Key Concepts and Formulas

• Property of Even Functions: For an even function $f(x)$ (where $f(-x) = f(x)$), the definite integral over a symmetric interval $[-a, a]$ is given by $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$.
• Definition of Absolute Value: $|u| = u$ if $u \ge 0$, and $|u| = -u$ if $u < 0$. This requires careful analysis of the sign of the expression inside the absolute value over the given interval.
• Fundamental Theorem of Calculus: $\int_a^b F'(x) dx = F(b) - F(a)$.

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Step-by-Step Solution

Step 1: Analyze the Integrand and Apply Symmetry Property

We are asked to evaluate the integral $I = \int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx}$. Let $f(x) = \left| {\pi - \left| x \right|} \right|$. We first check if $f(x)$ is an even function. $f(-x) = \left| {\pi - \left| {-x} \right|} \right|$ Since $\left| {-x} \right| = \left| x \right|$, we have: $f(-x) = \left| {\pi - \left| x \right|} \right| = f(x)$ Thus, $f(x)$ is an even function. The interval of integration is $[-\pi, \pi]$, which is symmetric about $x=0$. Therefore, we can use the property of even functions: $I = 2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|dx}$ Why this step? Recognizing the integrand as an even function and applying the symmetry property reduces the integration interval from $[-\pi, \pi]$ to $[0, \pi]$, simplifying the problem.

Step 2: Simplify the Absolute Value Expression

Now we focus on the integral $2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|dx}$. For the interval $0 \le x \le \pi$, we have $x \ge 0$. Therefore, $\left| x \right| = x$. Substituting this into the integrand: $I = 2\int\limits_0^\pi {\left| {\pi - x} \right|} dx$ Next, we analyze the expression inside the absolute value, $\pi - x$, over the interval $[0, \pi]$. When $0 \le x \le \pi$, we have $0 \le \pi - x \le \pi$. This means that $\pi - x$ is always non-negative in this interval. Therefore, $\left| {\pi - x} \right| = \pi - x$ for $x \in [0, \pi]$. The integral becomes: $I = 2\int\limits_0^\pi {\left( {\pi - x} \right)} dx$ Why this step? By analyzing the sign of the expression within the absolute value over the specific integration interval, we can remove the absolute value signs, transforming the integrand into a simpler polynomial form, which is directly integrable.

Step 3: Evaluate the Definite Integral

We now integrate the simplified expression: $I = 2 \left[ \int\limits_0^\pi {\pi \, dx} - \int\limits_0^\pi {x \, dx} \right]$ Using the power rule for integration ($\int x^n \, dx = \frac{x^{n+1}}{n+1}$) and the constant rule ($\int c \, dx = cx$): $I = 2 \left[ \pi x - \frac{x^2}{2} \right]_0^\pi$ Why this step? This is the standard integration process where we find the antiderivative of the simplified integrand.

Step 4: Apply the Fundamental Theorem of Calculus

Now, we evaluate the antiderivative at the upper and lower limits of integration and find the difference: $I = 2 \left[ \left( \pi (\pi) - \frac{(\pi)^2}{2} \right) - \left( \pi (0) - \frac{(0)^2}{2} \right) \right]$ $I = 2 \left[ \left( \pi^2 - \frac{\pi^2}{2} \right) - (0 - 0) \right]$ $I = 2 \left[ \frac{2\pi^2 - \pi^2}{2} \right]$ $I = 2 \left[ \frac{\pi^2}{2} \right]$ $I = \pi^2$ Why this step? This is the final step where we apply the Fundamental Theorem of Calculus to compute the definite value of the integral by substituting the limits of integration into the antiderivative.

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Common Mistakes & Tips

• Incorrect Absolute Value Handling: Failing to analyze the sign of the expression inside the absolute value over the integration interval is a common error. Always determine where the expression is positive and negative.
• Forgetting the Symmetry Property Factor: When using the even function property, remember to multiply the integral from $0$ to $a$ by $2$.
• Arithmetic Errors: Be meticulous when substituting the limits of integration and performing subtractions, as small calculation mistakes can lead to an incorrect final answer.

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Summary

The problem was solved by first recognizing the integrand as an even function and utilizing the symmetry property of definite integrals to halve the integration interval. Subsequently, the absolute value function was simplified by analyzing the sign of the expression within it over the new interval. Finally, standard integration techniques and the Fundamental Theorem of Calculus were applied to compute the definite value of the integral.

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