1. find the area of the region between the curves y = 9 - x² and y = x² + 1 from x = 0 to x = 3.

Explanation:

Step1: Determine the upper - lower curve

For (x\in[0,3]), we compare (y_1 = 9 - x^{2}) and (y_2=x^{2}+1). Let (f(x)=(9 - x^{2})-(x^{2}+1)=8 - 2x^{2}). When (x\in[0,2]), (f(x)\geq0), so (y = 9 - x^{2}) is above (y=x^{2}+1). When (x\in[2,3]), (y=x^{2}+1) is above (y = 9 - x^{2}).

Step2: Set up the integral for the area

The area (A=\int_{0}^{2}[(9 - x^{2})-(x^{2}+1)]dx+\int_{2}^{3}[(x^{2}+1)-(9 - x^{2})]dx). First integral: (\int_{0}^{2}(8 - 2x^{2})dx=\left[8x-\frac{2}{3}x^{3}\right]{0}^{2}=8\times2-\frac{2}{3}\times2^{3}=16-\frac{16}{3}=\frac{32}{3}). Second integral: (\int{2}^{3}(2x^{2}-8)dx=\left[\frac{2}{3}x^{3}-8x\right]_{2}^{3}=(\frac{2}{3}\times3^{3}-8\times3)-(\frac{2}{3}\times2^{3}-8\times2)=(18 - 24)-(\frac{16}{3}-16)=- 6-(\frac{16 - 48}{3})=-6+\frac{32}{3}=\frac{14}{3}).

Step3: Calculate the total area

(A=\frac{32}{3}+\frac{14}{3}=15.33\cdots = \frac{46}{3}).

Answer:

(\frac{46}{3})