What is `int_a^b [x] dx+ int_a^b [-x]dx` equal to where `[,]` is the greater integer function

Options

A. `b-a`
B. `a-b`
C. 0
D. `2(b-a)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx \]

where \([x]\) is the greatest integer function (also known as the floor function), and \([-x]\) is the greatest integer function applied to \(-x\).

### Step 1: Rewrite the integrals

We can combine the two integrals into one:

\[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx = \int_a^b \left( [x] + [-x] \right) \, dx \]

### Step 2: Analyze the expression \([x] + [-x]\)

For any real number \(x\), the greatest integer function \([x]\) gives the largest integer less than or equal to \(x\), and \([-x]\) gives the largest integer less than or equal to \(-x\).

The key observation here is that:

\[ [x] + [-x] = -1 \quad \text{for } x \text{ not an integer} \]

This is because if \(x\) is not an integer, then \([x] = n\) (where \(n\) is an integer) and \([-x] = -n - 1\), leading to:

\[ [x] + [-x] = n + (-n - 1) = -1 \]

### Step 3: Evaluate the integral

Since the expression \([x] + [-x] = -1\) holds for all \(x\) that are not integers, we can write:

\[ \int_a^b \left( [x] + [-x] \right) \, dx = \int_a^b -1 \, dx \]

Now we can compute this integral:

\[ \int_a^b -1 \, dx = -\int_a^b 1 \, dx = -[x]_a^b = -[b - a] = -(b - a) = a - b \]

### Conclusion

Thus, we find that:

\[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx = a - b \]

### Final Answer

The final result is:

\[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx = a - b \]

---