Let F : 3 , 5 → R be a twice differentiable function on 3 , 5 such that F x = e - x ∫ 3 x 3 t 2…

Given, 

$F\left(x\right)=e^{-x}{\int }_3^x\left(3t^2+2t+4F^{\text{'}}\left(t\right)\right)dt.$

So, $F\left(3\right)=e^{-3}{\int }_3^3\left(3t^2+2t+4F^{\text{'}}\left(t\right)\right)dt.$ 

$\Rightarrow F\left(3\right)=0$

Again, $e^{x}F\left(x\right)={\int }_3^x\left(3t^2+2t+4{ F}^{\text{'}}\left(t\right)\right)dt$

Differentiate both sides

$\Rightarrow e^{x}F\left(x\right)+e^x{F}^{\text{'}}\left(x\right)=3x^2+2x+4{ F}^{\text{'}}\left(x\right)$ (Newton Leibnitz Theorem for Definite Integral)

$\Rightarrow \left(e^x-4\right)\frac{dy}{dx}+e^{x}y=\left(3x^2+2x\right)$ (Use $y=F\left(x\right)$)

$\Rightarrow \frac{dy}{dx}+\frac{e^x}{\left(e^x-4\right)}y=\frac{\left(3x^2+2x\right)}{\left(e^x-4\right)}$

Here, integrating factor is $e^{\int \frac{e^x}{e^x-4}dx}=e^{\ln \left(e^x-4\right)}=e^x-4$

So, the solution is 

$y·\left(e^x-4\right)=\int \left(3x^2+2x\right)dx+c$

$\Rightarrow y·\left(e^x-4\right)=\left(x^3+x^2\right)+c$

Put $x=3, y=0\Rightarrow c=-36$

So, $F\left(x\right)=\frac{\left(x^3+x^2-36\right)}{\left(e^x-4\right)}$

$\Rightarrow F^{\text{'}}\left(x\right)=\frac{\left(3x^2+2x\right)\left(e^x-4\right)-\left(x^3+x^2-36\right)e^x}{{\left(e^x-4\right)}^2}$

Now put value of $x=4$ we will get

$F^{\text{'}}\left(4\right)=\frac{\left(3{\left(4\right)}^2+2\left(4\right)\right)\left(e^4-4\right)-\left({\left(4\right)}^3+{\left(4\right)}^2-36\right)e^4}{{\left(e^4-4\right)}^2}=\frac{12e^4-224}{{\left(e^4-4\right)}^2}$

So, $\alpha =12 \& \beta =4$

Hence, $\alpha +\beta =16$