + \int_{0}^{2} (4 – x^2) dx \right] \)
First: \( \int_{2}^{3} (x^2 – 4) dx = \left[ \dfrac{x^3}{3} – 4x \right]_{2}^{3} = \left( 9 – 12 \right) – \left( \dfrac{8}{3} – 8 \right) = -3 – \left( -\dfrac{16}{3} \right) = -3 + \dfrac{16}{3} = \dfrac{7}{3} \)
Second: \( \int_{0}^{2} (4 – x^2) dx = \left[ 4x – \dfrac{x^3}{3} \right]_{0}^{2} = 8 – \dfrac{8}{3} = \dfrac{16}{3} \)
Total area = \( 2\left[ \dfrac{7}{3} + \dfrac{16}{3} \right] = 2\left( \dfrac{23}{3} \right) = \dfrac{46}{3} \ \text{square units}\)