(1) The integral is straightforward; x ranges between two constants, and y ranges between two functions of x that don't intersect.
![\displaystyle\int_(-2)^1\int_(-x)^(x^2+2)\mathrm dy\,\mathrm dx](https://img.qammunity.org/2022/formulas/advanced-placement-ap/college/siyoaprcpnwvy2fmtozbtgjronzkkgf3eo.png)
(2) First find where the two curves intersect:
y ² - 4 = -3y
y ² + 3y - 4 = 0
(y + 4) (y - 1) = 0
y = -4, y = 1 → x = 12, x = -3
That is, they intersect at the points (-3, 1) and (12, -4). Since x ranges between two explicit functions of y, you can capture the area with one integral if you integrate with respect to x first:
![\displaystyle\int_(-4)^1\int_(y^2-4)^(-3y)\mathrm dx\,\mathrm dy](https://img.qammunity.org/2022/formulas/advanced-placement-ap/college/h7ispmzgq3hrgvkjitm9fj1o3qbg8s4r9r.png)
(3) No special tricks here, x is again bounded between two constants and y between two explicit functions of x.
![\displaystyle\int_1^5\int_0^{\frac1{x^2}}\mathrm dy\,\mathrm dx](https://img.qammunity.org/2022/formulas/advanced-placement-ap/college/5umz9oh63a5cvt150i28dp4ym8dle49eox.png)