Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) What is the equation of a circle of radius 2 centered at the origin? (b) Write a function, complete with domain, that gives the equation of the top half of a circle of radius 2 centered at the origin. (c) Let \(f(x)=\left\\{\begin{array}{ll}\sqrt{4-x^{2}} & \text { for }-2 \leq x \leq 0 \text { , } \\ 2 x & \text { for } x>0\end{array}\right.\) Evaluate \(\int_{-2}^{2} f(x) d x\)

Step-by-Step Solution

Answer

The equation of the circle is \(x^{2}+y^{2}=4\). The equation of the top half of the circle is \(y = \sqrt{4 - x^{2}}\) with a domain [-2,2]. The integrated value of \(f(x)\) is \(\pi + 4\)

Step 1: Find the Equation of a Circle

The general equation of a circle with center at the origin (0,0) is given by \(x^{2}+y^{2}=r^{2}\). Here, r = 2, so the equation of the circle becomes \(x^{2}+y^{2}=4\)

Step 2: Write the Equation of the Top Half of Circle

For a circle of radius 2 centered at the origin, its offshoot function representing the upper half is obtained by solving the circle equation for \(y = \sqrt{r^{2} - x^{2}}\). Here, \(r = 2\) and \(x ∈ [-2,2]\). Therefore, the function is \(y = \sqrt{4 - x^{2}}\) with domain [-2,2]

Step 3: Compute the Definite Integral of the Given Function

The provided function \(f(x)\) is a piecewise function defined on two intervals, [-2,0] and (0,2]. So, the integral of \(f(x)\) from -2 to 2 is the sum of the definite integrals over these two intervals. Thus, \(\int_{-2}^{2} f(x)\,dx = \int_{-2}^{0} \sqrt{4 - x^{2}}\,dx + \int_{0}^{2} 2x\,dx\). The first integral is the area under the upper semicircle of radius 2 and the second one can be calculated using the power rule for integration. The result is \((\frac{\pi}{2} * 2^{2}) + [x^{2}]_{0}^{2} = \pi + 4\).

Key Concepts

Equation of a Circle

Understanding the equation of a circle is a cornerstone of geometry and calculus. The standard form of a circle's equation centered at the origin (0, 0) is given as \(x^2 + y^2 = r^2\). Here, \(r\) represents the radius of the circle. For example, if you have a circle with a radius of 2, the equation becomes \(x^2 + y^2 = 4\). This equation encapsulates every point (x, y) that lies on the circle's perimeter, meaning any point that satisfies this equation is exactly 2 units from the center (0,0).
It's essential to remember that

• \(r\) is always the radius, which is a constant value, measured from the center out to any point on the circle.
• Changing the center of the circle will alter the equation, often resulting in a form like \((x - h)^2 + (y - k)^2 = r^2\) for a center at (h, k).

Definite Integral

In calculus, the definite integral is a fundamental concept representing the accumulation of quantities, often viewed as the area under a curve within a specified interval. For a given interval \([a, b]\), the definite integral of a function \(f(x)\) is denoted as \(\int_a^b f(x)\,dx\). This symbol signifies the total area above the x-axis and below the graph of \(f(x)\), possibly with some corrections if the curve dips below the x-axis.
The calculation involves

• identifying the limits, here \(-2\) and \(2\), indicating where to start and end the integration,
• understanding whether the function spans multiple pieces, which might necessitate integrating each part separately, and
• applying integration rules such as the power rule or trigonometric integrals depending on the form of \(f(x)\).

In the given exercise, the definite integral evaluates an area split between two functions over distinct intervals.

Piecewise Functions

Piecewise functions are functions that have different expressions based on various intervals of the independent variable, often the x-axis. They enable us to describe more complex behavior by piecing together simpler functions. For the presented function \(f(x)\), it is defined piecewise as:

• \(\sqrt{4-x^2}\) for \(-2 \leq x \leq 0\)
• \(2x\) for \(x > 0\)

This breakdown allows the function to exhibit specific characteristics over different sections of its domain.
When working with such functions, always pay special attention to:

• The intervals that dictate which part of the function to use,
• How each piece aligns with real-world scenarios, and
• Any continuity or differentiability conditions that may apply at the boundary points between pieces.

Domains of Functions

Every function has a 'domain', which refers to the all the x-values (inputs) for which the function is defined. Identifying the domain is crucial as it tells you the limitations for the input values.
For example, in the problem's context:

• The function describing the top half of a circle is only defined for \(x\) between \(-2\) and \(2\), inclusive, which mirrors the circle's diameter length on the x-axis.
• When dealing with piecewise functions, each "piece" might have its own domain defined within the broader domain of the entire function, as observed with \(-2 \leq x \leq 0\) and \(x > 0\) in the piecewise example \(f(x)\).

Understanding domains is important for ensuring valid inputs and anticipating function behavior across its range.