Calculus: Areas Between Curves and Integration Techniques

Foundations: Review of the Definite Integral

In Calculus I, the area under a curve is defined as a definite integral. This concept is derived through the following steps:

• The Region: Consider a region under a curve $f(x)$ on the interval $[a, b]$.

• Representative Rectangle:

  • We draw a rectangle within the region with a thickness of $\Delta x$.

  • A sample point $\bar{x}_i$ is chosen within each sub-interval.

  • The width of each rectangle is determined by $\Delta x = \frac{b-a}{n}$, where $n$ is the number of partitions (rectangles).

• Area of a Representative Rectangle: The height of the rectangle is specified by the function value at the sample point, $f(x_i)$. The area is calculated as:     $\text{Area of one rectangle} = f(x_i) \times \Delta x$

• Approximation by Summation: As we add multiple rectangles, we get an approximation of the total area:     $\text{Area} \approx \sum_{i=1}^{n} f(x_i) \Delta x$

• The Integral Definition: By taking the limit as the number of rectangles $n$ approaches infinity, the sum becomes a definite integral:     $\text{Area} = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_a^b f(x) \,dx$

• Differential Operator: The term $dx$ represents the differential operator associated with the integration process.

Determining the Area Between Curves

The logic used to find the area under a single curve can be extended to find the area bounded between two curves, $f(x)$ and $g(x)$.

• Methodology:

  • Draw a representative rectangle between the two curves.

  • The thickness remains $\Delta x$.

  • The height of the rectangle is no longer just $f(x_i)$, but the difference between the upper function value and the lower function value.

• Area of the Representative Rectangle:     $\text{Area} = (f(x_i) - g(x_i)) \Delta x$

• General Formula for Area ($S$): If we take the limit as $n \to \infty$, the area between the curves from $a$ to $b$ is:     $S = \int_a^b (f(x) - g(x)) \,dx$

• Requirement: This formula holds true as long as $f(x) \geq g(x)$ over the entire interval $[a, b]$. In simple terms:     $\text{Area} = \int_{\text{lower limit}}^{\text{upper limit}} (\text{upper function} - \text{lower function}) \,dx$

Example 1: Area Bounded by $y = x$ and $y = x^2$

• Identifying the Region: To find the limits of integration ($a$ and $b$), we solve for the points of intersection between the line and the parabola.

• Solving the System:

  • Set the equations equal: $x = x^2$

  • Rearrange: $x^2 - x = 0$

  • Factor: $x(x - 1) = 0$

  • Solutions: $x = 0$ and $x = 1$. Therefore, $a = 0$ and $b = 1$.

• Determining the Upper Function: Over the interval $[0, 1]$, the line $y = x$ is above the parabola $y = x^2$.

• Setting up the Integral:     $S = \int_0^1 (x - x^2) \,dx$

• Integration and Evaluation:

  • Antiderivative: $[\frac{1}{2}x^2 - \frac{1}{3}x^3]_0^1$

  • Calculation: $(\frac{1}{2}(1)^2 - \frac{1}{3}(1)^3) - (\frac{1}{2}(0)^2 - \frac{1}{3}(0)^3)$

  • Final Result: $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Example 2: Area Bounded by $y = x^3 - x$ and $y = 8x$

• Analysis of the Cubic Function:

  • Factor the cubic: $x(x^2 - 1) = x(x - 1)(x + 1)$.

  • The intercepts are $-1, 0, 1$.

• Finding Intersections with $y = 8x$:

  • Set equations equal: $x^3 - x = 8x$

  • Rearrange: $x^3 - 9x = 0$

  • Factor: $x(x^2 - 9) = x(x + 3)(x - 3) = 0$

  • Intersections occur at $x = -3, 0, 3$.

• Defining the Regions:

  • Region $S_1$ ($x ∈ [-3, 0]$): The cubic $x^3 - x$ is the upper function and $8x$ is the lower function.         $S_1 = \int_{-3}^0 (x^3 - x - 8x) \,dx = \int_{-3}^0 (x^3 - 9x) \,dx$

  • Region $S_2$ ($x ∈ [0, 3]$): The line $8x$ is the upper function and $x^3 - x$ is the lower function.         $S_2 = \int_0^3 (8x - (x^3 - x)) \,dx = \int_0^3 (9x - x^3) \,dx$

• Calculation and Symmetry:

  • $S_1 = 20.25$

  • $S_2 = 20.25$

  • The area is symmetric because both functions are odd functions.

• Total Area: $S = S_1 + S_2 = 20.25 + 20.25 = 40.5$

Example 3: Periodic Area Between $y = \sin(x)$ and $y = \sin(2x)$

• Visualizing the Graphs:

  • $y = \sin(x)$ has a period of $2\pi$ and an amplitude of $1$.

  • $y = \sin(2x)$ has its period adjusted by the coefficient $2$, resulting in a period of $\frac{2\pi}{2} = \pi$.

• Finding Intersection Points:

  • Set equal: $\sin(x) = \sin(2x)$.

  • Use the Double-Angle Identity: $\sin(2x) = 2\sin(x)\cos(x)$.

  • Equation: $\sin(x) = 2\sin(x)\cos(x)$.

  • Rearrange and Factor: $2\sin(x)\cos(x) - \sin(x) = 0 \Rightarrow \sin(x)(2\cos(x) - 1) = 0$.

  • Case 1 ($\sin(x) = 0$): $x = 0, \pi, 2\pi$.

  • Case 2 ($2\cos(x) - 1 = 0 \Rightarrow \cos(x) = \frac{1}{2}$): $x = \frac{\pi}{3}, \frac{5\pi}{3}$.

• Calculation of Specific Regions (from $0$ to $2\pi$):

  • $S_1$: Interval $[0, \frac{\pi}{3}]$         $S_1 = \int_0^{\pi/3} (\sin(2x) - \sin(x)) \,dx = [-\frac{1}{2}\cos(2x) + \cos(x)]_0^{\pi/3} = \frac{1}{4}$

  • $S_2$: Interval $[\frac{\pi}{3}, \pi]$         $S_2 = \int_{\pi/3}^{\pi} (\sin(x) - \sin(2x)) \,dx = \frac{9}{4}$

  • $S_3$: Interval $[\pi, \frac{5\pi}{3}]$         $S_3 = \int_{\pi}^{5\pi/3} (\sin(2x) - \sin(x)) \,dx = \frac{9}{4}$

  • $S_4$: Interval $[\frac{5\pi}{3}, 2\pi]$         $S_4 = \int_{5\pi/3}^{2\pi} (\sin(x) - \sin(2x)) \,dx = \frac{1}{4}$

• Summary: The infinite number of repeated enclosed regions have areas that are either $\frac{1}{4}$ or $\frac{9}{4}$.

Integrating with Respect to $y$

When functions are given as $x$ in terms of $y$, it is often easier to integrate along the y-axis (horizontal rectangles).

• Example Case: $x = -\frac{1}{2}y + 1$ (a line) and $x = \frac{1}{2}y^2$ (a right-opening parabola).

• Intersections:

  • Set equal: $-\frac{1}{2}y + 1 = \frac{1}{2}y^2$

  • Multiply by $2$: $-y + 2 = y^2$

  • Factor: $y^2 + y - 2 = (y + 2)(y - 1) = 0$

  • Interval: $y \in [-2, 1]$.

• Horizontal Slice Logic:

  • Representative rectangles have a height of $dy$.

  • Area is calculated as: $\int_{\text{bottom}}^{\text{top}} (\text{right function} - \text{left function}) \,dy$

• Setup:     $S = \int_{-2}^1 ((-\frac{1}{2}y + 1) - (\frac{1}{2}y^2)) \,dy$

• Result: The area calculated through these same integration techniques is $2.25$.

Example 5: Area of a Triangle Using Calculus

• Points of the Triangle: $(0, 0)$, $(2, 5)$, and $(5, 2)$.

• Finding Line Equations:

  • Line 1 ($L_1$): Connects $(0, 0)$ and $(2, 5)$. Slope $m = \frac{5}{2}$. Equation: $y = \frac{5}{2}x$.

  • Line 2 ($L_2$): Connects $(2, 5)$ and $(5, 2)$. Slope $m = \frac{2-5}{5-2} = -1$. Using point-slope: $y - 5 = -1(x - 2) \Rightarrow y = -x + 7$.

  • Line 3 ($L_3$): Connects $(0, 0)$ and $(5, 2)$. Slope $m = \frac{2}{5}$. Equation: $y = \frac{2}{5}x$.

• Dividing the Area into Regions:

  • Region $S_1$: Interval $[0, 2]$

    • Upper function: $L_1$ ($\frac{5}{2}x$). Lower function: $L_3$ ($\frac{2}{5}x$).         $S_1 = \int_0^2 (\frac{5}{2}x - \frac{2}{5}x) \,dx = 4.2$

  • Region $S_2$: Interval $[2, 5]$

    • Upper function: $L_2$ ($-x + 7$). Lower function: $L_3$ ($\frac{2}{5}x$).         $S_2 = \int_2^5 ((-x + 7) - \frac{2}{5}x) \,dx = 6.3$

• Total Triangle Area: $S = S_1 + S_2 = 4.2 + 6.3 = 10.5$.