Key Concepts and Formulas

• Greatest Integer Function: $[x]$ is the greatest integer less than or equal to $x$.
• Fractional Part Function: $\{x\} = x - [x]$, where $0 \le \{x\} < 1$.
• Properties of Definite Integrals: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$.
• Substitution Rule for Definite Integrals: If $u = g(x)$, then $\int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du$.

Step-by-Step Solution

Step 1: Analyze the integrand and the interval of integration. The integral is given by $I = {\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx$. The interval of integration is $[0, 2]$. The presence of $[x]$ suggests splitting the integral into subintervals where $[x]$ is constant. The fractional part is $(x - [x]) = \{x\}$.

Step 2: Split the integral based on the value of [x]. For $x \in [0, 2]$, the greatest integer $[x]$ can take values 0 and 1.

• On the interval $[0, 1)$, $[x] = 0$.
• On the interval $[1, 2)$, $[x] = 1$.
• At $x=2$, $[x]=2$. However, the upper limit is 2, and the interval $[1, 2)$ covers up to but not including 2. For the purpose of integration, the value at a single point does not affect the integral.

Therefore, we split the integral into two parts: $I = {\pi ^2}\left( \int_0^1 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx + \int_1^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx \right)$

Step 3: Evaluate the first integral (from 0 to 1). For $x \in [0, 1)$, $[x] = 0$. So, $(x - [x])^{[x]} = (x - 0)^0 = x^0 = 1$ (since $x$ is not 0 in the interval $(0, 1)$ and $0^0$ is typically taken as 1 in this context for integration, or more formally, as $x \to 0^+$, $x^0 \to 1$). Let $I_1 = \int_0^1 {\left( {\sin {{\pi x} \over 2}} \right)(1)} dx = \int_0^1 {\sin {{\pi x} \over 2}} dx$. To evaluate this, let $u = \frac{\pi x}{2}$. Then $du = \frac{\pi}{2} dx$, which means $dx = \frac{2}{\pi} du$. When $x=0$, $u = 0$. When $x=1$, $u = \frac{\pi}{2}$. $I_1 = \int_0^{\pi/2} {\sin(u)} \frac{2}{\pi} du = \frac{2}{\pi} \int_0^{\pi/2} {\sin(u)} du$ $I_1 = \frac{2}{\pi} [-\cos(u)]_0^{\pi/2} = \frac{2}{\pi} (-\cos(\frac{\pi}{2}) - (-\cos(0)))$ $I_1 = \frac{2}{\pi} (0 - (-1)) = \frac{2}{\pi} (1) = \frac{2}{\pi}$

Step 4: Evaluate the second integral (from 1 to 2). For $x \in [1, 2)$, $[x] = 1$. So, $(x - [x])^{[x]} = (x - 1)^1 = x - 1$. Let $I_2 = \int_1^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - 1)} dx$. This integral requires integration by parts. Let $u = x-1$ and $dv = \sin(\frac{\pi x}{2}) dx$. Then $du = dx$ and $v = \int \sin(\frac{\pi x}{2}) dx$. To find $v$, let $w = \frac{\pi x}{2}$, so $dw = \frac{\pi}{2} dx$, $dx = \frac{2}{\pi} dw$. $\int \sin(w) \frac{2}{\pi} dw = \frac{2}{\pi} (-\cos(w)) = -\frac{2}{\pi} \cos(\frac{\pi x}{2})$. So, $v = -\frac{2}{\pi} \cos(\frac{\pi x}{2})$.

Using integration by parts formula $\int u \, dv = uv - \int v \, du$: $I_2 = \left[ (x-1) \left(-\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right)\right) \right]_1^2 - \int_1^2 \left(-\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right)\right) dx$ $I_2 = \left[ -\frac{2}{\pi} (x-1) \cos\left(\frac{\pi x}{2}\right) \right]_1^2 + \frac{2}{\pi} \int_1^2 \cos\left(\frac{\pi x}{2}\right) dx$

Evaluate the first term: At $x=2$: $-\frac{2}{\pi} (2-1) \cos(\frac{\pi \cdot 2}{2}) = -\frac{2}{\pi} (1) \cos(\pi) = -\frac{2}{\pi} (-1) = \frac{2}{\pi}$. At $x=1$: $-\frac{2}{\pi} (1-1) \cos(\frac{\pi \cdot 1}{2}) = -\frac{2}{\pi} (0) \cos(\frac{\pi}{2}) = 0$. So, the first term is $\frac{2}{\pi} - 0 = \frac{2}{\pi}$.

Evaluate the second term: $\frac{2}{\pi} \int_1^2 \cos\left(\frac{\pi x}{2}\right) dx$. Let $k = \frac{\pi x}{2}$, $dk = \frac{\pi}{2} dx$, $dx = \frac{2}{\pi} dk$. When $x=1$, $k = \frac{\pi}{2}$. When $x=2$, $k = \pi$. $\frac{2}{\pi} \int_{\pi/2}^{\pi} \cos(k) \frac{2}{\pi} dk = \frac{4}{\pi^2} \int_{\pi/2}^{\pi} \cos(k) dk$ $= \frac{4}{\pi^2} [\sin(k)]_{\pi/2}^{\pi} = \frac{4}{\pi^2} (\sin(\pi) - \sin(\frac{\pi}{2}))$ $= \frac{4}{\pi^2} (0 - 1) = -\frac{4}{\pi^2}$

Combining the parts of $I_2$: $I_2 = \frac{2}{\pi} - \frac{4}{\pi^2}$

Step 5: Combine the results for $I_1$ and $I_2$ and multiply by $\pi^2$. The total integral is $I = {\pi ^2} (I_1 + I_2)$. $I = {\pi ^2} \left( \frac{2}{\pi} + \left(\frac{2}{\pi} - \frac{4}{\pi^2}\right) \right)$ $I = {\pi ^2} \left( \frac{2}{\pi} + \frac{2}{\pi} - \frac{4}{\pi^2} \right)$ $I = {\pi ^2} \left( \frac{4}{\pi} - \frac{4}{\pi^2} \right)$ Distribute $\pi^2$: $I = \pi^2 \cdot \frac{4}{\pi} - \pi^2 \cdot \frac{4}{\pi^2}$ $I = 4\pi - 4$ $I = 4(\pi - 1)$

Common Mistakes & Tips

• Handling $0^0$: In the interval $[0, 1)$, for $x=0$, we have $(0-0)^0 = 0^0$. For definite integrals, this is usually interpreted as 1. If the interval was $(0, 1]$, the term would not be problematic.
• Integration by Parts Errors: Carefully identify $u$ and $dv$. Ensure the differentiation and integration steps are correct. The limits of integration must be applied correctly at each stage.
• Sign Errors: Pay close attention to the signs, especially when integrating trigonometric functions and applying the fundamental theorem of calculus.

Summary

The problem required evaluating a definite integral involving the greatest integer function. We first broke down the integral into sub-intervals based on the constant values of $[x]$. For the interval $[0, 1)$, $[x]=0$, leading to a simple sine integral. For the interval $[1, 2)$, $[x]=1$, requiring integration by parts. After evaluating each part and combining them, we multiplied by the pre-factor $\pi^2$ to obtain the final result.

The final answer is $4(\pi - 1)$.

The final answer is \boxed{4(\pi - 1)}.