What is \(\displaystyle \rm \int_0^{8 \pi}|\sin x| d x\) equal to?
16
Understanding the Definite Integral of \(|\sin x|\)
We are asked to evaluate the definite integral \(\displaystyle \rm \int_0^{8 \pi}|\sin x| d x\). This involves integrating the absolute value of the sine function over a specific interval.
The function \(f(x) = |\sin x|\) is the absolute value of the sine function. Let's analyze the behavior of \(\sin x\) and \(|\sin x|\):
- The function \(\sin x\) is positive in the intervals \([0, \pi], [2\pi, 3\pi], [4\pi, 5\pi], \ldots\)
- The function \(\sin x\) is negative in the intervals \([\pi, 2\pi], [3\pi, 4\pi], [5\pi, 6\pi], \ldots\)
Therefore, \(|\sin x|\) can be written piecewise:
- \(|\sin x| = \sin x\) when \(\sin x \ge 0\)
- \(|\sin x| = -\sin x\) when \(\sin x < 0\)
The function \(|\sin x|\) is always non-negative. The graph of \(|\sin x|\) looks like a series of identical "humps" above the x-axis. Each hump corresponds to an interval where \(\sin x\) goes from 0 up to 1 and back down to 0, or from 0 down to -1 and back up to 0 (with the negative part flipped upwards by the absolute value).
Evaluating the Integral Over One Period of \(|\sin x|\)
The sine function \(\sin x\) has a period of \(2\pi\). However, the function \(|\sin x|\) has a period of \(\pi\). This is because the shape of the graph of \(|\sin x|\) from \([0, \pi]\) is repeated exactly from \([\pi, 2\pi]\), \([2\pi, 3\pi]\), and so on.
Let's calculate the integral of \(|\sin x|\) over one period, say from \(0\) to \(\pi\):
\(\displaystyle \int_0^{\pi} |\sin x| \, dx\)
In the interval \([0, \pi]\), \(\sin x \ge 0\), so \(|\sin x| = \sin x\).
\(\displaystyle \int_0^{\pi} |\sin x| \, dx = \int_0^{\pi} \sin x \, dx\)
Now, we find the antiderivative of \(\sin x\), which is \(-\cos x\).
\(\displaystyle \int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi}\)
Evaluate the antiderivative at the limits of integration:
\(\displaystyle [-\cos x]_0^{\pi} = (-\cos(\pi)) - (-\cos(0))\)
We know that \(\cos(\pi) = -1\) and \(\cos(0) = 1\).
\(\displaystyle = (-(-1)) - (-(1))\)
\(\displaystyle = 1 - (-1)\)
\(\displaystyle = 1 + 1 = 2\)
So, the integral of \(|\sin x|\) over one period (\(\pi\)) is 2.
Using Periodicity for the Definite Integral
The integral we need to evaluate is \(\displaystyle \int_0^{8 \pi}|\sin x| d x\). The interval of integration is \([0, 8\pi]\).
The function \(|\sin x|\) has a period of \(\pi\). The total length of the interval \([0, 8\pi]\) is \(8\pi\).
We can determine how many periods of length \(\pi\) fit into the interval \([0, 8\pi]\) by dividing the total length by the period length:
Number of periods = \(\displaystyle \frac{\text{Total interval length}}{\text{Period length}} = \frac{8\pi}{\pi} = 8\)
Since the function \(|\sin x|\) is periodic with period \(\pi\), the integral over the interval \([0, 8\pi]\) is simply the sum of the integrals over each period within that interval. Because the integral over one period is 2, the total integral is the number of periods multiplied by the integral over one period.
\(\displaystyle \int_0^{8 \pi}|\sin x| d x = \text{(Number of periods)} \times \int_0^{\pi}|\sin x| d x\)
\(\displaystyle \int_0^{8 \pi}|\sin x| d x = 8 \times 2\)
\(\displaystyle \int_0^{8 \pi}|\sin x| d x = 16\)
The definite integral of \(|\sin x|\) from 0 to \(8\pi\) is 16.
| Interval | Behavior of \(\sin x\) | \(|\sin x|\) | Integral over Interval |
|---|---|---|---|
| \([0, \pi]\) | \(\ge 0\) | \(\sin x\) | \(\int_0^\pi \sin x \, dx = 2\) |
| \([\pi, 2\pi]\) | \(\le 0\) | \(-\sin x\) | \(\int_\pi^{2\pi} (-\sin x) \, dx = [\cos x]_\pi^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 2\) |
| \([2\pi, 3\pi]\) | \(\ge 0\) | \(\sin x\) | \(\int_{2\pi}^{3\pi} \sin x \, dx = [-\cos x]_{2\pi}^{3\pi} = -\cos(3\pi) - (-\cos(2\pi)) = -(-1) - (-1) = 2\) |
| \(\ldots\) | \(\ldots\) | \(\ldots\) | \(\ldots\) |
| \([7\pi, 8\pi]\) | \(\le 0\) | \(-\sin x\) | \(\int_{7\pi}^{8\pi} (-\sin x) \, dx = [\cos x]_{7\pi}^{8\pi} = \cos(8\pi) - \cos(7\pi) = 1 - (-1) = 2\) |
Each interval of length \(\pi\) contributes 2 to the total integral. Since there are 8 such intervals from \(0\) to \(8\pi\), the total integral is \(8 \times 2 = 16\).
Summary of Steps to Evaluate the Definite Integral
- Identify the function being integrated: \(|\sin x|\).
- Recognize the interval of integration: \([0, 8\pi]\).
- Determine the period of the function \(|\sin x|\), which is \(\pi\).
- Calculate the definite integral of \(|\sin x|\) over one period, e.g., from \(0\) to \(\pi\). \(\int_0^\pi |\sin x| \, dx = 2\).
- Determine how many periods are in the given integration interval: \((8\pi - 0) / \pi = 8\).
- Multiply the integral over one period by the number of periods: \(8 \times 2 = 16\).
Thus, \(\displaystyle \rm \int_0^{8 \pi}|\sin x| d x = 16\).
Revision Table: Definite Integral of Absolute Sine
| Concept | Description | Application |
|---|---|---|
| Absolute Value Function | \(|f(x)|\) is \(f(x)\) if \(f(x)\ge 0\) and \(-f(x)\) if \(f(x) < 0\). | Used to define \(|\sin x|\) piecewise. |
| Periodicity of \(|\sin x|\) | \(|\sin(x+\pi)| = |\sin x|\) for all \(x\). Period is \(\pi\). | Allows breaking the integral \([0, 8\pi]\) into 8 intervals of length \(\pi\). |
| Definite Integral Property | \(\int_a^b f(x) \, dx = \sum_{i=1}^n \int_{x_{i-1}}^{x_i} f(x) \, dx\) if \(a=x_0 < x_1 < \ldots < x_n=b\). If \(f(x)\) is periodic with period \(T\), \(\int_a^{a+nT} f(x) \, dx = n \int_a^{a+T} f(x) \, dx\). | Applied to evaluate the integral over \(8\pi\) using the integral over \(\pi\). |
| Fundamental Theorem of Calculus | \(\int_a^b f(x) \, dx = F(b) - F(a)\) where \(F'(x) = f(x)\). | Used to evaluate the integral \(\int_0^\pi \sin x \, dx\). |
Additional Information: Properties of Absolute Value and Integration
Understanding the absolute value function is crucial when dealing with integrals like this. The absolute value always makes the output non-negative, effectively reflecting any negative parts of the original function's graph across the x-axis. For trigonometric functions like sine and cosine, taking the absolute value changes their period when the original function dips below the x-axis.
For example:
- \(\sin x\) has a period of \(2\pi\). \(|\sin x|\) has a period of \(\pi\).
- \(\cos x\) has a period of \(2\pi\). \(|\cos x|\) has a period of \(\pi\).
- \(\tan x\) has a period of \(\pi\). \(|\tan x|\) also has a period of \(\pi\).
When integrating a periodic function \(f(x)\) with period \(T\) over an interval whose length is an integer multiple of the period, say \([a, a+nT]\), the integral is \(n\) times the integral over one period \(\int_a^{a+T} f(x) \, dx\). In our case, the function \(|\sin x|\) has period \(\pi\), and the interval \([0, 8\pi]\) has length \(8\pi = 8 \times \pi\). So \(n=8\), and the starting point \(a=0\) works well with the period \(\pi\), which starts repeating its shape from 0.
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