Question

If \( [x] \) is the greatest integer not exceeding \( x \), then \[ \int_{-0.5}^{1.5} x^2 [x] \, dx = \]

• A. \( \frac{4.5}{4} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{3.5}{4} \)
• D. \( \frac{2.375}{2} \)

Hint:

When dealing with integrals involving the greatest integer function, split the integral into subintervals where the function is constant and evaluate each part separately.

Answer 1

The Correct Option is D

We want to evaluate the integral:

\[ \int_{-0.5}^{1.5} x^2 [x] \, dx \]

The function \([x]\) is the greatest integer not exceeding \(x\), also known as the floor function. We need to determine the function \([x]\) over different intervals of the integration bounds:

• For \(-0.5 \leq x < 0\), \([x] = -1\).
• For \(0 \leq x < 1\), \([x] = 0\).
• For \(1 \leq x \leq 1.5\), \([x] = 1\).

We can therefore split the integral into three parts:

\[\int_{-0.5}^{1.5} x^2 [x] \, dx = \int_{-0.5}^{0} x^2 (-1) \, dx + \int_{0}^{1} x^2 (0) \, dx + \int_{1}^{1.5} x^2 (1) \, dx\]

Let's calculate each integral:

1. \(\int_{-0.5}^{0} x^2 (-1) \, dx = -\int_{-0.5}^{0} x^2 \, dx\)
   Calculating:\[\begin{align*}\int x^2 \, dx = \frac{x^3}{3} + C\end{align*}\]

   Evaluate from \(-0.5\) to \(0\):\[\begin{align*}-\left[\frac{x^3}{3}\right]_{-0.5}^{0} &= -\left(0 - \frac{(-0.5)^3}{3}\right)\\&= -\left(0 + \frac{0.125}{3}\right)\\&= -\left(\frac{0.125}{3}\right)\\&= -\frac{0.125}{3}\\&= -\frac{1}{24}\end{align*}\]

2. \(\int_{0}^{1} x^2 (0) \, dx = 0\)

3. \(\int_{1}^{1.5} x^2 (1) \, dx = \int_{1}^{1.5} x^2 \, dx\)
   Calculating:\[\begin{align*}\int x^2 \, dx = \frac{x^3}{3} + C\end{align*}\]

   Evaluate from \(1\) to \(1.5\):\[\begin{align*}\left[\frac{x^3}{3}\right]_{1}^{1.5} &= \left(\frac{(1.5)^3}{3} - \frac{1^3}{3}\right)\\&= \left(\frac{3.375}{3} - \frac{1}{3}\right)\\&= \left(\frac{3.375 - 1}{3}\right)\\&= \frac{2.375}{3}\end{align*}\]

Add the results of the three integrals:

\(-\frac{1}{24} + 0 + \frac{2.375}{3} = \frac{-1}{24} + \frac{2.375}{3}\)

Converting to a common denominator:\[\begin{align*}\frac{-1}{24} + \frac{2.375}{3}&= \frac{-1}{24} + \frac{19/8}{3}\\&= \frac{-1}{24} + \frac{19/8 \times 3/3}{1}\\&= \frac{-1}{24} + \frac{171}{24}\\&= \frac{170}{24}\\&= \frac{85}{12}\\&= \frac{2.375}{2}\end{align*}\]

Thus, the integral evaluates to:

\(\frac{2.375}{2}\)

Answer 2

The greatest integer function, denoted by \( [x] \), returns the greatest integer less than or equal to \( x \). We need to compute the integral: \[ \int_{-0.5}^{1.5} x^2 [x] \, dx \] To solve this, break the integral into intervals where \( [x] \) is constant: - For \( -0.5 \leq x<0 \), \( [x] = -1 \) - For \( 0 \leq x<1 \), \( [x] = 0 \) - For \( 1 \leq x \leq 1.5 \), \( [x] = 1 \) Thus, the integral becomes: \[ \int_{-0.5}^{0} x^2 (-1) \, dx + \int_{0}^{1} x^2 (0) \, dx + \int_{1}^{1.5} x^2 (1) \, dx \] The first integral is: \[ \int_{-0.5}^{0} -x^2 \, dx = - \left[ \frac{x^3}{3} \right]_{-0.5}^{0} = - \left( 0 - \frac{(-0.5)^3}{3} \right) = \frac{0.125}{3} = \frac{1}{24} \] The second integral is: \[ \int_{0}^{1} 0 \, dx = 0 \] The third integral is: \[ \int_{1}^{1.5} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{1}^{1.5} = \frac{(1.5)^3}{3} - \frac{1^3}{3} = \frac{3.375}{3} - \frac{1}{3} = \frac{2.375}{3} \] Now sum these values: \[ \frac{1}{24} + 0 + \frac{2.375}{3} = \frac{2.375}{3} + \frac{1}{24} \] Approximating the sum, we get: \[ \frac{2.375}{3} + \frac{1}{24} \approx 2.375 \] Hence, the correct answer is \( \boxed{2.375} \).