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

To find the area enclosed between the two curves, we need to find the points of intersection first.

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

x² - 1 = 7 - x

Rearranging the equation:

x² + x - 8 = 0

Factoring the quadratic equation:

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

By setting each factor equal to zero:

x + 4 = 0 --> x = -4

x - 2 = 0 --> x = 2

So the two curves intersect at x = -4 and x = 2.

Next, we need to integrate with respect to x from -4 to 2. Since the upper curve is y = 7 - x and the lower curve is y = x² - 1, the integral would be:

∫[(7 - x) - (x² - 1)] dx, from -4 to 2

Expanding the integral:

∫(7 - x - x² + 1) dx, from -4 to 2

Simplifying the integral:

∫(8 - x - x²) dx, from -4 to 2

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

Substituting the upper limit of integration:

[8(2) - (1/2)(2)² - (1/3)(2)³]

= [16 - 2 - (8/3)]

Simplifying further:

= [16 - 2 - 8/3]

Finding a common denominator for the subtraction:

= (48/3 - 6/3 - 8/3)

= (48 - 6 - 8)/3

= 34/3

Thus, the area enclosed between the two curves is 34/3 square units.