Question

If \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of the integral \(\int _0^2x^2[x]\,dx\) is equal to

• A. \(\frac{8}{3}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{7}{3}\)
• D. \(0\)

Hint:

Split the interval at 1: the floor is 0 on [0,1) and 1 on [1,2).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The greatest integer function \([x]\) is 0 for \(0 \le x < 1\) and 1 for \(1 \le x < 2\). Split the integral at \(x = 1\).

Step 2: Split:
\[ \int_0^2 x^2[x]\,dx = \int_0^1 x^2 \cdot 0\,dx + \int_1^2 x^2 \cdot 1\,dx \]

Step 3: Evaluate:
\[ \int_1^2 x^2\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac13 = \frac73 \]

Step 4: Why the other options are wrong.
\(\frac83\) takes \(\int_0^2 x^2\) without the floor. \(\frac38\) is its reciprocal. 0 would hold only if \([x]\) were 0 on the whole interval.

Final Answer:
The integral equals \(\frac73\), option (C). \[ \boxed{\frac{7}{3}} \]