Reverse the order of the double integral

(y+1) dx dy, where -1<= y <= 4 and y - 4 <= x <= 4y-y^2

To reverse the order of integration in a double integral, we need to convert the given limits of integration into the new order.

The given limits are:
-1 <= y <= 4
y - 4 <= x <= 4y - y^2

First, let's draw the region of integration by plotting the given inequalities on a coordinate plane.

The first inequality, -1 <= y <= 4, represents a vertical strip from y = -1 to y = 4.

The second inequality, y - 4 <= x <= 4y - y^2, represents a parabolic shape. We can rewrite the inequalities as follows:
x >= y - 4
x <= 4y - y^2

To find the intersection points, we can set the right-hand sides of both inequalities equal to each other:
y - 4 = 4y - y^2

Rearranging the equation to standard quadratic form:
y^2 - 3y + 4 = 0

Solving the quadratic equation, we find the roots:
y = (3 ± √5)/2

Therefore, the intersection points are y = (3 + √5)/2 and y = (3 - √5)/2.

Now, we can visualize the region of integration as a vertical strip bounded by the parabolic curve and the vertical lines y = -1 and y = 4.

To reverse the order of integration, we need to express the limits of integration in terms of x instead of y.

Let's break down the region into two parts based on the intersection points we found:
1. When -1 <= y < (3 - √5)/2
2. When (3 - √5)/2 <= y <= 4

For the first part, the limit of x should be expressed in terms of y.
From y - 4 <= x, we have x >= y - 4. Since y < (3 - √5)/2, the corresponding limit becomes x >= (3 - √5)/2 - 4.

For the second part, the limit of x should be expressed in terms of y as well.
From x <= 4y - y^2, we have x <= 4y - y^2. Since y > (3 - √5)/2, the corresponding limit becomes x <= 4y - y^2.

Therefore, the reversed order of the given double integral becomes:
∫∫ (y + 1) dx dy, where x = (3 - √5)/2 - 4 to x = 4y - y^2, and y = -1 to y = 4.

Now you can evaluate the double integral using the reversed order of integration.