Question

\([\,.\,]\) represents a greatest integer function. If \[ \int_{\sqrt{3}}^{\sqrt{18}}[x]\,dx=a+b\sqrt{2}+c\sqrt{3}, \] then \[ a+b+c= \] is:

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

For greatest integer function integrals, split the interval at integer points because \([x]\) remains constant between two consecutive integers.

Answer 1

The Correct Option is D

Step 1: Identify the breakpoints of $[x]$.
On $[\sqrt{3}, \sqrt{18}]$: $\sqrt{3} \approx 1.73$, $\sqrt{18} = 3\sqrt{2} \approx 4.24$. So $[x]$ takes values 1, 2, 3, 4 in this range.

Step 2: Split the integral at integer points.
\[ \int_{\sqrt{3}}^{3\sqrt{2}} [x]\,dx = \int_{\sqrt{3}}^{2} 1\,dx + \int_{2}^{3} 2\,dx + \int_{3}^{4} 3\,dx + \int_{4}^{3\sqrt{2}} 4\,dx. \]

Step 3: Evaluate each piece.
$\int_{\sqrt{3}}^{2} 1\,dx = 2 - \sqrt{3}$. $\int_{2}^{3} 2\,dx = 2$. $\int_{3}^{4} 3\,dx = 3$. $\int_{4}^{3\sqrt{2}} 4\,dx = 4(3\sqrt{2}-4) = 12\sqrt{2} - 16$.

Step 4: Add all pieces.
Total $= (2 - \sqrt{3}) + 2 + 3 + (12\sqrt{2} - 16)$. $= 2 - \sqrt{3} + 2 + 3 + 12\sqrt{2} - 16$. $= -9 - \sqrt{3} + 12\sqrt{2}$. $= -9 + 12\sqrt{2} - \sqrt{3}$.

Step 5: Match with $a + b\sqrt{2} + c\sqrt{3}$.
$a = -9$, $b = 12$, $c = -1$.

Step 6: Compute $a + b + c$.
$a + b + c = -9 + 12 + (-1) = 2$.
\[ \boxed{2} \]