Find the area enclosed between the curve y=x²-1 and y=7-x

To find the area enclosed between two curves, we need to find the points of intersection and integrate the difference in y-values between the curves over that interval.

First, let's find the points of intersection by setting the two equations equal to each other:

x^2 - 1 = 7 - x

Rearranging this equation, we get:

x^2 + x - 8 = 0

We can solve this quadratic equation by factoring or using the quadratic formula:

(x + 4)(x - 2) = 0

So, we have two potential points of intersection: x = -4 and x = 2.

To determine which is the upper and lower bound for our integral, we need to look at the y-values of the curves at these x-values:

For y = x^2 - 1: At x = -4, y = (-4)^2 - 1 = 15 At x = 2, y = (2)^2 - 1 = 3

For y = 7 - x: At x = -4, y = 7 - (-4) = 11 At x = 2, y = 7 - 2 = 5

We can see that the curve y = x^2 - 1 is above y = 7 - x between -4 and 2.

Now, we can calculate the area by integrating the difference between the curves:

Area = ∫ (7 - x) - (x^2 - 1) dx from x = -4 to x = 2

Area = ∫ (8 - x - x^2) dx from x = -4 to x = 2

To find the integral of the function (8 - x - x^2), we simplify the expression and integrate:

Area = [8x - (x^2 / 2) - (x^3 / 3)] from x = -4 to x = 2

Now, we substitute the upper limit and lower limit into this expression:

Area = [8(2) - ((2)^2 / 2) - ((2)^3 / 3)] - [8(-4) - ((-4)^2 / 2) - ((-4)^3 / 3)]

Area = [16 - 2 - (8/3)] - [-32 - 8 - (-64/3)]

Area = 16 - 2 - (8/3) + 32 + 8 + (64/3)

Area = 50 + (64/3) - (8/3)

Area = (150 + 64 - 8) / 3

Area = 206 / 3

Therefore, the area enclosed between the curves y = x^2 - 1 and y = 7 - x is 206/3 square units.