Problem 18 - 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}{cl} \sec (x) & \text { if } 0 \leq x \leq \pi / 3 \\ 4 \cos (x) & \text { if } \pi / 3

Step-by-Step Solution

Answer

Area = \ln(2 + \sqrt{3}) - 2\sqrt{3}.

Step 1: Understand the Interval and Function

The function is defined on the interval \(I = [0, \pi]\)\, with two pieces: \(f(x) = \sec(x)\) for \(0 \leq x \leq \pi/3\) and \(f(x) = 4\cos(x)\) for \(\pi/3 < x \leq \pi\). We need to find the area under the function from \(x = 0\) to \(x = \pi\).

Step 2: Calculate the Area for the First Segment

For \(0 \leq x \leq \pi/3\), the area under \(f(x) = \sec(x)\) is given by the integral \int_{0}^{\pi/3} \sec(x)\,dx\. The antiderivative of \(\sec(x)\) is \ln|\sec(x) + \tan(x)|\, so the integral becomes: \\int_{0}^{\pi/3} \sec(x)\,dx = \left[\ln|\sec(x) + \tan(x)|\right]_{0}^{\pi/3}\. Substituting the limits yields \ln|2 + \sqrt{3}| - \ln|1| = \ln(2 + \sqrt{3})\.

Step 3: Calculate the Area for the Second Segment

For \(\pi/3 < x \leq \pi\), the area under \(f(x) = 4\cos(x)\) is given by the integral \int_{\pi/3}^{\pi} 4\cos(x)\,dx\. The antiderivative of \(\cos(x)\) is \sin(x)\, so the integral becomes: \\int_{\pi/3}^{\pi} 4\cos(x)\,dx = 4\left[\sin(x)\right]_{\pi/3}^{\pi}\. Calculating this gives us \4(0 - \frac{\sqrt{3}}{2}) = -2\sqrt{3}\.

Step 4: Combine Both Areas

Add the results from both regions: \Area = \ln(2 + \sqrt{3}) - 2\sqrt{3}\. This sum represents the total area under the function between the vertical lines \(x = 0\) and \(x = \pi\).

Key Concepts

Piecewise Functions

Piecewise functions are fascinating mathematical tools that allow us to define a function with different rules over different parts of a domain. This means we can have a single function express complex behaviors without straying into complicated definitions. In our example, the function is defined differently on two subintervals from \(x = 0\) to \(x = \pi\).

• When \(0 \leq x \leq \frac{\pi}{3}\), the function is defined as \(f(x) = \sec(x)\), which is the reciprocal of the cosine function.
• For \(\frac{\pi}{3} < x \leq \pi\), the function switches to \(f(x) = 4\cos(x)\), which scales the cosine function by 4.

This approach lets us model a variety of real-world scenarios where a function's behavior changes over different parts of its domain. Recognizing where these changes occur is crucial when performing calculations like integration.

Definite Integrals

The concept of definite integrals revolves around finding the accumulated value of a function over a specific interval. In simpler terms, it calculates the total area under the curve of a specified part of the function. For piecewise functions, we need to evaluate each part of the function separately over its respective interval.

• For our first interval, from \(x = 0\) to \(x = \frac{\pi}{3}\), we calculate the integral of \(\sec(x)\). The result is obtained using the antiderivative \(\ln|\sec(x) + \tan(x)|\).
• The integral on the second interval, from \(x = \frac{\pi}{3}\) to \(x = \pi\), involves \(4\cos(x)\), with the antiderivative \(4\sin(x)\).

Calculating these definite integrals independently and then summing them helps us find the total area enclosed by the curve. Each integral assesses the function’s properties in its defined segment.

Area Under a Curve

Finding the area under a curve is a common problem in calculus, as it visually and mathematically shows how much "space" a function occupies over a specific range on a graph. This concept involves integration, where we compute it section by section for piecewise functions.

• In practice, this means splitting our function into understandable segments, as demonstrated with \(f(x)\) being split across two distinct intervals.
• By computing the definite integral over each segment, we determine how much each part contributes to the total area.

Let's put it all together: the area under \(f(x) = \sec(x)\) from \(x = 0\) to \(x = \frac{\pi}{3}\) calculates to \(\ln(2 + \sqrt{3})\); from \(x = \frac{\pi}{3}\) to \(x = \pi\) under \(f(x) = 4\cos(x)\), it results in \(-2\sqrt{3}\). Together, they depict the entire area the curve encompasses between \(x=0\) and \(x=\pi\). This method of summing areas gives us a comprehensive understanding of the function's footprint over its domain.