Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(\int_{a}^{b} f(x) d x\), where a and \(b\) are the left and right end points for which f is defined, by using the Interval Additive Property and the appropriate area formulas from plane geometry. Begin by graphing the given function. $$ f(x)=\left\\{\begin{array}{ll} \sqrt{1-x^{2}} & \text { if } 0 \leq x \leq 1 \\ x-1 & \text { if } 1

Step-by-Step Solution

Answer

\(\frac{\pi}{4} + \frac{1}{2}\)

Step 1: Understand the Given Piecewise Function

The function \( f(x) \) is piecewise defined. The first part, \( \sqrt{1-x^2} \), is valid for \( 0 \leq x \leq 1 \), resembling a semicircle on the upper half of the unit circle centered at the origin. The second part, \( x-1 \), is a linear function valid for \( 1 < x \leq 2 \). We will need to consider these intervals separately for integration.

Step 2: Graph the Function

For \( 0 \leq x \leq 1 \), graph the semicircle given by \( y = \sqrt{1-x^2} \), which reaches a maximum at \( x=0 \) with \( y=1 \) and is \( y=0 \) at \( x=1 \). For \( 1 < x \leq 2 \), graph the line \( y = x-1 \), starting at the point \( (1, 0) \) and ending at \( (2, 1) \). These graphs will guide the integration process.

Step 3: Integrate Over [0, 1] Using Geometry

The region under \( y = \sqrt{1-x^2} \) from \( 0 \) to \( 1 \) is a quarter-circle of radius 1. The area of this quarter-circle can be computed as \( \frac{1}{4} \times \pi \times 1^2 = \frac{\pi}{4} \). This result comes from the area formula for a circle, \( \pi r^2 \), applied to a quarter of the circle.

Step 4: Integrate Over (1,2] Using Geometry

The integral over \( y = x - 1 \) from \( 1 \) to \( 2 \) forms a right triangle under the line segment. The base runs from \( x=1 \) to \( x=2 \) (length 1), and the height from \( y=0 \) to \( y=1 \) (height 1). The area of the triangle is \( \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \).

Step 5: Add the Areas

Using the Interval Additive Property, add the areas from both intervals to find the total area under the curve from \( 0 \) to \( 2 \). This is \( \frac{\pi}{4} + \frac{1}{2} \).

Key Concepts

Piecewise Functions

Piecewise functions are essential in mathematics for modeling situations where a rule changes over different intervals. In our case, the function \( f(x) \) is made up of two parts, each defined over a specific domain interval:

• \( \sqrt{1-x^2} \) is applicable when \( 0 \leq x \leq 1 \) and represents a semicircle.
• \( x-1 \) applies when \( 1 < x \leq 2 \) and forms a simple linear equation or line.

Understanding these intervals is crucial because each segment of the function is analyzed separately. When solving problems with piecewise functions, always pay attention to where each part is defined. This ensures accurate calculations and interpretations.
When dealing with definite integrals of piecewise functions, we handle each segment independently and then combine the results, respecting the boundaries where the function's definition changes.

Area Under a Curve

Calculating the area under a curve gives us the integral of a function over a specified interval. For piecewise functions, it's important to understand that the area calculation needs to be adapted to each segment of the function separately.
In the given problem, for \( 0 \leq x \leq 1 \), the curve is part of a semicircle. Graphically, this segment's area can be computed using the known area formula for a circle: \( \pi r^2 \). Since it's a quarter of a full circle with radius 1, the area is \( \frac{\pi}{4} \).
Meanwhile, for the interval \( 1 < x \leq 2 \), the graph of \( x-1 \) is linear and can be handled using the geometry of triangles. The area of the right triangle present is found using \( \frac{1}{2} \times \text{base} \times \text{height} \), resulting in an area of \( \frac{1}{2} \).

• This approach teaches that visualizing functions geometrically can simplify integration efforts significantly.
• It also emphasizes breaking down complex functions into understandable components.

Geometric Integration

Geometric integration utilizes geometric shapes to find definite integrals, making it particularly useful for functions that form recognizably simple shapes.
Recognizing parts of a curve as circles, rectangles, or triangles can allow us to skip algebraic integration for a more visual and intuitive method.
In our exercise:

• The semicircle \( \sqrt{1-x^2} \) was handled by recognizing it as a quarter-circle, using its area formula \( \frac{1}{4} \pi \times \text{radius}^2 \).
• The linear segment \( x-1 \) for \( 1 < x \leq 2 \) was treated as a right triangle, simplifying the calculation to \( \frac{1}{2} \times \text{base} \times \text{height} \).

This method is not only simpler but sometimes the only practical approach if the function is complex. It also provides valuable insights into the characteristics of functions by analyzing their shapes.