Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each of the following, sketch a graph of the indicated region and write a de nite integral (or, if you prefer, the sum and/or differences of de nite integrals) that gives the area of the region. (a) The area between the horizontal line \(y=4\) and the parabola \(y=x^{2}\) (b) The area between the line \(y=x+1\) and the parabola \(y=x^{2}-1\)

Step-by-Step Solution

Answer

For the area between \(y=4\) and \(y=x^{2}\), the integral is \(\int_{-2}^{2} (4 - x^{2}) dx\). For the area between \(y=x+1\) and \(y=x^{2}-1\), the integral is \(\int_{-1}^{2}(x+1 - (x^{2}-1)) dx\).

Step 1: Sketch Graph for Area (a)

Plot the curves \(y=4\) and \(y=x^{2}\) on the same Cartesian plane. The region of interest is where \(y=x^{2}\) is below \(y=4\).

Step 2: Find Intersection Points for Area (a)

To find the limits of integration, solve \(4 = x^{2}\) to get \(x = 2\) and \(x = -2\). These are the x-values where the two functions intersect.

Step 3: Calculate Area (a)

Integrate the difference between the functions \(4 - x^{2}\) from \(-2\) to \(2\). This result will give the area between the two curves. In mathematical terms: \(\int_{-2}^{2} (4 - x^{2}) dx\).

Step 4: Sketch Graph for Area (b)

Plot the curves \(y=x+1\) and \(y=x^{2}-1\) on the same Cartesian plane. The region of interest is where \(y=x+1\) is above \(y=x^{2}-1\).

Step 5: Find Intersection Points for Area (b)

To find the limits of integration, solve \(x+1 = x^{2}-1\) to get \(x = -1\) and \(x = 2\). These are the x-values where the two functions intersect.

Step 6: Calculate Area (b)

Integrate the absolute difference between the functions \(x+1\) and \(x^{2}-1\) from \(-1\) to \(2\). This result will give the area between the two curves. In mathematical terms: \(\int_{-1}^{2}(x+1 - (x^{2}-1)) dx\).

Key Concepts

Intersection Points

Intersection points are where two or more curves meet at the same coordinate on a graph. Identifying these points is essential for determining where one function ends and another begins across the axis of interest. This step helps define the limits of integration necessary for calculating areas between curves. For example, consider the line \(y = 4\) and the parabola \(y = x^2\). To find the intersection points, you set their equations equal: \(4 = x^2\). Solving this equation involves finding the values of \(x\) that satisfy this equality. In this case, solving \(4 = x^2\) results in \(x = 2\) and \(x = -2\). Each intersection point marks the boundary for the region where one function is positioned above or below the other. Finding intersection points is equally important when dealing with other pairs of functions, such as \(y = x + 1\) and \(y = x^2 - 1\). Setting these equal \(x + 1 = x^2 - 1\), we simplify and find \(x = -1\) and \(x = 2\). These points determine the segment of the graph we are interested in for calculating the respective areas.

Area Between Curves

Calculating the area between curves involves determining the space that lies between two or more functions graphed on the same axes. This area represents how much space is enclosed between the curves, typically expressed in square units. Here's a simple guide on how to find these areas:

• First, identify which function is on top and which is on the bottom along the selected interval.
• Calculate the difference between the top function and the bottom function across the interval of interest, which is determined by the intersection points.

For instance, with the set of functions \(y = 4\) and \(y = x^2\), \(y = 4\) is always above \(y = x^2\) between the intersection points \(-2\) and \(2\). The area can then be expressed with the definite integral \(\int_{-2}^{2} (4 - x^2) \, dx\), where the integrand \((4 - x^2)\) represents the vertical distance between the curves. Likewise, for \(y = x + 1\) and \(y = x^2 - 1\), you calculate the area by integrating the difference \(x + 1\) and \(x^2 - 1\) from \(x = -1\) to \(x = 2\). Since the line \(y = x + 1\) spans above the parabola \(y = x^2 - 1\) in this range, you perform \(\int_{-1}^{2} (x + 1 - (x^2 - 1)) \, dx\). The result of these integrations provides the total area between each pair of curves.

Integral Bounds

Integral bounds are the limits between which you evaluate a definite integral for finding the area between curves. These are numerical values, often determined by the intersection points of the curve expressions, marking where to start and stop the integration. By determining where the curves intersect, you also find the x-values for your bounds. For instance, with curves \(y = 4\) and \(y = x^2\), the bounds are \(-2\) and \(2\). Therefore, you integrate from \(x = -2\) to \(x = 2\). In another example, where the curves are \(y = x + 1\) and \(y = x^2 - 1\), solving for their intersection similarly provides bounds \(x = -1\) and \(x = 2\). These bounds ensure that you're calculating the area only over the section of the graph where both functions interact. It is crucial to use the correct integral bounds to ensure an accurate calculation of areas. Using incorrect limits will extend or reduce the interval beyond the actual region between the curves, resulting in miscalculation.