The area of the region {(x,y): x2≤y≤8−x2,y≤7} is

The area of the region can be found by integrating the functions that define the boundaries of the region.

The region is defined by three inequalities:

1. y ≥ x^2
2. y ≤ 8 - x^2
3. y ≤ 7

The intersection of these inequalities defines the region.

First, we need to find the points of intersection of the curves y = x^2 and y = 8 - x^2. We set them equal to each other and solve for x:

x^2 = 8 - x^2 2x^2 = 8 x^2 = 4 x = -2, 2

So, the points of intersection are (-2, 4) and (2, 4).

Next, we need to find the points of intersection of the curves y = x^2 and y = 7. We set them equal to each other and solve for x:

x^2 = 7 x = -sqrt(7), sqrt(7)

So, the points of intersection are (-sqrt(7), 7) and (sqrt(7), 7).

Now, we can set up the integral to find the area of the region. The area is given by the integral of the upper function minus the lower function, from the leftmost point of intersection to the rightmost point of intersection.

The upper function is y = min(8 - x^2, 7) and the lower function is y = x^2.

So, the area A is given by:

A = ∫ from -2 to 2 of (min(8 - x^2, 7) - x^2) dx + ∫ from 2 to sqrt(7) of (7 - x^2) dx

This integral can be computed using standard techniques of integration.