Problem 66 - University Calculus: Early Transcendentals

Question

Find the areas of the regions enclosed by the lines and curves in Exercises \(63-72\). $$y=x^{2}-2 x \quad \text { and } \quad y=x$$

Step-by-Step Solution

Answer

The area is 4.5 square units.

Step 1: Find Intersection Points

To find the intersection points of the curves, set the equations equal to each other: \(x^2 - 2x = x\). Rearranging gives \(x^2 - 3x = 0\). Factor out \(x\) to get \(x(x-3) = 0\), which results in intersection points at \(x = 0\) and \(x = 3\).

Step 2: Determine the Area to Integrate

The area between the two curves from \(x = 0\) to \(x = 3\) can be calculated by integrating the difference between the two functions: \(\int_{0}^{3} (x - (x^2 - 2x)) \, dx\). Simplify the integrand to \(\int_{0}^{3} (3x - x^2) \, dx\).

Step 3: Integrate the Function

Calculate the integral \(\int_{0}^{3} (3x - x^2) \, dx\). First, find the antiderivative: \(\frac{3}{2}x^2 - \frac{1}{3}x^3\). Evaluate this from 0 to 3: \[\left(\frac{3}{2}(3)^2 - \frac{1}{3}(3)^3\right) - \left(\frac{3}{2}(0)^2 - \frac{1}{3}(0)^3\right)\].

Step 4: Evaluate the Definite Integral

Calculate: \[\frac{3}{2}(9) - \frac{1}{3}(27) = \frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2} = 4.5\]. The area of the region enclosed by the curves is \(4.5\).

Key Concepts

Definite Integral

Understanding the concept of a **definite integral** is crucial when you want to find the area between curves. A definite integral provides a way to calculate the accumulation of values, often representing the area under a curve when integrated over a certain interval. Unlike indefinite integrals which provide antiderivatives, definite integrals have upper and lower limits which signify specific points on the x-axis.When calculating the area between two curves, you subtract the lower curve from the upper curve within the limits of integration. This effectively gives you the vertical distance between the two curves at any given x-value within the interval. By accumulating these distances through integration, you can find the total area enclosed.In the context of the given exercise, we calculated the definite integral from 0 to 3 of the difference between the linear equation \(y = x\) and the quadratic equation \(y = x^2 - 2x\). This involved:

• Setting up the integral with the correct functions and limits.
• Simplifying the integrand, \((3x - x^2)\).
• Integrating term-by-term to find the antiderivative.
• Evaluating the integral at the upper and lower limits to find the area.

Thus, using a definite integral helps transform the problem of finding the area into a manageable calculation.

Intersection Points

Finding the **intersection points** between two curves is a critical step when determining the area enclosed by them. These points give you the boundaries of integration, the x-values where the curves meet and switch dominance, ensuring you correctly set your limits for the definite integral.To find intersection points, you set the equations of the curves equal to each other and solve for x. This essentially finds where both curves share the same y-value, indicating they intersect. In our given problem:

• We equaled \(x^2 - 2x\) to \(x\), the two provided functions.
• We rearranged the equation to \(x^2 - 3x = 0\), a standard quadratic form.
• Factoring the equation yielded solutions at \(x = 0\) and \(x = 3\).

These points, \(x = 0\) and \(x = 3\), defined the interval over which we integrated. Without these crucial points, it would be impossible to accurately calculate the area between the curves as specified.

Area Between Curves

Calculating the **area between curves** is a fascinating application of definite integrals, allowing you to determine the size of regions enclosed by different mathematical functions. The process involves comparing two curves over a specified interval and measuring the total space between them.Once the intersection points are determined, the next steps are well-defined:

• Identify which curve is above the other in your interval. This affects the order in the integral because \(\int (f(x) - g(x)) \, dx\) is used, where \(f(x)\) is the upper curve.
• Calculate the definite integral of this difference across the intersection points.
• Fully evaluate the resulting expression after finding antiderivatives.

In the exercise, the linear function \(y = x\) was above the quadratic function \(y = x^2 - 2x\) throughout the interval from 0 to 3, resulting in the integral \(\int_{0}^{3} (3x - x^2) \, dx\). Once integrated, this expression gave us the total area of 4.5 square units, representing the measure of the region enclosed by the two curves.