Step 1: Find the key meeting points.
The parabola $y=x^2+x$ meets the line $y=x$ where $x^2+x=x$, i.e. $x=0$. It meets $y=2$ where $x^2+x-2=0$, i.e. $x=1$ (taking the positive side). So the parabola goes from $(0,0)$ to $(1,2)$.
Step 2: Picture the region.
The region is closed by the parabola, the line $y=x$, the line $x=1$, and the line $y=2$. It is easiest to use horizontal strips (in terms of $y$).
Step 3: Get $x$ from the parabola.
Solving $y=x^2+x$ for the right branch gives $x=\dfrac{-1+\sqrt{1+4y}}{2}$.
Step 4: Lower band, $0\le y\le1$.
Here the strip runs from the parabola (left) to the line $x=y$ (right). Its area is $\displaystyle\int_0^1\Big(y-\tfrac{-1+\sqrt{1+4y}}{2}\Big)dy=\dfrac{13}{12}-\dfrac{5\sqrt5}{12}$.
Step 5: Upper band, $1\le y\le2$.
Here the right edge is $x=1$ and the left edge is the parabola. Its area is $\displaystyle\int_1^2\Big(1-\tfrac{-1+\sqrt{1+4y}}{2}\Big)dy=-\dfrac34+\dfrac{5\sqrt5}{12}$.
Step 6: Add the two bands.
The $\sqrt5$ pieces cancel, leaving
\[ \left(\tfrac{13}{12}-\tfrac{5\sqrt5}{12}\right)+\left(-\tfrac34+\tfrac{5\sqrt5}{12}\right)=\frac{13}{12}-\frac{9}{12}=\frac{1}{3}. \]
\[ \boxed{\dfrac{1}{3}} \]