Let $f:[0,4]\to R$ be a differentiable function. There exist $\alpha$ and $\beta$ in the open interval $(0,2)$ such that ${\int }_0^{4}f\left(t\right)dt=$

$\text{Let}F\left(x\right)={\int }_0^{x^2}f\left(t\right)dt\text{for}x\in [0,2]$

$F\left(x\right)$ is continuous and differentiable on $[0,2]$ and $F^{\mathrm{\prime }}\left(x\right)=2xf\left(x^2\right)$ .
Using LMVT for $F\left(x\right)$ on $[0,1]$ and $[1,2]$there exist $\alpha \in \left(\right(0,1)\text{and}\beta \in ((1,2)$
such that $F^{\mathrm{\prime }}\left(\alpha \right)=F\left(1\right)-F\left(0\right)$ and $F^{\mathrm{\prime }}\left(\beta \right)=F\left(2\right)-F\left(1\right)$
Adding $F\left(2\right)-F\left(0\right)=F^{\mathrm{\prime }}\left(\alpha \right)+F^{\mathrm{\prime }}\left(\beta \right)$

$=2\alpha f\left({\alpha }^2\right)+2\beta f\left({\beta }^2\right)$