Key Concepts and Formulas

• Definition of Absolute Value: $|f(x)| = f(x)$ if $f(x) \ge 0$, and $|f(x)| = -f(x)$ if $f(x) < 0$.
• Splitting Definite Integrals: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$ for $a < c < b$.
• Periodicity of Absolute Value of Sine: If $f(x)$ has period $T$, then $|f(x)|$ might have a smaller period. Specifically, $|\sin(ax)|$ has a period of $\pi/|a|$.
• Integral Property over Multiple Periods: $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ for a periodic function $f(x)$ with period $T$ and positive integer $n$.
• Standard Integral: $\int \sin(ax) dx = -\frac{\cos(ax)}{a} + C$.

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Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_0^\pi {|{{\sin }\,}2x|dx}$

Method 1: Splitting the Integral based on the Sign of $\sin 2x$

Step 1: Determine the intervals where $\sin 2x$ is positive and negative within $[0, \pi]$. To remove the absolute value, we need to know the sign of $\sin 2x$. Let $u = 2x$. As $x$ ranges from $0$ to $\pi$, $u$ ranges from $2 \times 0 = 0$ to $2 \times \pi = 2\pi$. The sine function, $\sin u$, is:

• Positive for $u \in (0, \pi)$. This corresponds to $0 < 2x < \pi$, which means $0 < x < \pi/2$. In this interval, $|\sin 2x| = \sin 2x$.
• Negative for $u \in (\pi, 2\pi)$. This corresponds to $\pi < 2x < 2\pi$, which means $\pi/2 < x < \pi$. In this interval, $|\sin 2x| = -\sin 2x$.
• Zero at $u = 0, \pi, 2\pi$, which corresponds to $x = 0, \pi/2, \pi$.

Step 2: Split the integral at the points where the sign of $\sin 2x$ changes. The sign changes at $x = \pi/2$. So, we split the integral into two parts: $I = \int\limits_0^{\pi/2} {|\sin 2x|dx} + \int\limits_{\pi/2}^\pi {|\sin 2x|dx}$ Applying the sign analysis from Step 1: $I = \int\limits_0^{\pi/2} {\sin 2x \,dx} + \int\limits_{\pi/2}^\pi {(-\sin 2x) \,dx}$

Step 3: Evaluate the first integral. $\int\limits_0^{\pi/2} {\sin 2x \,dx} = \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2}$ $= \left( -\frac{\cos(2 \cdot \pi/2)}{2} \right) - \left( -\frac{\cos(2 \cdot 0)}{2} \right)$ $= \left( -\frac{\cos \pi}{2} \right) - \left( -\frac{\cos 0}{2} \right)$ Since $\cos \pi = -1$ and $\cos 0 = 1$: $= \left( -\frac{-1}{2} \right) - \left( -\frac{1}{2} \right) = \frac{1}{2} + \frac{1}{2} = 1$

Step 4: Evaluate the second integral. $\int\limits_{\pi/2}^\pi {(-\sin 2x) \,dx} = \left[ \frac{\cos 2x}{2} \right]_{\pi/2}^\pi$ $= \left( \frac{\cos(2 \cdot \pi)}{2} \right) - \left( \frac{\cos(2 \cdot \pi/2)}{2} \right)$ $= \left( \frac{\cos 2\pi}{2} \right) - \left( \frac{\cos \pi}{2} \right)$ Since $\cos 2\pi = 1$ and $\cos \pi = -1$: $= \left( \frac{1}{2} \right) - \left( \frac{-1}{2} \right) = \frac{1}{2} + \frac{1}{2} = 1$

Step 5: Add the results of the two integrals. $I = 1 + 1 = 2$

Method 2: Using Periodicity (More Efficient Approach)

Step 1: Determine the period of the integrand $|\sin 2x|$. The function $\sin(2x)$ has a period of $T = \frac{2\pi}{|2|} = \pi$. The function $|\sin(2x)|$ has a period of $T' = \frac{T}{2} = \frac{\pi}{2}$. This is because $|\sin(\theta + \pi)| = |-\sin(\theta)| = |\sin(\theta)|$, and replacing $\theta$ with $2x$ gives $|\sin(2x + \pi)| = |\sin(2x)|$. Thus, $|\sin(2(x + \pi/2))| = |\sin(2x + \pi)| = |\sin(2x)|$.

Step 2: Apply the periodicity property for integrals. The interval of integration is $[0, \pi]$. The period of $|\sin 2x|$ is $\pi/2$. The interval $[0, \pi]$ is exactly twice the period, i.e., $\pi = 2 \times (\pi/2)$. We can use the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$. Here, $n=2$ and $T=\pi/2$. $I = \int_0^{\pi} {|\sin 2x|dx} = 2 \int_0^{\pi/2} {|\sin 2x|dx}$

Step 3: Simplify the absolute value in the new interval. For $x \in [0, \pi/2]$, the value of $2x$ is in $[0, \pi]$. In this interval, $\sin(2x) \ge 0$. Therefore, $|\sin 2x| = \sin 2x$ for $x \in [0, \pi/2]$. The integral becomes: $I = 2 \int_0^{\pi/2} {\sin 2x \,dx}$

Step 4: Evaluate the simplified integral. $I = 2 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2}$ $I = 2 \left[ \left( -\frac{\cos(2 \cdot \pi/2)}{2} \right) - \left( -\frac{\cos(2 \cdot 0)}{2} \right) \right]$ $I = 2 \left[ \left( -\frac{\cos \pi}{2} \right) - \left( -\frac{\cos 0}{2} \right) \right]$ Since $\cos \pi = -1$ and $\cos 0 = 1$: $I = 2 \left[ \left( -\frac{-1}{2} \right) - \left( -\frac{1}{2} \right) \right]$ $I = 2 \left[ \frac{1}{2} + \frac{1}{2} \right] = 2 [1] = 2$

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Common Mistakes & Tips

• Incorrectly determining the period: Remember that the period of $|\sin(ax)|$ is $\pi/|a|$, not $2\pi/|a|$.
• Forgetting to split the integral: When dealing with absolute values, it's crucial to identify the points where the argument of the absolute value becomes zero and split the integral accordingly.
• Sign errors in integration: Double-check the signs when integrating $\sin(ax)$ and when evaluating the definite integral.

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Summary

The integral $\int\limits_0^\pi {|{{\sin }\,}2x|dx}$ involves an absolute value function. We can solve this by either splitting the integral into sub-intervals where $\sin 2x$ is positive or negative, or by utilizing the periodicity of $|\sin 2x|$. Both methods lead to the same result. Method 2, using periodicity, is generally more efficient. The period of $|\sin 2x|$ is $\pi/2$. The integration interval $[0, \pi]$ is twice this period. Evaluating the integral over one period $[0, \pi/2]$ and multiplying by 2 yields the final answer.

The final answer is $\boxed{2}$.