Double integration 
Evaluate
  $$
 \iint \limits_R(2xy+9) \;\mathrm{d}A
$$ 
  where $R$ is the region bounded by $y=x^2$ and $y=x+2$.

I have drawn my picture and have come up with my regions from $\sqrt{y}$ to $(y-2)$ for $\mathrm{d}x$ and $0$ to $4$ for $\mathrm{d}y$, but I seem to be computing these wrong though. Or are my bounds wrong?
 A: In order to find the bounds we have to find the values of $x$ for which $y=x^2$ and $y=x+2$ are equivalent. To do this set the equations equal to each and solve the quadratic $x^2=x+2$:
$$x^2-x-2=0 \implies (x-2)(x+1)=0 \implies x=-1,2 \implies y=1,4$$
Now it is a choice of whether you want to integrate over $y$ or $x$ first. By the looks of it you want to use $x$ first, but I highly recommend integrating over $y$ first:
$$\int_{-1}^2 \int_{x^2}^{x+2} (2xy+9)dydx = \int_{-1}^2 \Big( xy^2+9y \Big) \Big|_{x^2}^{x+2} dx = \int_{-1}^2 \Big( (x(x+2)^2+9(x+2)) - (x^5+9x^2) \Big) dx = \int_{-1}^2 (-x^5+x^3-5x^2+13x+18) dx = \frac{207}{4}$$
Notice that integrating over $y$ first led me to saying that the bottom curve is always the same on the interval and so is the top curve. Equivalently we can integrate over $x$ first to attain the same solution, but it will take more work since there is a split required in the bounds:
$$\int_0^1 \int_{-\sqrt{y}}^\sqrt{y} (2xy+9y)dxdy + \int_1^4 \int_{y-2}^\sqrt{y} (2xy+9y)dxdy$$
Specifically it is the left curve that switches at $y=1$. Below this line it is defined by $x=-\sqrt{y}$ and above by $x=y-2$. It is up to you whether you want to evaluate two double integrals or one, but the first method is definitely the easier choice.
