Question

The value of the integral, ∫ from 1 to 3 of [x² - 2x - 2] dx, where [x] denotes the greatest integer less than or equal to x, is :

• A. -5
• B. -4
• C. -√2 - √3 - 1
• D. -√2 - √3 + 1

Hint:

To integrate $[f(x)]$, break the interval into sub-intervals where $f(x)$ stays between two consecutive integers.

Answer 1

The Correct Option is C

Step 1: Let $f(x) = x^2 - 2x - 2 = (x-1)^2 - 3$.
Step 2: For $x \in [1, 3]$, $(x-1)^2$ ranges from $0$ to $4$. Thus $f(x)$ ranges from $-3$ to $1$.
Step 3: Critical points where $f(x)$ is an integer: - $f(x) = -3 \Rightarrow (x-1)^2=0 \Rightarrow x=1$. - $f(x) = -2 \Rightarrow (x-1)^2=1 \Rightarrow x=2$. - $f(x) = -1 \Rightarrow (x-1)^2=2 \Rightarrow x=1+\sqrt{2}$. - $f(x) = 0 \Rightarrow (x-1)^2=3 \Rightarrow x=1+\sqrt{3}$. - $f(x) = 1 \Rightarrow (x-1)^2=4 \Rightarrow x=3$.
Step 4: Integrate by parts: $\int_1^2 (-3)dx + \int_2^{1+\sqrt{2}} (-2)dx + \int_{1+\sqrt{2}}^{1+\sqrt{3}} (-1)dx + \int_{1+\sqrt{3}}^3 (0)dx$.
Step 5: $-3(1) - 2(\sqrt{2}-1) - 1(\sqrt{3}-\sqrt{2}) = -3 - 2\sqrt{2} + 2 - \sqrt{3} + \sqrt{2} = -1 - \sqrt{2} - \sqrt{3}$.