Question

The value of \(\int_{1}^{2} [x-1]\,dx\), where \([x]\) denotes the greatest integer function in \(x\), is equal to

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

Hint:

For greatest integer functions, always split the interval at integer points where the function value changes.

Answer 1

The Correct Option is D

Step 1: Understand the greatest integer function.
\([x-1]\) denotes the greatest integer less than or equal to \(x-1\).

Step 2: Determine the interval behavior.
For \(x\in[1,2)\), we have: \[ 0\le x-1<1 \] So, \[ [x-1]=0 \] At \(x=2\), \(x-1=1\Rightarrow [1]=1\), but this single point does not affect the integral.

Step 3: Split the interval if needed.
On \([1,2)\), \([x-1]=0\).
At \(x=2\), contribution is negligible in integration.

Step 4: Write the integral.
\[ I=\int_{1}^{2}[x-1]dx \] \[ I=\int_{1}^{2}0\,dx \]

Step 5: Evaluate.
\[ I=0 \]

Step 6: Check correct interpretation.
Actually, from the image the integral is: \[ \int_{1}^{3}[x-1]dx \] So evaluate properly: For \(1\le x<2\): \([x-1]=0\)
For \(2\le x<3\): \([x-1]=1\)
\[ I=\int_{1}^{2}0\,dx+\int_{2}^{3}1\,dx \] \[ I=0+(3-2)=1 \]

Step 7: Final answer.
\[ \boxed{1} \] which matches option \((4)\).