Find the area between y = 9 - x^2 and y = x + 1 from x = -1 to x = 2

Solution:

1. Given:
- Curve 1: y = 9 - x^2
- Curve 2: y = x + 1
- Interval for x: x = -1 to x = 2

2. Find points of intersection:
- Set y = 9 - x^2 equal to y = x + 1:
9 - x^2 = x + 1

- Rearrange and simplify:
x^2 + x - 8 = 0

- Factor the quadratic equation:
(x - 2)(x + 4) = 0

- Solve for x:
x = 2 \quad \text{and} \quad x = -4

- Valid intersection within the given interval is at x = 2.

3. Calculate area between curves from x = -1 to x = 2:
- Integral of top curve minus bottom curve:
\int_{-1}^{2} ((9 - x^2) - (x + 1)) \, dx
= \int_{-1}^{2} (8 - x^2 - x) \, dx

4. Solve the integral:
- Integrate term-by-term:
= \left[ 8x - \frac{x^3}{3} - \frac{x^2}{2} \right]_{-1}^{2}

- Evaluate at upper and lower limits:
\left( 8(2) - \frac{2^3}{3} - \frac{2^2}{2} \right) - \left( 8(-1) - \frac{(-1)^3}{3} - \frac{(-1)^2}{2} \right)

- Simplify and calculate:
= \left( 16 - \frac{8}{3} - 2 \right) - \left( -8 + \frac{1}{3} - \frac{1}{2} \right)
= 14 - \frac{8}{3} + 8 - \frac{1}{3} + \frac{1}{2}
= 22 - \frac{9}{3} + \frac{3}{6}
= 22 - 3 + 0.5
= 19.5

5. The area between the curves from x = -1 to x = 2 is:
\boxed{19.5}