What is \(\mathop \smallint \nolimits_a^b \left[ x \right]dx + \mathop \smallint \nolimits_a^b \left[ { - x} \right]dx\) equal to, where [.] is the greatest integer function?
a - b and 0
Solving the Definite Integral of Greatest Integer Function
The question asks us to evaluate the definite integral \(\mathop \smallint \nolimits_a^b \left[ x \right]dx + \mathop \smallint \nolimits_a^b \left[ { - x} \right]dx\), where \([x]\) represents the greatest integer function.
First, let's analyze the integrand, which is the sum of the greatest integer of \(x\) and the greatest integer of \(-x\). We need to understand the property of \(\left[ x \right] + \left[ { - x} \right]\).
- Case 1: \(x\) is an integer. Let \(x = n\), where \(n\) is an integer. Then \(\left[ x \right] = \left[ n \right] = n\) and \(\left[ { - x} \right] = \left[ { - n} \right] = -n\). So, \(\left[ x \right] + \left[ { - x} \right] = n + (-n) = 0\).
- Case 2: \(x\) is not an integer. Let \(x\) be a non-integer. Then there exists an integer \(n\) such that \(n < x < n+1\). From \(n < x < n+1\), we get \(\left[ x \right] = n\). Also, multiplying the inequality by -1, we get \(-(n+1) < -x < -n\). Therefore, the greatest integer less than or equal to \(-x\) is \(-(n+1)\), so \(\left[ { - x} \right] = -(n+1)\). So, \(\left[ x \right] + \left[ { - x} \right] = n + (-(n+1)) = n - n - 1 = -1\).
Combining these two cases, we find that:
\[ \left[ x \right] + \left[ { - x} \right] = \begin{cases} 0 & \text{if } x \in \mathbb{Z} \\ -1 & \text{if } x \notin \mathbb{Z} \end{cases} \]
Now, let's consider the given integral:
\[ \mathop \smallint \nolimits_a^b \left[ x \right]dx + \mathop \smallint \nolimits_a^b \left[ { - x} \right]dx \]
Using the property of integrals that the sum of integrals is the integral of the sum (provided the limits are the same), we can write this as:
\[ \mathop \smallint \nolimits_a^b \left( \left[ x \right] + \left[ { - x} \right] \right)dx \]
Let \(f(x) = \left[ x \right] + \left[ { - x} \right]\). As we determined, \(f(x)\) is equal to -1 for all non-integer values of \(x\). The value of \(f(x)\) is 0 only at integer points.
For definite integrals, the value of the integral of a function is not affected by the value of the function at a finite number of points or even a countably infinite number of points (like the integers). The integral is determined by the value of the function over intervals.
Since \(f(x) = -1\) for all \(x\) in the interval \((a, b)\) except possibly at integer points, the integral of \(f(x)\) from \(a\) to \(b\) is the same as the integral of the constant function \(-1\) from \(a\) to \(b\).
\[ \mathop \smallint \nolimits_a^b \left( \left[ x \right] + \left[ { - x} \right] \right)dx = \mathop \smallint \nolimits_a^b (-1) dx \]
Evaluating the integral of the constant function \(-1\):
\[ \mathop \smallint \nolimits_a^b (-1) dx = [-x]_a^b = (-b) - (-a) = -b + a = a - b \]
Thus, the value of the given integral is \(a-b\).
Understanding the Option "a - b and 0"
Our calculation shows the integral is equal to \(a-b\) for any real numbers \(a\) and \(b\). The correct option is given as "a - b and 0". This phrasing might seem confusing as it suggests two possible values. However, the value of the integral is uniquely determined by \(a\) and \(b\), and that value is \(a-b\).
The phrasing "a - b and 0" likely means that the expression \(a-b\) is the general result, and 0 is highlighted as a possible value that \(a-b\) can take. Specifically, when \(a=b\), the limits of integration are the same, and the integral from \(a\) to \(a\) is always 0. In this case, \(a-b = a-a = 0\). So, when \(a=b\), the integral is 0, which is consistent with the formula \(a-b\).
Therefore, the value of the integral is \(a-b\), and 0 is included in the option because it is the specific value of \(a-b\) when \(a=b\). The option essentially states the general form of the answer and notes a special case where the answer is 0.
Summary of the Solution
- The integrand \(\left[ x \right] + \left[ { - x} \right]\) is equal to 0 if \(x\) is an integer and -1 if \(x\) is not an integer.
- The definite integral of a function that is -1 almost everywhere in an interval \([a, b]\) is \(\mathop \smallint \nolimits_a^b (-1) dx\).
- \(\mathop \smallint \nolimits_a^b (-1) dx = a - b\).
- The value of the given integral is \(a-b\).
- The option "a - b and 0" includes the general result \(a-b\) and highlights the value 0, which occurs when \(a=b\).
| Value of \(x\) | \(\left[ x \right]\) | \(\left[ { - x} \right]\) | \(\left[ x \right] + \left[ { - x} \right]\) |
|---|---|---|---|
| Integer (\(n\)) | \(n\) | \(-n\) | \(0\) |
| Non-integer (\(n < x < n+1\)) | \(n\) | \(-\left( {n+1} \right)\) | \(-1\) |
Revision Table: Definite Integral and Greatest Integer Function
Here is a quick review of the concepts used:
| Concept | Description | Relevance to Problem |
|---|---|---|
| Greatest Integer Function \([x]\) | Largest integer less than or equal to \(x\). | The core function in the integrand. |
| Property: \(\left[ x \right] + \left[ { - x} \right]\) | Equals 0 for integers, -1 for non-integers. | Simplifies the integrand significantly. |
| Definite Integral | Represents the signed area under a curve between limits \(a\) and \(b\). | The operation requested in the problem. |
| Integral of a constant \(c\) | \(\mathop \smallint \nolimits_a^b c \, dx = c(b-a)\). | Used because the integrand is constant almost everywhere. |
| Integral property: \(\mathop \smallint \nolimits_a^b (f(x) + g(x))dx\) | \(\mathop \smallint \nolimits_a^b f(x)dx + \mathop \smallint \nolimits_a^b g(x)dx\). | Used to combine the two initial integrals. |
| Integral over a single point | \(\mathop \smallint \nolimits_a^a f(x)dx = 0\). | Explains why the result is 0 when \(a=b\). |
Additional Information on Greatest Integer Function Integrals
Integrals involving the greatest integer function or fractional part function (\(\{x\} = x - [x]\)) often require splitting the interval of integration at integer points. For example, to calculate \(\mathop \smallint \nolimits_0^3 \left[ x \right]dx\), you would split it as:
\[ \mathop \smallint \nolimits_0^3 \left[ x \right]dx = \mathop \smallint \nolimits_0^1 \left[ x \right]dx + \mathop \smallint \nolimits_1^2 \left[ x \right]dx + \mathop \smallint \nolimits_2^3 \left[ x \right]dx \]
In each interval \([n, n+1)\), \(\left[ x \right] = n\). So the integral becomes:
\[ \mathop \smallint \nolimits_0^1 0 \, dx + \mathop \smallint \nolimits_1^2 1 \, dx + \mathop \smallint \nolimits_2^3 2 \, dx \] \[ = 0 \cdot (1-0) + 1 \cdot (2-1) + 2 \cdot (3-2) = 0 + 1 + 2 = 3 \]
However, in our problem, the integrand \(\left[ x \right] + \left[ { - x } \right]\) simplifies to a function that is piecewise constant with values 0 or -1. Because the function is -1 everywhere except at integer points, the total integral value is determined by the length of the interval multiplied by -1, which is \(a-b\). The points where the function is 0 do not contribute to the integral value.
The final answer is \(\boxed{a - b \text{ and } 0}\)
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