A calculus problem by Akshay Yadav

It's a delight to do calculations of this question without a calculator :-).....
First let us find the point of intersection of line and curve in third quadrant which is given by: $\begin{aligned}~~&x^3-x+1=x\\&\implies x^3-2x+1=0\\&\implies(x-1)(x^2+x-1)=0\\&\implies x=\dfrac{-1-\sqrt 5}{2}=-\phi\end{aligned}$ $(\because ~x<0)$ Area of shaded part is given by:- $\mathfrak{T}=\int_{-\phi}^0\left(x^3-2x+1\right)\, dx+\int_{0}^{0.4}\left(-x^3-2x+1\right)\, dx$ $\large=\left(\dfrac{x^4}{4}-x^2+x\right)|_{-\phi}^0+\left(\dfrac{-x^4}{4}-x^2+x\right)|_{0}^{\frac{2}{5}}$ $=\left(\phi^2-\phi-\frac{\phi^4}{4}\right)+\dfrac{146}{625}$ Putting $\phi=\dfrac{1+\sqrt 5}{2}$ and simplification gives:- $\Large\mathfrak{T}=\dfrac{6793}{5000}+\dfrac{5\sqrt5}{8}$ $\therefore 6793+5000+5+8=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{11806}}}}}}$

Note:-I have used everywhere the fact that $|x^3|=x^3$ when $x>0$ and $|x^3|=-x^3$ when $x<0$ .