Key Concepts and Formulas

• Definition of Absolute Value: $|u| = u$ if $u \ge 0$, and $|u| = -u$ if $u < 0$. This requires splitting integrals at points where the argument of the absolute value is zero.
• Definition of Greatest Integer Function: $[x]$ is the greatest integer less than or equal to $x$. This function is piecewise constant, and its integral can be computed by summing integrals over intervals where it is constant.
• Properties of Definite Integrals:
  • $\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$.
  • If $f(x)$ is an even function, $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.
  • If $f(x)$ is an odd function, $\int_{-a}^a f(x) dx = 0$.
  • $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$.
• Integral of $[x]$: For an integer $k$, $\int_k^{k+1} [x] dx = \int_k^{k+1} k dx = k(k+1-k) = k$.

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Step-by-Step Solution: Part 1 - Finding the value of 'a'

We are given the equation: $\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$ with the condition $a > 2$.

Step 1: Analyze the integrand $\left| x \right| + \left| {x - 2} \right|$ and split the integral. The critical points for the absolute value functions are $x=0$ and $x=2$. Since $a > 2$, the interval of integration $[-a, a]$ contains both $0$ and $2$. We need to split the integral based on the signs of $x$ and $x-2$.

For $x \in [-a, 0)$: $|x| = -x$ and $|x-2| = -(x-2) = 2-x$. For $x \in [0, 2)$: $|x| = x$ and $|x-2| = -(x-2) = 2-x$. For $x \in [2, a]$: $|x| = x$ and $|x-2| = x-2$.

Thus, the integrand can be written as: $\left| x \right| + \left| {x - 2} \right| = \begin{cases} -x + (2-x) = 2 - 2x & \text{if } x < 0 \\ x + (2-x) = 2 & \text{if } 0 \le x < 2 \\ x + (x-2) = 2x - 2 & \text{if } x \ge 2 \end{cases}$

Step 2: Split the definite integral $\int_{-a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx$ based on the intervals. Since the integrand has different forms on $[-a, 0)$, $[0, 2)$, and $[2, a]$, we split the integral: $\int_{-a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = \int_{-a}^0 {\left( 2 - 2x \right)} dx + \int_0^2 {\left( 2 \right)} dx + \int_2^a {\left( 2x - 2 \right)} dx$

Step 3: Evaluate each integral.

• First integral: $\int_{-a}^0 {\left( 2 - 2x \right)} dx = \left[ 2x - x^2 \right]_{-a}^0 = (2(0) - 0^2) - (2(-a) - (-a)^2) = 0 - (-2a - a^2) = 2a + a^2$

• Second integral: $\int_0^2 {\left( 2 \right)} dx = \left[ 2x \right]_0^2 = 2(2) - 2(0) = 4$

• Third integral: $\int_2^a {\left( 2x - 2 \right)} dx = \left[ x^2 - 2x \right]_2^a = (a^2 - 2a) - (2^2 - 2(2)) = a^2 - 2a - (4 - 4) = a^2 - 2a$

Step 4: Sum the results and solve for 'a'. Adding the results from Step 3: $(2a + a^2) + 4 + (a^2 - 2a) = 22$ $2a^2 + 4 = 22$ $2a^2 = 18$ $a^2 = 9$ Since $a > 2$, we take the positive square root: $a = 3$

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Step-by-Step Solution: Part 2 - Evaluating the second integral

Now we need to evaluate: $\int\limits_{ - a}^a {\left( {x + \left[ x \right]} \right)} dx$ Substitute the value $a=3$ found in Part 1: $\int\limits_{ - 3}^3 {\left( {x + \left[ x \right]} \right)} dx$

Step 5: Split the integral based on the properties of the greatest integer function. The greatest integer function $[x]$ changes its value at integer points. The interval of integration is $[-3, 3]$. We split the integral at each integer: $\int_{-3}^3 {\left( {x + \left[ x \right]} \right)} dx = \int_{-3}^{-2} (x + [-3]) dx + \int_{-2}^{-1} (x + [-2]) dx + \int_{-1}^0 (x + [-1]) dx + \int_0^1 (x + [0]) dx + \int_1^2 (x + [1]) dx + \int_2^3 (x + [2]) dx$

Step 6: Evaluate each sub-integral.

• For $x \in [-3, -2)$, $[x] = -3$: $\int_{-3}^{-2} (x - 3) dx = \left[ \frac{x^2}{2} - 3x \right]_{-3}^{-2} = \left( \frac{(-2)^2}{2} - 3(-2) \right) - \left( \frac{(-3)^2}{2} - 3(-3) \right) = (2 + 6) - \left( \frac{9}{2} + 9 \right) = 8 - \frac{27}{2} = -\frac{11}{2}$

• For $x \in [-2, -1)$, $[x] = -2$: $\int_{-2}^{-1} (x - 2) dx = \left[ \frac{x^2}{2} - 2x \right]_{-2}^{-1} = \left( \frac{(-1)^2}{2} - 2(-1) \right) - \left( \frac{(-2)^2}{2} - 2(-2) \right) = \left( \frac{1}{2} + 2 \right) - (2 + 4) = \frac{5}{2} - 6 = -\frac{7}{2}$

• For $x \in [-1, 0)$, $[x] = -1$: $\int_{-1}^0 (x - 1) dx = \left[ \frac{x^2}{2} - x \right]_{-1}^0 = \left( \frac{0^2}{2} - 0 \right) - \left( \frac{(-1)^2}{2} - (-1) \right) = 0 - \left( \frac{1}{2} + 1 \right) = -\frac{3}{2}$

• For $x \in [0, 1)$, $[x] = 0$: $\int_0^1 (x + 0) dx = \int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$

• For $x \in [1, 2)$, $[x] = 1$: $\int_1^2 (x + 1) dx = \left[ \frac{x^2}{2} + x \right]_1^2 = \left( \frac{2^2}{2} + 2 \right) - \left( \frac{1^2}{2} + 1 \right) = (2 + 2) - \left( \frac{1}{2} + 1 \right) = 4 - \frac{3}{2} = \frac{5}{2}$

• For $x \in [2, 3)$, $[x] = 2$: $\int_2^3 (x + 2) dx = \left[ \frac{x^2}{2} + 2x \right]_2^3 = \left( \frac{3^2}{2} + 2(3) \right) - \left( \frac{2^2}{2} + 2(2) \right) = \left( \frac{9}{2} + 6 \right) - (2 + 4) = \frac{21}{2} - 6 = \frac{9}{2}$

Step 7: Sum the results of the sub-integrals. $-\frac{11}{2} - \frac{7}{2} - \frac{3}{2} + \frac{1}{2} + \frac{5}{2} + \frac{9}{2} = \frac{-11 - 7 - 3 + 1 + 5 + 9}{2} = \frac{-21 + 15}{2} = \frac{-6}{2} = -3$

Alternatively, we can use the property that $\int_{-a}^a f(x) dx = \int_{-a}^a (x + [x]) dx = \int_{-a}^a x dx + \int_{-a}^a [x] dx$. Since $x$ is an odd function, $\int_{-a}^a x dx = 0$. So, $\int_{-a}^a (x + [x]) dx = \int_{-a}^a [x] dx$.

For $a=3$, we need to calculate $\int_{-3}^3 [x] dx$. $\int_{-3}^3 [x] dx = \int_{-3}^{-2} (-3) dx + \int_{-2}^{-1} (-2) dx + \int_{-1}^0 (-1) dx + \int_0^1 (0) dx + \int_1^2 (1) dx + \int_2^3 (2) dx$ $= (-3)(-2 - (-3)) + (-2)(-1 - (-2)) + (-1)(0 - (-1)) + (0)(1-0) + (1)(2-1) + (2)(3-2)$ $= (-3)(1) + (-2)(1) + (-1)(1) + 0 + (1)(1) + (2)(1)$ $= -3 - 2 - 1 + 0 + 1 + 2 = -3$

Common Mistakes & Tips

• Incorrectly splitting intervals: Ensure all critical points of absolute value and greatest integer functions within the integration limits are used to split the integral.
• Sign errors with negative numbers: Be careful when evaluating expressions involving negative numbers, especially when squaring them or applying the greatest integer function.
• Using symmetry incorrectly: While $\int_{-a}^a x dx = 0$ is correct, applying it to the entire integrand $x + [x]$ without separating the terms might lead to errors if $[x]$ is not odd.

Summary

The problem was solved in two parts. First, we used the given definite integral involving absolute values to determine the value of $a$, which was found to be $3$. In the second part, we used this value of $a$ to evaluate the integral of $x + [x]$ from $-a$ to $a$. By splitting the integral into sub-intervals based on the behavior of the greatest integer function and summing the results, we found the final value. The property of odd functions was also used to simplify the integral of $x$.

The final answer is \boxed{-3}.