Evaluate f(x)dx (limits -5 to 5) , where f(x)= min ( {x-1} , {x+1} ) for all x€R , {.} denotes fractional part function

To evaluate the integral of the function \( f(x) = \min(\{x - 1\}, \{x + 1\}) \) from -5 to 5, we first need to understand the behavior of the fractional part function, denoted by \(\{x\}\). The fractional part of a number \( x \) is defined as \(\{x\} = x - \lfloor x \rfloor\), where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \( x \). This means that \(\{x\}\) always lies in the interval [0, 1). Let's break down the function \( f(x) \) and then evaluate the integral.

Understanding the Function f(x)

We need to analyze \( f(x) = \min(\{x - 1\}, \{x + 1\}) \). To do this, we will look at the fractional parts of \( x - 1 \) and \( x + 1 \).

• For \( x - 1 \), the fractional part is \(\{x - 1\} = x - 1 - \lfloor x - 1 \rfloor = x - 1 - (n - 1) = x - n\) if \( n \leq x < n + 1 \) for some integer \( n \).
• For \( x + 1 \), the fractional part is \(\{x + 1\} = x + 1 - \lfloor x + 1 \rfloor = x + 1 - (n + 1) = x - n\) if \( n \leq x < n + 1 \).

Thus, we can see that both \(\{x - 1\}\) and \(\{x + 1\}\) yield the same expression \( x - n \) in the interval \( [n, n + 1) \). However, we need to consider how these fractional parts behave across different intervals.

Behavior Across Intervals

Let’s analyze the function \( f(x) \) over the intervals defined by integers. The critical points occur at integer values where the behavior of the fractional parts changes. We can evaluate \( f(x) \) in the following intervals:

• For \( x \in [-5, -4) \): Here, \( \{x - 1\} = x - (-5) = x + 5 \) and \( \{x + 1\} = x - (-4) = x + 4 \). Thus, \( f(x) = \min(x + 5, x + 4) = x + 4 \).
• For \( x \in [-4, -3) \): \( f(x) = \min(x + 4, x + 3) = x + 3 \).
• For \( x \in [-3, -2) \): \( f(x) = \min(x + 3, x + 2) = x + 2 \).
• For \( x \in [-2, -1) \): \( f(x) = \min(x + 2, x + 1) = x + 1 \).
• For \( x \in [-1, 0) \): \( f(x) = \min(x + 1, x) = x \).
• For \( x \in [0, 1) \): \( f(x) = \min(x, x + 1) = x \).
• For \( x \in [1, 2) \): \( f(x) = \min(x, x + 1) = x \).
• For \( x \in [2, 3) \): \( f(x) = \min(x, x + 1) = x \).
• For \( x \in [3, 4) \): \( f(x) = \min(x, x + 1) = x \).
• For \( x \in [4, 5) \): \( f(x) = \min(x, x + 1) = x \).

Setting Up the Integral

Now that we have defined \( f(x) \) over the intervals, we can set up the integral:

\[ \int_{-5}^{5} f(x) \, dx = \int_{-5}^{-4} (x + 4) \, dx + \int_{-4}^{-3} (x + 3) \, dx + \int_{-3}^{-2} (x + 2) \, dx + \int_{-2}^{-1} (x + 1) \, dx + \int_{-1}^{0} x \, dx + \int_{0}^{5} x \, dx \]

Calculating Each Integral

Let’s calculate each integral one by one:

• \(\int_{-5}^{-4} (x + 4) \, dx = \left[ \frac{x^2}{2} + 4x \right]_{-5}^{-4} = \left( \frac{(-4)^2}{2} + 4(-4) \right) - \left( \frac{(-5)^2}{2} + 4(-5) \right) = (8 - 16) - \left( \frac{25}{2} - 20 \right) = -8 + \frac{15}{2} = -\frac{1}{2}\)

• \(\int_{-4}^{-3} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-4}^{-3} = \left( \frac{(-3)^2}{2} + 3(-3) \right) - \left( \frac{(-4)^2}{2} + 3(-4) \right) = \left( \frac{9}{2} - 9 \right) - \left( 8 - 12 \right) = -\frac{9}{2} + 4 = -\frac{1}{2}\)

• \(\int_{-3}^{-2} (x + 2)