Question

$\int_{\frac{1}{2}}^{\frac{3}{2}}\{\frac{1}{2}(|x-3|+|1-x|-4)\}dx$ equals-

• A. $-\frac{3}{2}$
• B. $\frac{9}{8}$
• C. $-\frac{1}{4}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{9}{8}$

Answer 2

The problem requires evaluating the definite integral $\int_{\frac{1}{2}}^{\frac{3}{2}}\{\frac{1}{2}(|x-3|+|1-x|-4)\}dx$, where $\{.\}$ denotes the fractional part function.

The fractional part function $\{y\}$ is defined as $y - \lfloor y \rfloor$, and its range is $[0, 1)$. Since the integrand is a fractional part, it must be non-negative. The interval of integration $[\frac{1}{2}, \frac{3}{2}]$ has a positive length. Therefore, the value of the integral must be non-negative. This eliminates options (A), (C), and (D) as they are negative.

Let's analyze the expression inside the fractional part function: $f(x) = \frac{1}{2}(|x-3|+|1-x|-4)$. The interval of integration is $[\frac{1}{2}, \frac{3}{2}]$. We need to consider the behavior of the absolute value functions within this interval.

1. $|x-3|$: For $x \in [\frac{1}{2}, \frac{3}{2}]$, $x$ is always less than $3$. Thus, $x-3$ is negative, and $|x-3| = -(x-3) = 3-x$.

2. $|1-x|$: The expression $1-x$ changes sign at $x=1$. Since $1$ is within our interval $[\frac{1}{2}, \frac{3}{2}]$, we must split the integral at $x=1$.

We divide the integral into two parts: $\int_{\frac{1}{2}}^{1} \dots dx$ and $\int_{1}^{\frac{3}{2}} \dots dx$.

Case 1: $x \in [\frac{1}{2}, 1]$ In this interval: $|x-3| = 3-x$ $|1-x| = 1-x$ (since $x \le 1$, $1-x \ge 0$)

Substitute these into $f(x)$: $f(x) = \frac{1}{2}((3-x) + (1-x) - 4)$ $f(x) = \frac{1}{2}(4 - 2x - 4)$ $f(x) = \frac{1}{2}(-2x) = -x$.

Now we need to find the fractional part of $f(x)$, which is $\{-x\}$. For $x \in [\frac{1}{2}, 1]$, the value of $-x$ is in the interval $[-1, -\frac{1}{2}]$. The fractional part $\{y\}$ is $y - \lfloor y \rfloor$.

• If $y = -1$ (i.e., $x=1$), then $\{-1\} = -1 - \lfloor -1 \rfloor = -1 - (-1) = 0$.
• If $y \in (-1, -\frac{1}{2}]$ (i.e., $x \in [\frac{1}{2}, 1)$), then $\lfloor y \rfloor = -1$. So, $\{y\} = y - (-1) = y+1$. Substituting back $y=-x$, we get $\{-x\} = -x+1$. Thus, for $x \in [\frac{1}{2}, 1]$, the integrand is $1-x$.

The integral over this interval is: $\int_{\frac{1}{2}}^{1} (1-x) dx = \left[x - \frac{x^2}{2}\right]_{\frac{1}{2}}^{1}$ $= \left(1 - \frac{1^2}{2}\right) - \left(\frac{1}{2} - \frac{(1/2)^2}{2}\right)$ $= \left(1 - \frac{1}{2}\right) - \left(\frac{1}{2} - \frac{1/4}{2}\right)$ $= \frac{1}{2} - \left(\frac{1}{2} - \frac{1}{8}\right)$ $= \frac{1}{2} - \frac{3}{8} = \frac{4}{8} - \frac{3}{8} = \frac{1}{8}$.

Case 2: $x \in [1, \frac{3}{2}]$ In this interval: $|x-3| = 3-x$ $|1-x| = -(1-x) = x-1$ (since $x \ge 1$, $1-x \le 0$)

Substitute these into $f(x)$: $f(x) = \frac{1}{2}((3-x) + (x-1) - 4)$ $f(x) = \frac{1}{2}(3 - x + x - 1 - 4)$ $f(x) = \frac{1}{2}(2 - 4)$ $f(x) = \frac{1}{2}(-2) = -1$.

Now we need to find the fractional part of $f(x)$, which is $\{-1\}$. $\{-1\} = -1 - \lfloor -1 \rfloor = -1 - (-1) = 0$. Thus, for $x \in [1, \frac{3}{2}]$, the integrand is $0$.

The integral over this interval is: $\int_{1}^{\frac{3}{2}} 0 dx = 0$.

Total Integral The total integral is the sum of the integrals from the two cases: $\int_{\frac{1}{2}}^{\frac{3}{2}}\{\frac{1}{2}(|x-3|+|1-x|-4)\}dx = \int_{\frac{1}{2}}^{1} (1-x) dx + \int_{1}^{\frac{3}{2}} 0 dx$ $= \frac{1}{8} + 0 = \frac{1}{8}$.

There appears to be a discrepancy between the calculated answer ($\frac{1}{8}$) and the provided options. Options (A), (C), and (D) are negative, which is impossible for an integral of a non-negative function. Option (B) is $\frac{9}{8}$. Given that negative options are impossible, and assuming there is a correct answer among the options, option (B) is the only plausible choice, despite the calculated value being $\frac{1}{8}$. This suggests a potential error in the question statement or the provided options. However, if forced to choose from the given options and knowing negative values are incorrect, $\frac{9}{8}$ is the only remaining possibility.