Area Between Two Parabolas Using Definite Integration

Concept Overview

This problem tests the application of definite integration to find the area of a region bounded by two curves. Specifically, it focuses on the scenario where both curves are parabolas. The key steps involve finding the points of intersection of the parabolas, which will serve as the limits of integration, and then integrating the difference between the 'upper' and 'lower' curves with respect to $x$.

Solution:

Step 1: Identify the equations of the two parabolas. Let the equations of the two parabolas be $y = f(x)$ and $y = g(x)$. For example, we might have $y = ax^2 + bx + c$ and $y = px^2 + qx + r$.

Step 2: Find the points of intersection of the two parabolas. To find the points where the parabolas intersect, we set their equations equal to each other: $f(x) = g(x)$ This will result in a quadratic equation in $x$. Solving this equation will give us the $x$-coordinates of the intersection points. Let these $x$-coordinates be $x = \alpha$ and $x = \beta$, where $\alpha < \beta$. These will be our limits of integration.

Step 3: Determine which parabola is 'upper' and which is 'lower' in the interval $[\alpha, \beta]$. In the interval between the intersection points, one parabola will lie above the other. To determine this, we can pick a test value of $x$ between $\alpha$ and $\beta$ (e.g., $x = (\alpha + \beta)/2$) and substitute it into both equations. The equation that yields a larger $y$-value corresponds to the 'upper' parabola, and the one yielding a smaller $y$-value corresponds to the 'lower' parabola. Alternatively, by sketching the graphs or analyzing the leading coefficients of the parabolas, we can often determine which is upper and which is lower. Let's assume $f(x) \ge g(x)$ for all $x$ in $[\alpha, \beta]$.

Step 4: Set up the definite integral for the area. The area $A$ enclosed between the two parabolas from $x = \alpha$ to $x = \beta$ is given by the integral of the difference between the upper curve and the lower curve: $A = \int_{\alpha}^{\beta} [f(x) - g(x)] dx$ This formula represents the sum of infinitesimal vertical strips of height $(f(x) - g(x))$ and width $dx$, integrated over the interval $[\alpha, \beta]$.

Step 5: Evaluate the definite integral. Calculate the indefinite integral of $[f(x) - g(x)]$. Let this be $F(x)$. Then, apply the Fundamental Theorem of Calculus: $A = F(\beta) - F(\alpha)$ This step involves standard integration techniques for polynomials.

Example: Find the area enclosed between the parabolas $y = x^2$ and $y = 2x - x^2$.

Step 1: The equations are $y = x^2$ and $y = 2x - x^2$.

Step 2: Find intersection points: $x^2 = 2x - x^2$ $2x^2 - 2x = 0$ $2x(x - 1) = 0$ So, $x = 0$ and $x = 1$. Thus, $\alpha = 0$ and $\beta = 1$.

Step 3: Determine upper and lower curves. Let's test $x = 0.5$: For $y = x^2$, $y = (0.5)^2 = 0.25$. For $y = 2x - x^2$, $y = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$. Since $0.75 > 0.25$, $y = 2x - x^2$ is the upper curve and $y = x^2$ is the lower curve in the interval $[0, 1]$. So, $f(x) = 2x - x^2$ and $g(x) = x^2$.

Step 4: Set up the integral: $A = \int_{0}^{1} [(2x - x^2) - x^2] dx$ $A = \int_{0}^{1} (2x - 2x^2) dx$

Step 5: Evaluate the integral: $A = \left[ x^2 - \frac{2x^3}{3} \right]_{0}^{1}$ $A = \left( (1)^2 - \frac{2(1)^3}{3} \right) - \left( (0)^2 - \frac{2(0)^3}{3} \right)$ $A = \left( 1 - \frac{2}{3} \right) - (0)$ $A = \frac{1}{3}$

Key Takeaways:

• The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by $\int_a^b |f(x) - g(x)| dx$.
• For enclosed areas, the limits of integration ($a$ and $b$) are the $x$-coordinates of the intersection points of the curves.
• It is crucial to identify which function is greater (the 'upper' curve) over the interval of integration to correctly set up the integrand as (upper function - lower function).

Answer: $\boxed{\frac{1}{3}}$