Question

Let $[t]$ denote the largest integer less than or equal to $t$. If \[ \int_0^1 \left(\left[x^2\right] + \left\lfloor \frac{x^2}{2} \right\rfloor\right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}, \] where $a, b, c \in \mathbb{Z}$, then $a + b + c$ is equal to ________.

Answer 1

Correct Answer: 23

Breaking the integral into intervals based on values of \( x \):
\[\int_0^3 \left( \lfloor x^2 \rfloor + \left\lfloor \frac{x^2}{2} \right\rfloor \right) dx\]
\[= \int_0^1 0 \, dx + \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{\sqrt{5}} 3 \, dx + \int_{\sqrt{5}}^{\sqrt{6}} 4 \, dx + \int_{\sqrt{6}}^{\sqrt{7}} 5 \, dx\]
\[+ \int_{\sqrt{7}}^{\sqrt{8}} 6 \, dx + \int_{\sqrt{8}}^3 7 \, dx\]
After evaluating each integral, we get:
\[a = 31, \quad b = -6, \quad c = -2\]
Thus,
\[a + b + c = 31 - 6 - 2 = 23\]