Tricky integration!

$\begin{aligned} I & = \int_{-3}^3 x^8\left \{x^{11}\right \} dx \\ & = \int_{-3}^3 x^8\left( x^{11} - \left \lfloor x^{11} \right \rfloor \right) dx \\ & = {\color{#3D99F6} \int_{-3}^3 x^{19} dx} - \int_{-3}^3 x^8 \left \lfloor x^{11} \right \rfloor dx & \small \color{#3D99F6} \text{Note that } x^{19} \text{ is odd.} \\ & = {\color{#3D99F6} 0} - {\color{#D61F06} \int_{-3}^0 x^8 \left \lfloor x^{11} \right \rfloor dx} - \int_0^3 x^8 \left \lfloor x^{11} \right \rfloor dx & \small \color{#D61F06} \text{Replace }x \text{ with }-x \\ & = - {\color{#D61F06} \int_0^3 x^8 \left \lfloor -x^{11} \right \rfloor dx} - \int_0^3 x^8 \left \lfloor x^{11} \right \rfloor dx \\ & = - \int_0^3 x^8 \left({\color{#3D99F6}\left \lfloor -x^{11} \right \rfloor + \left \lfloor x^{11} \right \rfloor} \right) dx & \small \color{#3D99F6} \text{Note that } \lfloor x \rfloor + \lfloor -x \rfloor = - 1 \text{ for non-integer} x \\ & = \int_0^3 x^8 \ dx = \frac {x^9}9 \ \bigg|_0^3 = \boxed{2187} \end{aligned}$

You have to use two properties here . One of integration and other of fractional part. $\displaystyle\int_{a}^{b}f(x) {\,dx}$ = $\displaystyle\int_{a}^{b}f(a+b-x) {\,dx}$ and

$\langle x \rangle$ $+$ $\langle -x \rangle$ $=$ 1