Question

In Exercises \(15-18,\) find the average value of the function on the interval without integrating, by appealing to the geometry of the region between the graph and the \(x\) -axis. $$f ( \theta ) = \tan \theta , \quad \left[ - \frac { \pi } { 4 } , \frac { \pi } { 4 } \right]$$

Short answer

The average value of the function \(f(\theta) = \tan(\theta)\) on the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\) is \(\frac{1}{2}\).

Step-by-step solution

Identify the geometric shape formed by the function

On the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\), the function \(f(\theta) = \tan(\theta)\) forms a triangle. Its vertices are the points where the function intersects the \(x\)-axis and the points at the ends of the interval.

Calculate the base and height of the triangle

The base of the triangle is the length of the interval, which is \( \frac{\pi}{4} - ( -\frac{\pi}{4}) = \frac{\pi}{2}\). The height of the triangle is the maximum value that the function takes on the interval, which happens at \(\frac{\pi}{4}\). So, the height is \(f(\frac{\pi}{4}) = \tan(\frac{\pi}{4}) = 1\).

Calculate the area of the triangle

The area of a triangle is given by the formula \(\frac{1}{2} \times \text{base} \times \text{height}\). Substituting the values from Step 2, the area is \(\frac{1}{2} \times \frac{\pi}{2} \times 1 = \frac{\pi}{4}\).

Find the average value of the function

The average value of a function over an interval can be found by dividing the area under the curve (in this case, the area of the triangle) by the length of the interval. The length of the interval is \(\frac{\pi}{2}\), so the average value is \(\frac{\frac{\pi}{4}}{\frac{\pi}{2}} = \frac{1}{2}\).

Key concepts

Geometry and Integration

The beautiful synergy between geometry and integration is seen in various mathematical problems, and it often simplifies the process of finding solutions. Integration, a core concept in calculus, is fundamentally about adding up an infinite number of infinitesimally small quantities to find things like the area under a curve. In certain problems, however, the shape formed by the curve and the axes can be a simple, familiar geometric figure, such as a rectangle or, as in our given exercise, a triangle.

In cases like these, rather than performing a complex integration, we can appeal to geometry to calculate areas. The area under a curve over a given interval can represent the total sum (integral) of the values of the function across that interval. When the integral curve forms a simple geometric shape, we can employ geometric formulas to calculate this area directly. This approach not only makes the problem more tangible but also serves as a fantastic reminder of how interconnected different areas of mathematics are.

Area of a Triangle

The area of a triangle is a fundamental concept in geometry that arises in many areas of mathematics and science. It's given by the simple formula \[ A = \frac{1}{2} \times \text{base} \times \text{height} \. \] This formula can be visualized as taking the rectangle that encompasses the triangle and cutting it in half, along the height.

In the case of our exercise, where the function \(f(\theta) = \tan(\theta)\) creates a right-angled triangle, calculating its area becomes straightforward when we identify the base and height. These measurements are directly related to the function's behavior on the given interval. The ability to connect the behavior of a trigonometric function to the area of a geometric figure exemplifies how geometric intuition can simplify and solve calculus problems without performing actual integration.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are the backbone of many mathematical subjects, including geometry, trigonometry, calculus, and even complex areas such as Fourier analysis.

The function \(\tan(\theta)\), as seen in our exercise, is the ratio of sine to cosine and therefore provides a relationship connecting angles to ratios of side lengths in a right triangle. These functions exhibit interesting symmetries and periodic behaviors that can be immensely useful in various mathematical applications. For instance, the tangent function has a period of \(\pi\), meaning that it repeats its values every \(\pi\) units along the \(\theta\) axis.

Understanding the properties of trigonometric functions can aid in visualizing the graph of the function and, as a result, determining the shape it forms within a given interval. This understanding is crucial when calculating areas geometrically as it forms the bridge between abstract functions and concrete shapes.