We have,

$y=x^2+x+10$

$\Rightarrow x^2+x+\frac{1}{4}=y-10+\frac{1}{4}$

$\Rightarrow {\left(x+\frac{1}{2}\right)}^2=\left(y-\frac{39}{4}\right)$

Latusrectum of parabola is $1.$

$\therefore$ Length of chord is also $1.$

Area is maximum when chord is latusrectum.

Area of shaded region$=$Area of rectangle $OABC-2{\int }_{-1/2}^0$parabola

$=10-2{\int }_{\frac{-1}{2}}^0\left(x^2+x+10\right)dx$

$=10-2{\left[\frac{x^3}{3}+\frac{x^2}{2}+10\right]}_{-\frac{1}{2}}^0$

$=10-2\left[\frac{-1}{24}+\frac{1}{8}-\frac{10}{2}\right]=\frac{1}{6}$