Integration with Square Roots

Let $t = \sqrt{9 + x^2}$ , so we have $dt = \frac{1}{2\sqrt{9+x^2}}(2x) dx = \frac{xdx}{\sqrt{9+x^2}}$ $\Rightarrow dx = \frac{\sqrt{9+x^2}dt}{x}$ . The limits get transformed to $t = \sqrt{9+0^2} = 3$ and $t = \sqrt{9 + 4^2} = 5$ . Substituting the above into the integral, we have $\begin{aligned} \int_0^4 x^3\sqrt{9+x^2} dx &= \int_3^5 t^2(t^2-9)dt \\ &= \left[\frac{t^5}{5} - 3t^3\right]_3^5 \\ &= 250 - \left(-\frac{129}{4}\right) \\ &= \frac{1412}{5} \\ \end{aligned}$

Thus $a = 282.4$ , hence $\lfloor a \rfloor = 282$ .