Problem 48 - Calculus Early Transcendentals

Question

Use the most efficient strategy for computing the area of the following regions. The region bounded by \(y=x^{2}-4,4 y-5 x-5=0,\) and \(y=0,\) for \(y \geq 0\)

Step-by-Step Solution

Answer

Question: Determine the area of the region bounded by the equations \(y=x^2-4\), \(4y-5x-5=0\), and \(y=0\). Answer: The area of the region bounded by the given equations is 2 square units.

Step 1: Sketch and identify region

We start by sketching the three given curves on the same coordinate axis: 1. Quadratic equation: \(y=x^2-4\) 2. Linear equation: \(4y-5x-5 = 0\) or \(y=\frac{5}{4}x-\frac{5}{4}\) 3. Horizontal line: \(y=0\) Now, we can visualize the enclosed region in which we need to find the area. Moreover, we can see that the area to be found is in the first quadrant, since we know that \(y \geq 0\).

Step 2: Find intersection points

To determine the interval for which we need to find the area, we need to find the intersection points of the given curves. Let's find the intersection point between the quadratic equation and the linear equation. To do this, we set \(y=x^2-4\) equal to \(y=\frac{5}{4}x-\frac{5}{4}\) and solve for \(x\). This gives us the following equation: \(x^2-4=\frac{5}{4}x-\frac{5}{4}\) We now need to solve for \(x\) by rearranging the equation and factoring if necessary: \(x^2-\frac{5}{4}x-3=0\) Let's use the quadratic formula to find the roots of this equation, since it is a quadratic equation: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) Here, \(a=1\), \(b=-\frac{5}{4}\), and \(c=-3\). Plugging these values into the formula, we find two values for \(x\): \(x=\frac{(\frac{5}{4})+\sqrt{(\frac{5}{4})^2-4(1)(-3)}}{2(1)}=2\) and \(x=\frac{(\frac{5}{4})-\sqrt{(\frac{5}{4})^2-4(1)(-3)}}{2(1)}=-1\) Now, we need to find the corresponding \(y\) values for each of these \(x\) values. We can find the y-values by plugging in these x-values into any of the original equations. We'll use the quadratic equation here. For \(x=2\), \(y=2^2-4=0\) For \(x=-1\), \(y=(-1)^2-4=-3\) However, since we are only interested in the region for \(y\geq 0\), we will disregard the point \((-1,-3)\). So, the area we need to find lies between the intersection points of the quadratic and linear equations where \(x=2\).

Step 3: Integrate to find the area

To find the area of the region bounded by these curves, we will use integration. Since we know that both the quadratic and linear equations intersect at the point \((2,0)\), we can set up an integration to find the area between these curves: \(A=\int_{a}^{b} (top - bottom)dx\) In this case, "top" refers to the higher (linear) equation, and "bottom" refers to the lower (quadratic) equation. The integration limits are determined by the intersection point \(x=2\). Note that one of the boundaries is the x-axis where \(y=0\). The area we want to find is in first quadrant, so we need to take into account that the y values of the quadratic equation are negative in this region. Thus, our integral becomes: \(A=\int_{0}^{2} \left( \frac{5}{4}x-\frac{5}{4}-(x^2-4) \right) dx\) Next, we simplify the integrand: \(A=\int_{0}^{2} \left(\frac{5}{4}x-\frac{5}{4}-x^2+4\right) dx\) Now, we integrate each term of the integrand with respect to \(x\): \(A=\left[\frac{5}{8}x^2-\frac{5}{4}x-\frac{1}{3}x^3+4x\right]_{0}^{2}\) Finally, we evaluate the integral by plugging in the limits: \(A=\left[\frac{5}{8}(2)^2-\frac{5}{4}(2)-\frac{1}{3}(2)^3+4(2)\right]\) After simplification, we find: \(A=2\) Thus, the area of the region bounded by the given curves is 2 square units.

Key Concepts

Integration

Integration is a fundamental concept in calculus that allows us to find areas, volumes, central points, and many other helpful quantities. The integral of a function gives us the accumulation of quantities represented by the function. Think of integration as the reverse process of differentiation. Here, we integrate to find the area bounded by the given curves.

In general, integration involves taking a function and summing it over an interval. For instance, if you want to determine the area under the curve of a function, you integrate that function between the defined limits. The process involves:

• Identifying your integrand (the function to be integrated).
• Establishing your limits of integration (where you start and stop summing).
• Computing the integral using those limits. The result is usually the net area if the region incorporates negatives.

Whether calculating a single integral, like in our problem, or a more complex double or triple integral, the procedure remains fundamentally about summing a function over a specified domain.

Area between curves

Finding the area between curves is a practical application of integration. This process helps determine the space trapped between two or more curves on a graph. In our scenario, calculating the area involves finding the region enclosed by the quadratic and linear equations.

To determine this area, you need to identify:

• Top Curve: This is the curve represented by the function positioned above the other. In our example, it's the linear function.
• Bottom Curve: This is the curve that lies below the upper curve. Here, it's the quadratic function.
• Integration Limits: These are determined by the points where the curves intersect, as identified by solving the equations simultaneously.

Once these components are defined, you set up an integral where you subtract the "bottom" function from the "top" and integrate over the limits given by the intersections. This process allows you to efficiently calculate the desired area, taking into account the portions of the curves involved.

Intersection points

Intersection points are key to many calculus problems, especially those involving area calculations between curves. These points represent where two or more curves meet within a given coordinate system. Identifying them is crucial to correctly setting the bounds for integration.

To find the intersection points, follow these steps:

• Set the equations of the functions equal to each other. This algebraic step determines where the curves cross each other.
• Solve for the variable in question, typically "x" in a Cartesian plane.
• Validate these solutions by substituting back into either of the original equations to find the corresponding "y" values.
• Ensure only the points applicable to the problem constraints (like "y ≥ 0" in this situation) are considered.

In our case, resolving these intersection points provided the specific interval for integration, ensuring accuracy in calculating the area between the specified curves. Intersection points serve as the anchor for understanding how two graphs interrelate spatially.