Areas Between Curves Detailed Study Notes

Extending the Integral: Beyond calculating the area under a single curve relative to the x-axis, integrals can be used to determine the area precisely trapped between two or more functions.
Visualizing the Region: Consider two functions, $y = f(x)$ and $y = g(x)$ . The goal is to find the area of the region bounded by their graphs.
The Subtraction Principle:
- The area under the higher function ( $f(x)$ , for example) encapsulates the total area down to the x-axis.
- The area under the lower function ( $g(x)$ ) is a subset of that total area.
- By subtracting the smaller area from the larger area, you are left with the area trapped specifically between the two curves.
The Representative Integral: If two functions intersect at points $x = a$ and $x = b$ , and $f(x)$ is the top function throughout that interval, the area is defined as: $\text{Area} = \int_{a}^{b} [f(x) - g(x)] \, dx$

Top Minus Bottom: Rather than strictly using $f$ and $g$ , it is more practical to frame the formula as: $\text{Area} = \int_{a}^{b} [\text{Top Function}(x) - \text{Bottom Function}(x)] \, dx$
Importance of Intersection: The limits of integration ( $a$ and $b$ ) are the x-coordinates where the functions intersect. Solving for these points is almost always the necessary first step if they are not provided.

Functions: $y = x^2$ and $y = 2x$ .
Solving for Intersection: Set the equations equal to each other to find the bounds:
- $x^2 = 2x$
- $x^2 - 2x = 0$
- $x(x - 2) = 0$
- Solutions: $x = 0$ and $x = 2$ .
Determining Top vs. Bottom: Graphing these functions reveals that between $x = 0$ and $x = 2$ , the line $y = 2x$ sits above the parabola $y = x^2$ .
Integration Setup: $\text{Area} = \int_{0}^{2} (2x - x^2) \, dx$
Evaluation:
- Antiderivative: $x^2 - \frac{x^3}{3}$
- Plug in limits: $(2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
- Calculation: $4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$
Result: The area is $\frac{4}{3}$ . This value is positive because it denotes a physical region rather than a "signed area."

Functions: $y = 2^x$ , $y = 4$ , and $x = 1$ .
Visualizing the Region: The region is a triangular shape bounded on the top by the horizontal line $y = 4$ , on the bottom by the exponential curve $y = 2^x$ , and on the left by the vertical line $x = 1$ .
Intersection Points:
- Where $2^x$ intersects $y = 4$ : $2^x = 4 \implies x = 2$ . Point: $(2, 4)$ .
- Where $x = 1$ intersects $y = 4$ : Point $(1, 4)$ .
- Where $x = 1$ intersects $y = 2^x$ : $y = 2^1 = 2$ . Point $(1, 2)$ .
Integration Setup: The area goes from $x = 1$ to $x = 2$ . The top function is $y = 4$ and the bottom is $y = 2^x$ . $\text{Area} = \int_{1}^{2} (4 - 2^x) \, dx$
Evaluation:
- Antiderivative: $4x - \frac{2^x}{\ln(2)}$
- Plug in limits: $(4(2) - \frac{2^2}{\ln(2)}) - (4(1) - \frac{2^1}{\ln(2)})$
- Simplification: $(8 - \frac{4}{\ln(2)}) - (4 - \frac{2}{\ln(2)})$
- Final Answer: $4 - \frac{2}{\ln(2)}$

Complex Interactions: Sometimes curves intersect multiple times, causing the "top" function to switch. In these cases, the total area must be calculated in separate pieces.
Example Functions: $y = 4x - x^2$ (parabola) and $y = x^3 - x^2$ (cubic).
Solution for Intersections:
- $x^3 - x^2 = 4x - x^2$
- $x^3 - 4x = 0$
- $x(x^2 - 4) = 0$
- $x(x - 2)(x + 2) = 0$
- Intersections occur at $x = -2, 0, 2$ . Points: $(-2, -12), (0, 0), (2, 4)$ .
Testing Top/Bottom Without a Detailed Graph:
- Interval $(-2, 0)$ : Pick $x = -1$ . Cubic evaluates to $-2$ ; Parabola evaluates to $-5$ . Since $-2 > -5$ , the Cubic is the top function.
- Interval $(0, 2)$ : Pick $x = 1$ . Cubic evaluates to $0$ ; Parabola evaluates to $3$ . Since $3 > 0$ , the Parabola is the top function.
Total Area Calculation: $\text{Area} = \int_{-2}^{0} [(x^3 - x^2) - (4x - x^2)] \, dx + \int_{0}^{2} [(4x - x^2) - (x^3 - x^2)] \, dx$ $\text{Area} = \int_{-2}^{0} (x^3 - 4x) \, dx + \int_{0}^{2} (4x - x^3) \, dx$
Evaluation:
- First Integral: $[\frac{x^4}{4} - 2x^2]_{-2}^{0} = (0) - (\frac{16}{4} - 2(4)) = -(4 - 8) = 4$
- Second Integral: $[2x^2 - \frac{x^4}{4}]_{0}^{2} = (2(4) - \frac{16}{4}) - (0) = 8 - 4 = 4$
- Total: $4 + 4 = 8$ .

Theoretical Shift: Instead of using vertical rectangles ( $dx$ ) with width $\Delta x$ , area can be approximated using horizontal rectangles ( $dy$ ) with width determined by the difference between an x-value on the right and an x-value on the left.
Formula: $\text{Area} = \int_{c}^{d} [\text{Rightmost Function}(y) - \text{Leftmost Function}(y)] \, dy$
When to Use $dy$ : It is often easier when the "top" and "bottom" relationships change frequently across the x-axis, but the "right" and "left" relationships remain constant across the y-axis.
Example Case: $y = \sqrt{x}$ , $y = 6 - x$ , and $y = 1$ .
- Finding Intersections:
  - $\sqrt{x} = 6 - x \implies x = (6 - x)^2 \implies x = 36 - 12x + x^2$
  - $x^2 - 13x + 36 = 0 \implies (x - 9)(x - 4) = 0$
  - Check for extraneous solutions: $x = 9 \rightarrow \sqrt{9} = 3$ vs $6 - 9 = -3$ (No). $x = 4 \rightarrow \sqrt{4} = 2$ vs $6 - 4 = 2$ (Yes). Intersection at $(4, 2)$ .
- The x-method (Split): Requires splitting the area at $x = 4$ into two integrals from $x = 1$ to $x = 4$ and from $x = 4$ to $x = 5$ .
- The y-method (Single Integral):
  - Rewrite functions as functions of $y$ : $x = y^2$ and $x = 6 - y$ .
  - Bounds: The bottom line is at $y = 1$ ; the top intersection is at $y = 2$ .
  - Setup: $\int_{1}^{2} [(6 - y) - y^2] \, dy$
  - Evaluation: $[6y - \frac{y^2}{2} - \frac{y^3}{3}]_{1}^{2} = (12 - 2 - \frac{8}{3}) - (6 - 0.5 - \frac{1}{3}) = \frac{13}{6}$ .

Curves: $x = y^2$ and $y = 1.5x - 1.5$ .
Intersections: Substituting $x = y^2$ into the line equation and solving yields $y = 3, y = -1$ . Points are $(9, 3)$ and $(1, -1)$ .
Method A: dy Integral (Easier)
- Rewrite line as $x = 2y + 3$ (Note: specific algebra provided in context).
- Limits: $y \in [-1, 3]$ .
- Setup: $\int_{-1}^{3} [(2y + 3) - y^2] \, dy$
- Result: $\frac{32}{3}$ .
Method B: dx Integral (Harder)
- Requires splitting at $x = 1$ .
- From $x = 0$ to $x = 1$ : The area is between the top radical $y = \sqrt{x}$ and the bottom radical $y = -\sqrt{x}$ .
- From $x = 1$ to $x = 9$ : The area is between the top radical $y = \sqrt{x}$ and the bottom line $y = 1.5x - 1.5$ .
- Setup: $\int_{0}^{1} (\sqrt{x} - (-\sqrt{x})) \, dx + \int_{1}^{9} (\sqrt{x} - (1.5x - 1.5)) \, dx$
- Result: Yields the same value of $\frac{32}{3}$ but with significantly more complex calculation steps.

Question: What if I cannot draw the picture accurately or at all?
Response: You do not need a perfect picture. The algebraic solution for intersection points dictates where the area starts and stops. By picking test points within those intervals and plugging them into the original functions, you can mathematically determine which function is on top and which is on bottom without a visual aid.