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: Split the interval according to greatest integer values.
We have \[ \sqrt{3}\lt 2,\qquad \sqrt{18}=3\sqrt{2}\gt 4 \] So, \[ \int_{\sqrt{3}}^{\sqrt{18}}[x]\,dx = \int_{\sqrt{3}}^{2}1\,dx+\int_{2}^{3}2\,dx+\int_{3}^{4}3\,dx+\int_{4}^{3\sqrt{2}}4\,dx \]

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

Step 3: Add all terms.
\[ \int_{\sqrt{3}}^{\sqrt{18}}[x]\,dx = 2-\sqrt{3}+2+3+12\sqrt{2}-16 \] \[ =12\sqrt{2}-\sqrt{3}-9 \] Comparing with \[ a+b\sqrt{2}+c\sqrt{3}, \] we get \[ a=-9,\quad b=12,\quad c=-1 \] Therefore, \[ a+b+c=-9+12-1=2 \]

Step 4: Final conclusion.
Hence, \[ \boxed{2} \]