Question

The value of $I = \int_{0}^{1.5} \left\lfloor x^2 \right\rfloor dx$, where [ ] denotes the greatest integer function, is:

• A. $2 - \sqrt{2}$
• B. $\sqrt{2}$
• C. $5 \sqrt{2}$
• D. $3 - 2 \sqrt{2}$

Answer 1

Answer: (A)

$2 - \sqrt{2}$

Answer 2

The greatest integer function $[x^2]$ takes different values depending on $x^2$.

Over $0 \leq x \leq \sqrt{2}$, we split the integral into ranges where $[x^2]$ is constant:

For $0 \leq x < 1$, $x^2 < 1$ and $[x^2] = 0$. The contribution to the integral is:

$\int_{0}^{1} [x^2]dx = \int_{0}^{1} 0 \, dx = 0.$

For $1 \leq x \leq \sqrt{2}$, $1 \leq x^2 < 2$ and $[x^2] = 1$. The contribution to the integral is:

$\int_{1}^{\sqrt{2}} [x^2]dx = \int_{1}^{\sqrt{2}} 1 \, dx = (\sqrt{2} - 1).$

Adding these results:

$I = 0 + (\sqrt{2} - 1) = \sqrt{2} - 1.$

Thus, the value of $I$ is $2 - \sqrt{2}$.