Problem 15 - Calculus Single Variable

Question

A function \(f\) is defined piecewise on an interval \(I=[a, b] .\) Find the area of the region that is between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\left\\{\begin{array}{ll} \sin (x) & \text { if } 0 \leq x \leq \pi / 4 \\ \cos (x) & \text { if } \pi / 4

Step-by-Step Solution

Answer

The area is \(1 - \sqrt{2}\).

Step 1: Understand the Problem and Identify the Intervals

We need to find the area under the curve of the piecewise function given on the interval \([0, \pi]\). The function is defined as \(f(x) = \sin(x)\) for \(0 \leq x \leq \frac{\pi}{4}\) and \(f(x) = \cos(x)\) for \(\frac{\pi}{4} < x \leq \pi\). This means we have two separate areas to consider.

Step 2: Calculate the Area for the First Interval

For \(0 \leq x \leq \frac{\pi}{4}\), the function is \(f(x) = \sin(x)\). The area under this curve from \(x = 0\) to \(x = \frac{\pi}{4}\) is given by the integral: \[A_1 = \int_{0}^{\frac{\pi}{4}} \sin(x) \, dx.\] To compute this, we find the antiderivative of \(\sin(x)\), which is \(-\cos(x)\), and evaluate it: \[A_1 = \left[-\cos(x)\right]_{0}^{\frac{\pi}{4}} = \left(-\cos\left(\frac{\pi}{4}\right)\right) - \left(-\cos(0)\right) = - \frac{\sqrt{2}}{2} + 1.\] Thus, \(A_1 = 1 - \frac{\sqrt{2}}{2}.\)

Step 3: Calculate the Area for the Second Interval

For \(\frac{\pi}{4} < x \leq \pi\), the function is \(f(x) = \cos(x)\). The area under this curve from \(x = \frac{\pi}{4}\) to \(x = \pi\) is given by the integral: \[A_2 = \int_{\frac{\pi}{4}}^{\pi} \cos(x) \, dx.\] To compute this, we find the antiderivative of \(\cos(x)\), which is \(\sin(x)\), and evaluate it: \[A_2 = \left[\sin(x)\right]_{\frac{\pi}{4}}^{\pi} = \sin(\pi) - \sin\left(\frac{\pi}{4}\right) = 0 - \frac{\sqrt{2}}{2}.\] Thus, \(A_2 = - \frac{\sqrt{2}}{2}.\)

Step 4: Add Both Areas for Total Area

To find the total area under the piecewise function, add the areas from both intervals: \[\text{Total Area} = A_1 + A_2 = \left(1 - \frac{\sqrt{2}}{2}\right) + \left(- \frac{\sqrt{2}}{2}\right).\] Simplifying this expression gives: \[\text{Total Area} = 1 - \sqrt{2}.\]

Step 5: Write the Final Answer

The total area between the vertical lines \(x = 0\) and \(x = \pi\), between the graph of the function and the \(x\) -axis, is \(1 - \sqrt{2}\).

Key Concepts

Piecewise Functions

Piecewise functions are intriguing as they allow us to define different expressions for different parts of their domain. In other words, a piecewise function can be seen as a combination of several sub-functions, each applying to a specific interval on the x-axis. This approach provides flexibility in modeling scenarios that change behavior at certain points.

For instance, in this particular exercise, the function \( f(x) \) is defined piecewise over the interval \([0, \pi]\). There are two parts:

• \( f(x) = \sin(x) \) for \( 0 \leq x \leq \frac{\pi}{4} \)
• \( f(x) = \cos(x) \) for \( \frac{\pi}{4} < x \leq \pi \)

This means that the function switches its behavior at the point \( x = \frac{\pi}{4} \). Understanding how these intervals are applied is key to solving problems involving piecewise functions, as it enables us to calculate different areas or solve for different segments independently.

Trigonometric Functions

Trigonometric functions such as sine and cosine are central to the study of periodic phenomena. These functions are particularly useful when dealing with waveforms, circular motion, and oscillating systems.

In this exercise, we encounter the sine function \( \sin(x) \) and the cosine function \( \cos(x) \), both popular trigonometric functions. They have specific characteristics:

• \( \sin(x) \) varies between -1 and 1, starting from 0 at \( x = 0 \), reaching 1 at \( x = \frac{\pi}{2} \), and returning to 0 at \( x = \pi \).
• \( \cos(x) \) similarly varies between -1 and 1, beginning at 1 at \( x = 0 \), moving down to 0 at \( x = \frac{\pi}{2} \), and then to -1 at \( x = \pi \).

Understanding the behavior of these trigonometric functions over an interval helps in calculating areas and solving integrals associated with wave patterns and vibrations.

Area Under Curve

Calculating the area under a curve is a common application of definite integrals in calculus. It allows us to determine the total "accumulated" value of a function over an interval, often representing physical quantities like distance, area, or volume.

In this problem, finding the area under the piecewise curve consists of two main steps:

• Find the integral of \( \sin(x) \) over \([0, \frac{\pi}{4}]\).
• Find the integral of \( \cos(x) \) over \([\frac{\pi}{4}, \pi]\).

The definite integral gives the exact area beneath each segment, calculated as:

• \( A_1 = \int_{0}^{\frac{\pi}{4}} \sin(x) \, dx = 1 - \frac{\sqrt{2}}{2} \)
• \( A_2 = \int_{\frac{\pi}{4}}^{\pi} \cos(x) \, dx = -\frac{\sqrt{2}}{2} \)

By adding these areas \((A_1 + A_2)\), we derive the total area under the curve from \( x = 0 \) to \( x = \pi \), which in this case is \(1 - \sqrt{2}\). This process highlights the importance of integrating across piecewise domains to determine total areas efficiently.