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Magazine Chapter 16: Problem 62
\(\int_{0}^{\pi}|1+2 \cos x| d x\) is equal to (A) \(\frac{\pi}{3}-2 \sqrt{3}\) (B) \(\frac{\pi}{3}-\sqrt{3}\) (C) \(\frac{\pi}{3}+\sqrt{3}\) (D) \(\frac{\pi}{3}+2 \sqrt{3}\)
Short Answer
Expert verified
(D) \(\frac{\pi}{3} + 2\sqrt{3}\)
Step by step solution
01
Understand the Absolute Value
The integral is from 0 to 蟺 of the function \(|1 + 2 \cos x|\). The absolute value makes this more complex as it changes the function depending on whether \(1 + 2 \cos x\) is positive or negative.
02
Find the Critical Points
Solve \(1 + 2 \cos x = 0\) to find where the expression changes sign. This gives \(\cos x = -\frac{1}{2}\), which occurs at \(x = \frac{2\pi}{3}\) and in this interval, \(x = \frac{4\pi}{3}\). However, \(\frac{4\pi}{3}\) is outside the interval [0, \(\pi\)], so only \(\frac{2\pi}{3}\) needs to be considered.
03
Split the Integral
Since \(1 + 2\cos x\) changes sign at \(x = \frac{2\pi}{3}\), split the integral from 0 to 蟺 into two parts: \( \int_{0}^{\frac{2\pi}{3}} |1 + 2 \cos x| \, dx + \int_{\frac{2\pi}{3}}^{\pi} |1 + 2 \cos x| \, dx \).
04
Evaluate the Integral from 0 to 2蟺/3
For \(x \in \left[0, \frac{2\pi}{3}\right]\), \(1 + 2\cos x\) is positive. Thus, \(|1 + 2\cos x| = 1 + 2\cos x\) here. Calculate the integral: \(\int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx\).
05
Integrate 1 and 2cos(x)
The integral of 1 from 0 to \(\frac{2\pi}{3}\) is \(\frac{2\pi}{3}\). The integral of \(2\cos x\) is \(2\sin x\). Evaluate it from 0 to \(\frac{2\pi}{3}\): \(2 \sin(\frac{2\pi}{3}) - 2 \sin(0) = 2 \left(\frac{\sqrt{3}}{2}\right) - 0 = \sqrt{3}\). Thus, this part of the integral is \(\frac{2\pi}{3} + \sqrt{3}\).
06
Evaluate the Integral from 2蟺/3 to 蟺
For \(x \in \left[\frac{2\pi}{3}, \pi\right]\), \(1 + 2\cos x\) is negative. Thus, \(|1 + 2\cos x| = -(1 + 2\cos x)\). Calculate the integral: \(-\int_{\frac{2\pi}{3}}^{\pi} (1 + 2 \cos x) \, dx\).
07
Integrate -1 and -2cos(x)
The integral of -1 from \(\frac{2\pi}{3}\) to \(\pi\) is \(-\left(\pi - \frac{2\pi}{3}\right) = -\frac{\pi}{3}\). The integral of \(-2\cos x\) is \(-2\sin x\). Evaluate it from \(\frac{2\pi}{3}\) to \(\pi\): \(-2 \sin(\pi) + 2 \sin(\frac{2\pi}{3}) = -0 + \left(2\frac{\sqrt{3}}{2}\right) = \sqrt{3}\). Thus, this part of the integral is \(-\frac{\pi}{3} + \sqrt{3}\).
08
Combine the Integral Results
Add the results from the two integrals: \(\left(\frac{2\pi}{3} + \sqrt{3}\right) + \left(-\frac{\pi}{3} + \sqrt{3}\right) = \frac{\pi}{3} + 2\sqrt{3}\). So, the total integral is \(\frac{\pi}{3} + 2\sqrt{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value
When dealing with the definite integral \(\int_{0}^{\pi}|1+2 \cos x| \, dx\), the presence of an absolute value makes things a bit more complex. The absolute value essentially affects how we approach the function: it mandates that whatever is inside it must always be non-negative.
- If a function inside the absolute value is positive or zero, the absolute value doesn't change it.
- If the function is negative, the absolute value flips its sign, making it positive.
Critical Points
To find the critical points for the expression \(1 + 2 \cos x\), we set it equal to zero: \(1 + 2 \cos x = 0\). Solving this, we arrive at \(\cos x = -\frac{1}{2}\).
The cosine function \(\cos x\) equals \(-\frac{1}{2}\) at particular values of \(x\). Within the range \([0, \pi]\), this occurs at \(x = \frac{2\pi}{3}\).
The cosine function \(\cos x\) equals \(-\frac{1}{2}\) at particular values of \(x\). Within the range \([0, \pi]\), this occurs at \(x = \frac{2\pi}{3}\).
- Finding critical points is key to understanding where our function changes sign.
- These points determine where we split the integral to consider the impact of the absolute value.
Cosine Function
The cosine function, \(\cos x\), is periodic and oscillates between -1 and 1. For the function \(1 + 2\cos x\), the value of \(\cos x\) significantly impacts the entire expression.
- In the interval \([0, \frac{2\pi}{3}]\), \(\cos x\) starts from 1 (at \(x=0\)) and decreases, but it is always such that \(1 + 2\cos x\) remains positive.
- In the interval \([\frac{2\pi}{3}, \pi]\), \(\cos x\) continues to decrease further, resulting in \(1 + 2\cos x\) becoming negative.
Splitting Integrals
Splitting an integral is a useful technique when a function alters characteristics across its domain, often due to critical points or discontinuities. In this exercise, we split the integral of \(|1 + 2 \cos x|\) because it changes sign at \(x = \frac{2\pi}{3}\).
To handle the absolute value effectively:
To handle the absolute value effectively:
- Split the integral at the critical point into two separate parts: \(\int_{0}^{\frac{2\pi}{3}}|1 + 2\cos x|\, dx + \int_{\frac{2\pi}{3}}^{\pi}|1 + 2\cos x|\, dx\).
- For \(x\) in \([0, \frac{2\pi}{3}]\), \(|1 + 2\cos x|\) remains \(1 + 2\cos x\).
- For \(x\) in \([\frac{2\pi}{3}, \pi]\), \(|1 + 2\cos x|\) becomes \(- (1 + 2\cos x)\) as it is negative.
