Question

Let \([t]\) denote the greatest integer \(\leq t\). Then the value of \(8 \cdot \int_{-1/2}^{1} ([2x] + |x|) \, dx\) is _________
 

Hint:

Always look for symmetry or net areas in GIF integrals. In this case, the negative area of \([2x]\) from \(-0.5\) to \(0\) perfectly cancelled the positive area from \(0.5\) to \(1\).

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
The integral can be split into two parts: \(\int [2x] dx\) and \(\int |x| dx\). We integrate step functions and piecewise functions by breaking the interval at points of discontinuity or definition changes.
Step 2: Detailed Explanation:
Let \(I = \int_{-1/2}^{1} ([2x] + |x|) \, dx\).
1. Evaluating \(\int_{-1/2}^{1} [2x] dx\):
The value of \([2x]\) changes at \(2x = 0, 1 \implies x = 0, 1/2\).
\[ \int_{-1/2}^{0} -1 dx + \int_{0}^{1/2} 0 dx + \int_{1/2}^{1} 1 dx = [-1(0 - (-1/2))] + 0 + [1(1 - 1/2)] = -1/2 + 1/2 = 0 \]
2. Evaluating \(\int_{-1/2}^{1} |x| dx\):
\[ \int_{-1/2}^{0} -x dx + \int_{0}^{1} x dx = \left[ -\frac{x^2}{2} \right]_{-1/2}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{1} \]
\[ = [0 - (-1/8)] + [1/2 - 0] = 1/8 + 1/2 = 5/8 \]
3. Total Integral:
\[ I = 0 + 5/8 = 5/8 \]
The required value is \(8 \cdot I = 8 \cdot (5/8) = 5\).
Step 3: Final Answer:
The value is 5.