Key Concepts and Formulas

1. Periodicity of $|\sin x|$: The function $|\sin x|$ is periodic with period $\pi$. This means $|\sin(x+\pi)| = |\sin x|$.
2. Property of Definite Integrals over Multiple Periods: If $f(x)$ is periodic with period $T$, then $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ for any positive integer $n$.
3. Handling Absolute Value Functions: For an integral of $|f(x)|$, we must consider the sign of $f(x)$ over the integration interval. If $f(x) \ge 0$, $|f(x)| = f(x)$. If $f(x) < 0$, $|f(x)| = -f(x)$.
4. Fundamental Theorem of Calculus: $\int_a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_0^{10\pi } {\left| {\sin x} \right|dx}$

Step 1: Identify the Periodicity and Simplify the Integral

The integrand is $f(x) = |\sin x|$. We know that $\sin x$ has a period of $2\pi$. Let's examine the behavior of $|\sin x|$:

• For $x \in [0, \pi]$, $\sin x \ge 0$, so $|\sin x| = \sin x$.
• For $x \in [\pi, 2\pi]$, $\sin x \le 0$, so $|\sin x| = -\sin x$.
• Since $\sin(x+\pi) = -\sin x$, we have $|\sin(x+\pi)| = |-\sin x| = |\sin x|$. This shows that the period of $|\sin x|$ is $T = \pi$.

The given integral is from $0$ to $10\pi$. We can write the upper limit as $10T$, where $T=\pi$. Using the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ with $n=10$ and $T=\pi$: $I = 10 \int\limits_0^\pi {\left| {\sin x} \right|dx}$ Why this step? This step leverages the periodicity of the integrand to reduce the upper limit of integration from $10\pi$ to $\pi$, making the problem much simpler to solve.

Step 2: Remove the Absolute Value by Considering the Sign of $\sin x$

Now we need to evaluate $\int\limits_0^\pi {\left| {\sin x} \right|dx}$. In the interval $[0, \pi]$, the sine function, $\sin x$, is non-negative ($\sin x \ge 0$). Therefore, for $x \in [0, \pi]$, we have $|\sin x| = \sin x$. Substituting this into our integral expression: $I = 10 \int\limits_0^\pi {\sin x \,dx}$ Why this step? To perform the integration, we must eliminate the absolute value. By analyzing the sign of $\sin x$ in the interval $[0, \pi]$, we can replace $|\sin x|$ with $\sin x$.

Step 3: Evaluate the Definite Integral

We now evaluate the integral $\int\limits_0^\pi {\sin x \,dx}$. The antiderivative of $\sin x$ is $-\cos x$. Applying the Fundamental Theorem of Calculus: $I = 10 \left[ -\cos x \right]_0^\pi$ $I = 10 \left( (-\cos(\pi)) - (-\cos(0)) \right)$ We know that $\cos(\pi) = -1$ and $\cos(0) = 1$. $I = 10 \left( (-(-1)) - (-1) \right)$ $I = 10 \left( 1 + 1 \right)$ $I = 10 \times 2$ $I = 20$ Why this step? This is the core calculation of the definite integral using the antiderivative and evaluating it at the limits of integration to find the exact value.

Common Mistakes & Tips

• Incorrect Period: A common error is to assume the period of $|\sin x|$ is $2\pi$. Always remember or derive that the period of $|\sin x|$ and $|\cos x|$ is $\pi$.
• Ignoring Absolute Value: Failing to address the absolute value by not considering the sign of the function over the interval will lead to an incorrect result. If the sign changes within the interval, the integral must be split.
• Calculation Errors: Simple arithmetic errors when evaluating trigonometric functions at specific angles (like $\cos(0)$ or $\cos(\pi)$) can lead to the wrong final answer.

Summary

The integral $\int_0^{10\pi} |\sin x| dx$ was solved by first recognizing that the integrand $|\sin x|$ is periodic with period $\pi$. This allowed us to use the property of definite integrals over multiples of the period to transform the integral into $10 \int_0^\pi |\sin x| dx$. Within the interval $[0, \pi]$, $\sin x$ is non-negative, so $|\sin x| = \sin x$. We then evaluated the definite integral of $\sin x$ from $0$ to $\pi$, which resulted in $2$. Multiplying by $10$, we obtained the final answer of $20$.

The final answer is $\boxed{20}$.