Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-22, 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)= \begin{cases}\sqrt{1-x^{2}} & \text { if } 0 \leq x \leq 1 \\ x-1 & \text { if } 1

Step-by-Step Solution

Answer

The calculated integral is \( \frac{1}{4}\pi + \frac{1}{2} \).

Step 1: Sketch the Graph

First, let's analyze the function and sketch its graph. The function \( f(x) = \sqrt{1 - x^2} \) for \( 0 \leq x \leq 1 \) represents a semicircle in the first quadrant with radius 1, centered at the origin. For \( 1 < x \leq 2 \), the function \( f(x) = x - 1 \) represents a linear function, a straight line starting at \( (1, 0) \) and going up to \( (2, 1) \). Draw this graph to visualize the integrated regions.

Step 2: Calculate Area under the Semicircle

Identify that the area under \( f(x) = \sqrt{1 - x^2} \) from \( x = 0 \) to \( x = 1 \) is a quarter of a circle with radius 1. The area of a full circle is \( \pi r^2 \), which is \( \pi \cdot 1^2 = \pi \). So, the area under the semicircle is \( \frac{1}{4} \pi \).

Step 3: Calculate Area under the Line

Now, consider the linear function \( f(x) = x - 1 \) from \( x = 1 \) to \( x = 2 \). This forms a right triangle with a base 1 (from \( x = 1 \) to \( x = 2 \)) and height 1 (from value 0 to 1). The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \).

Step 4: Use the Interval Additive Property

Apply the interval additive property which states that the definite integral over an interval can be expressed as the sum of integrals over subintervals. Thus, the total area from \( x = 0 \) to \( x = 2 \) is the sum of the areas calculated. Add the area under the semicircle and the line: \( \frac{1}{4}\pi + \frac{1}{2} \).

Key Concepts

Interval Additive Property

The interval additive property is a fundamental principle used to compute definite integrals over a composite interval. It allows breaking down a single definite integral into smaller, more manageable sub-integrals.
This is particularly helpful when dealing with functions that change behavior or form over different intervals. To apply this property, consider a function defined over a combined range, such as \(a\) to \(c\). This interval can be divided into smaller parts, \([a, b]\) and \([b, c]\).
Here is how the property is applied:

• Calculate the integral over the first segment \(\int_a^b f(x) \, dx\).
• Calculate the integral over the second segment \(\int_b^c f(x) \, dx\).
• Sum these two results to get the total integral over the range \(\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\).

This process simplifies working with complex functions and is highly advantageous in problems like calculating areas under curves where different geometric shapes are involved.

Semicircle Area Calculation

Calculating the area under a semicircle within specific bounds is a common exercise in calculus, especially when the function represents a segment of a circle. A semicircle is half of a circle, and when considering a quadrant, it reduces further to a quarter.
For the function \(f(x) = \sqrt{1 - x^2}\), from \(x = 0\) to \(x = 1\), it's the equation of a semicircle centered at the origin with radius 1.The area of a complete circle with radius \(r\) is \(\pi r^2\). Since radius \(r = 1\), the area of the whole circle is \(\pi \).
Because we're considering a quarter of that circle (first quadrant), the area under the curve is \(\frac{1}{4} \pi\):

• Identify the function's form and its parameters (like radius).
• Calculate the whole shape's area, then determine the fraction of that shape needed.

This approach builds a foundation for handling more complex area computations involving circular shapes.

Linear Function Area

Determining the area under a linear function can often be simplified to computing the area of basic geometric shapes. In this scenario, for the linear function \(f(x) = x - 1\) over the interval \(x = 1\) to \(x = 2\), we deal with a right triangle.
The straight line represents a diagonal cutting through the coordinate plane.
Here's how to find the area under the line:

• Mark the base of the triangle from \(x = 1\) to \(x = 2\), yielding a length of 1.
• Locate the height from the y-value at \(x = 1\) (which is 0) to the y-value at \(x = 2\) (which is 1), providing a height of 1.
• The triangle area is calculated as \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).

Understanding how to relate simple geometric shapes aids in mastering the integration of linear functions.

Graph Sketching in Calculus

Sketching the graph of a function is a crucial step in understanding and solving problems related to integration. These visual aids help interpret the area under the curve and the behavior of the function over specified intervals.
For our composite function, it's essential to sketch two segments:- The semicircle represents \(f(x) = \sqrt{1-x^2}\), which is part of a circle with a radius of 1. It is limited to the range \(0 \leq x \leq 1\).- The linear function \(f(x) = x - 1\) forms a straight line from \(x = 1\) to \(x = 2\).Steps to sketch effectively:

• Identify key points where the function transitions—like endpoints of intervals.
• Label intersection points with axes and any significant symmetry or shape, such as circular parts.
• Accurately represent the scale and relationships between lines and curves.

Through practice, graph sketching becomes a powerful tool for visualizing mathematical information that enhances numerical calculations and conceptual understanding.