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

To find the area enclosed between the curves y = x² - 1 and y = 7 - x, we need to find the points of intersection first.

Setting the two equations equal to each other, we have:

x² - 1 = 7 - x

Rearranging, we get:

x² + x - 8 = 0

Factoring, we have:

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

So the solutions are x = 2 and x = -4.

To find the y-values corresponding to these x-values, we substitute them into either equation.

For y = x² - 1:

When x = 2, y = 2² - 1 = 4 - 1 = 3.

When x = -4, y = (-4)² - 1 = 16 - 1 = 15.

Now we have the points of intersection as (2, 3) and (-4, 15).

To find the area enclosed between the curves, we integrate the positive difference between the two curves.

∫[a,b] (7 - x - (x² - 1)) dx

Setting up the integral with the limits of integration [a, b], where a = -4 and b = 2, we have:

A = ∫[-4, 2] (7 - x - (x² - 1)) dx

Expanding the expression, we get:

A = ∫[-4, 2] (8 - x - x²) dx

Simplifying further, we have:

A = ∫[-4, 2] (-x² - x + 8) dx

Integrating this expression, we get:

A = [(-1/3)x³ - (1/2)x² + 8x] from -4 to 2

Evaluating at the limits of integration, we have:

A = [(-1/3)(2)³ - (1/2)(2)² + 8(2)] - [(-1/3)(-4)³ - (1/2)(-4)² + 8(-4)]

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

A = -8/3 - 2 + 16 - 64/3 - 8 + 32

A = -16/3 - 10 + 16 -64/3 + 24

Simplifying further, we have:

A = -32/3 - 10 - 64/3 + 40

A = -96/3 - 10 + 40

A = -32 - 10 + 40

A = -2

However, we are looking for the area which is always positive, so we take the absolute value of the result.

Therefore, the area enclosed between the curves y = x² - 1 and y = 7 - x is 2 square units.