$\text{ If the function }\mathrm{f}:[0,16]\to \mathrm{R}\text{ is differentiable. If }0<\alpha <1\text{ and }$$1<\beta <2,\text{ then }{\int }_0^{16}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\text{ is cqual to }-$

$\mathrm{I}={\int }_0^{16}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$

$\text{ Consider }g\left(x\right)={\int }_0^{x^4}f\left(t\right)dt\Rightarrow g\left(0\right)=0$

$\text{ LMVT for }\mathrm{g}\text{ in }[0,1]\text{ gives, some }\alpha \in (0,1)\text{ such that }$

$\frac{\mathrm{g}\left(1\right)-\mathrm{g}\left(0\right)}{1-0}={\mathrm{g}}^{\text{'}}\left(\alpha \right) \dots \dots \dots$

$\text{ Similarly, LMVT in }[1,2]\text{ gives, some }\beta \in (1,2)\text{ such }$

$\text{ that }\frac{\mathrm{g}\left(2\right)-\mathrm{g}\left(1\right)}{2-1}={\mathrm{g}}^{\text{'}}\left(\beta \right)\dots \dots \dots .\left(2\right)$

$\text{ Eq. }\left(1\right)+\text{ Eq. }\left(2\right)$

$g\text{'}\left(\alpha \right)+g\text{'}\left(\beta \right)=g\left(2\right)-\underset{\text{zero }}{\underbrace{g\left(0\right)}};\text{ but }g\text{'}\left(x\right)=f\left(x^4\right)·4x^3$

$\therefore 4\left[{\alpha }^3\mathrm{f}\left({\alpha }^4\right)+{\beta }^3\mathrm{f}\left({\beta }^4\right)\right]={\int }_0^{16}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$