Q249 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let $f(x) = [x^2]$, where $[\cdot]$ is the greatest integer function.
What is $\int_{\sqrt{2}}^{2} f(x) \, dx$ equal to?
Let $f(x) = [x^2]$, where $[\cdot]$ is the greatest integer function.
What is $\int_{\sqrt{2}}^{2} f(x) \, dx$ equal to?
- Correct answerA$6 - \sqrt{3} - 2\sqrt{2}$
- B$6 - \sqrt{3} - \sqrt{2}$
- C$6 - \sqrt{3} + 2\sqrt{2}$
- D$6 + \sqrt{3} - 2\sqrt{2}$
Solution
The given integral is $\int_{\sqrt{2}}^{2} [x^2] \, dx$.
As $x$ varies from $\sqrt{2}$ to $2$, $x^2$ varies from $2$ to $4$. The greatest integer function $[x^2]$ changes its value when $x^2$ is an integer. The integer values of $x^2$ in this interval are $2, 3,$ and $4$, which correspond to $x = \sqrt{2}, \sqrt{3},$ and $2$.
We split the integral at $x = \sqrt{3}$:
For $x \in (\sqrt{2}, \sqrt{3})$, $2 < x^2 < 3 \Rightarrow [x^2] = 2$
For $x \in (\sqrt{3}, 2)$, $3 < x^2 < 4 \Rightarrow [x^2] = 3$
The integral can be written as:
$\int_{\sqrt{2}}^{2} [x^2] \, dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$
Evaluating the integrals:
$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2[x]_{\sqrt{2}}^{\sqrt{3}} = 2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$
$\int_{\sqrt{3}}^{2} 3 \, dx = 3[x]_{\sqrt{3}}^{2} = 3(2 - \sqrt{3}) = 6 - 3\sqrt{3}$
Adding the two values:
$\int_{\sqrt{2}}^{2} [x^2] \, dx = (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3})$
$\int_{\sqrt{2}}^{2} [x^2] \, dx = 6 - \sqrt{3} - 2\sqrt{2}$
Answer: $6 - \sqrt{3} - 2\sqrt{2}$
As $x$ varies from $\sqrt{2}$ to $2$, $x^2$ varies from $2$ to $4$. The greatest integer function $[x^2]$ changes its value when $x^2$ is an integer. The integer values of $x^2$ in this interval are $2, 3,$ and $4$, which correspond to $x = \sqrt{2}, \sqrt{3},$ and $2$.
We split the integral at $x = \sqrt{3}$:
For $x \in (\sqrt{2}, \sqrt{3})$, $2 < x^2 < 3 \Rightarrow [x^2] = 2$
For $x \in (\sqrt{3}, 2)$, $3 < x^2 < 4 \Rightarrow [x^2] = 3$
The integral can be written as:
$\int_{\sqrt{2}}^{2} [x^2] \, dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$
Evaluating the integrals:
$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2[x]_{\sqrt{2}}^{\sqrt{3}} = 2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$
$\int_{\sqrt{3}}^{2} 3 \, dx = 3[x]_{\sqrt{3}}^{2} = 3(2 - \sqrt{3}) = 6 - 3\sqrt{3}$
Adding the two values:
$\int_{\sqrt{2}}^{2} [x^2] \, dx = (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3})$
$\int_{\sqrt{2}}^{2} [x^2] \, dx = 6 - \sqrt{3} - 2\sqrt{2}$
Answer: $6 - \sqrt{3} - 2\sqrt{2}$
Hence, the correct answer is option: 'A'.