$\displaystyle{x^2}$ Laid Down On The Floor

$\displaystyle\int_{1}^{2} x^{\lfloor x^{2} \rfloor}+\lfloor x^{2} \rfloor^{x} dx$

$\displaystyle\int_{1}^{\sqrt{2}} x^{\lfloor x^{2} \rfloor}+\lfloor x^{2} \rfloor^{x} dx+\displaystyle\int_{\sqrt{2}}^{\sqrt{3}} x^{\lfloor x^{2} \rfloor}+\lfloor x^{2} \rfloor^{x} dx+\displaystyle\int_{\sqrt{3}}^{2} x^{\lfloor x^{2} \rfloor}+\lfloor x^{2} \rfloor^{x} dx$

$=\displaystyle\int_{1}^{\sqrt{2}} (x+1) dx+\displaystyle\int_{\sqrt{2}}^{\sqrt{3}} (x^{2}+2^{x}) dx+\displaystyle\int_{\sqrt{3}}^{2} x^{3}+3^{x} dx$

$=\bigg[\frac{x^{2}}{2}+x\bigg]_{1}^{\sqrt{2}}+\bigg[\frac{x^{3}}{3}+\frac{2^{x}}{ln 2}\bigg]_{\sqrt{2}}^{\sqrt{3}}+\bigg[\frac{x^{4}}{4}+\frac{3^{x}}{ln 3}\bigg]_{\sqrt{3}}^{2}$

$=\frac{\sqrt{2}}{3}+\sqrt{3}+\frac{2^{\sqrt{3}}-2^{\sqrt{2}}}{ln 2}+\frac{9-3^{\sqrt{3}}}{ln 3}+\frac{5}{4}$

Hence answer is 5+4+3+2+3+3+2+2+9+3+3= $\boxed{39}$

Integrate (x+1) from 1 to root 2 + Integrate (x^2 + 2^x) from root 2 to root 3 + Integrate (x^3 + 3^x) from root 3 to 2 Simplify & add Answer is 39