Question

The value of \(\int\limits_{-1}^1x^2 [x] dx\) is:

• A. \(\frac{1}{3}\)
• B. \(\frac{2}{3}\)
• C. 1
• D. \(-\frac{1}{3}\)

Answer 1

The Correct Option is D

To evaluate the integral \(\int\limits_{-1}^1x^2 [x] dx\), we need to analyze the expression involving the floor function \([x]\). The integration range is from \(-1\) to \(1\), and the floor function \([x]\) implies step changes at integers within this range. We split the integral into separate intervals based on where \([x]\) changes value:

\(\int\limits_{-1}^1x^2 [x] dx = \int\limits_{-1}^0x^2(-1)dx + \int\limits_{0}^1x^2(0)dx\).

Evaluate the first integral:
\(\int\limits_{-1}^0x^2(-1)dx = -\int\limits_{-1}^0x^2dx\).

Find the antiderivative of \(x^2\):
\(\frac{x^3}{3}\).

Evaluate from \(-1\) to \(0\):
\(-\left[\frac{x^3}{3}\right]_{-1}^{0} = -\left(\frac{0^3}{3} - \frac{(-1)^3}{3}\right) = -\left(0 - \frac{-1}{3}\right) = -\left(\frac{1}{3}\right) = -\frac{1}{3}\).

Evaluate the second integral:
\(\int\limits_{0}^1x^2(0)dx = 0\).

Combine the results:
\(\int\limits_{-1}^1x^2[x]dx = -\frac{1}{3} + 0 = -\frac{1}{3}\).

Thus, the value of the integral is \(-\frac{1}{3}\).