Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the area bounded below by the \(x\) -axis, on the left by the \(y\) -axis, and above by \(y=x^{2}+1\) and \(y=-x+6\).

Step-by-Step Solution

Answer

The area of the region is approximately \(8.5417\) square units.

Step 1: Identify the intersection points

The two given functions are \(y=x^{2}+1\) and \(y=-x+6\). The first step should be setting these equal to each other in order to find the \(x\)-values at which the two curves intersect. Thus, we find \(x^{2}+1=-x+6\). Reorder the equation to \(x^{2}+x-5=0\). Solving this quadratic equation, \(x\) values are found to be \(2\) and \(-2.5\).

Step 2: Identify the bounded region

Next, draw a rough sketch of two curves and the axes to visualize the region. Identifying which function is the upper function and which is the lower is necessary for setting up the integrals. If we substitute \(x=0\) in both functions, we find that \(y=-x+6\) is above the curve \(y=x^{2}+1\) . So, the region is bounded by \(y=-x+6\) on the top, \(y=x^{2}+1\) on the bottom, \(x=2.5\) on the right and \(x=0\) on the left.

Step 3: Set up and compute the integral

Now that we've identified the upper and lower functions as well as the bounds, we can set up the integral: \(\int_0^{2.5} ((-x+6)-(x^{2}+1)) dx\). This computes the area of the region. To calculate this definite integral, find the anti-derivatives of each function, evaluate them at the two \(x\) boundaries, and subtract the results. The result of this integral gives the area of the region.

Step 4: Evaluate the integral and find the area of the region

The antiderivative of \(-x+6\) is \(-\frac{1}{2}x^2+6x\), and the antiderivative of \(x^{2}+1\) is \(\frac{1}{3}x^{3}+x\). So, \(\int_0^{2.5} ((-x+6)-(x^{2}+1)) dx = [-\frac{1}{2}x^{2}+6x - (\frac{1}{3}x^{3}+x)] \Bigg|_{0}^{2.5}\) , which simplifies to \([-\frac{1}{2}(2.5)^{2}+(6)(2.5) - (\frac{1}{3}(2.5)^{3}+(2.5))]\) and evaluates to \(8.5417\) square units.

Key Concepts

Intersection Points

Understanding intersection points is essential in finding the area bounded by curves. In this exercise, we have two curves:

• The quadratic function: \(y = x^2 + 1\)
• The linear function: \(y = -x + 6\)

The intersection points are where these two functions have the same value. To find them, we set the two equations equal to each other: \(x^2 + 1 = -x + 6\). This equation is a quadratic in form, and solving it reveals the x-values where the curves intersect. By rearranging it, we get: \(x^2 + x - 5 = 0\). This is a classic quadratic equation, and solving it yields \(x = 2\) and \(x = -2.5\) as the intersection points. Knowing these points helps in drawing the region to be calculated, which is essential for setting up the integral.

Definite Integral

The definite integral is a crucial tool for finding the area under curves. In this example, we aim to find the area under the intersection of two functions on a specific interval.

• Upper function: \(y = -x + 6\)
• Lower function: \(y = x^2 + 1\)

By evaluating the integral of the difference between the upper function and lower function over the interval from their intersection points, we compute the area between them. Our integral looks like this: \[ \int_{0}^{2.5} ((-x+6) - (x^2+1)) \, dx \] Without definite integrals, we'd struggle to find the accurate enclosed area, as integration allows for summing up the infinite, tiny sections of area under a curve within a range.

Quadratic Equation

Quadratic equations arise when we try to find intersection points involving at least one quadratic curve. Their standard form is \(ax^2 + bx + c = 0\). In this exercise, we encounter a quadratic equation when equating the two given functions: \(x^2 + 1 = -x + 6\). Rearranging gives us \(x^2 + x - 5 = 0\), a classic quadratic equation. Solving it can be achieved by various methods, including factoring, completing the square, or the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This quadratic reveals the intersection points, necessary for setting up the bounded region we're interested in when using the definite integral. Recognizing a quadratic equation and knowing how to solve it is foundational for understanding intersections in bounding curve areas.

Antiderivative

The antiderivative, or indefinite integral, is the reverse process of differentiation. It allows us to find a function which, when derived, gives the original function. When calculating the area between curves, the antiderivative is used in definite integrals. Here, we determined antiderivatives for:

• \(-x + 6\) becomes \(-\frac{1}{2}x^2 + 6x\)
• \(x^2 + 1\) becomes \(\frac{1}{3}x^3 + x\)

Using these, our integral calculation becomes: \([ -\frac{1}{2}x^2 + 6x - (\frac{1}{3}x^3 + x)] \Bigg|_0^{2.5}\). Evaluating the antiderivatives at the boundaries of the interval \([0, 2.5]\) gives the precise area of the bounded region. Understanding antiderivatives is key to solving many integration problems, particularly when calculating areas.