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

Question

Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

Answer 1

The greatest integer function $[x^2]$ exhibits varying values based on $x^2$.

Within the range $0 \leq x \leq \sqrt{2}$, the integral is segmented into intervals where $[x^2]$ remains constant:

For $0 \leq x<1$, $x^2<1$, thus $[x^2] = 0$. The integral's contribution is:

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

For $1 \leq x \leq \sqrt{2}$, $1 \leq x^2<2$, so $[x^2] = 1$. The integral's contribution is:

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

Summing these contributions:

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

Therefore, the value of $I$ is $\sqrt{2} - 1$.

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