Determine the area of the region bounded by y= x^2+1, y=x+3

To determine the area of the region bounded by the curves y = x^2 + 1 and y = x + 3, we need to find the points of intersection first. To find these points, we set the two equations equal to each other:

$$x^2 + 1 = x + 3$$

Simplifying this equation, we get:

$$x^2 - x - 2 = 0$$

This equation can be factored as:

$$(x - 2)(x + 1) = 0$$

Setting each factor to zero and solving for x, we find that x = 2 and x = -1.

Next, we need to integrate the difference between the two curves (y = x^2 + 1 and y = x + 3) with respect to x over the interval [-1, 2]. The area can be calculated using the formula:

$$A = \int_{-1}^{2} (x + 3 - (x^2 + 1)) \ dx$$

Simplifying the integrand, we have:

$$A = \int_{-1}^{2} (-x^2 + x + 2) \ dx$$

Integrating the expression, we get:

$$A = \left[ -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x \right]_{-1}^{2}$$

Evaluating the definite integral, we find:

$$A = \left( -\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + 2(2) \right) - \left( -\frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2(-1) \right)$$

Simplifying further,

$$A = \left( -\frac{8}{3} + 2 + 4 \right) - \left( \frac{1}{3} + \frac{1}{2} - 2 \right)$$

$$\text{Answer:} \ A = \frac{9}{2}$$

Therefore, the area of the region bounded by y = x^2 + 1 and y = x + 3 is 9/2 square units.