question let the region r be the area enclosed by the function f(x)=e^x + 1, the horizontal line y = 0 and…

Explanation:

Step1: Find the side - length of the square cross - section

The height of the function (y = f(x)=e^{x}+1) above the (x) - axis ((y = 0)) gives the side - length (s) of each square cross - section perpendicular to the (x) - axis. So, (s=e^{x}+1).

Step2: Set up the volume formula

The volume (V) of a solid with square cross - sections perpendicular to the (x) - axis over the interval ([a,b]) is given by (V=\int_{a}^{b}s^{2}dx), where (a = 0), (b = 2), and (s=e^{x}+1). So, (V=\int_{0}^{2}(e^{x}+1)^{2}dx).

Step3: Expand the integrand

Expand ((e^{x}+1)^{2}) using the formula ((a + b)^{2}=a^{2}+2ab + b^{2}). Here, (a = e^{x}) and (b = 1), so ((e^{x}+1)^{2}=e^{2x}+2e^{x}+1).

Step4: Integrate term - by - term

(\int_{0}^{2}(e^{2x}+2e^{x}+1)dx=\int_{0}^{2}e^{2x}dx+2\int_{0}^{2}e^{x}dx+\int_{0}^{2}1dx). For (\int_{0}^{2}e^{2x}dx), let (u = 2x), (du=2dx). When (x = 0), (u = 0); when (x = 2), (u = 4). So, (\int_{0}^{2}e^{2x}dx=\frac{1}{2}\int_{0}^{4}e^{u}du=\frac{1}{2}[e^{u}]{0}^{4}=\frac{1}{2}(e^{4}-1)). (2\int{0}^{2}e^{x}dx=2[e^{x}]{0}^{2}=2(e^{2}-1)). (\int{0}^{2}1dx=[x]_{0}^{2}=2).

Step5: Calculate the volume

(V=\frac{1}{2}(e^{4}-1)+2(e^{2}-1)+2). (V=\frac{e^{4}}{2}-\frac{1}{2}+2e^{2}-2 + 2). (V=\frac{e^{4}}{2}+2e^{2}-\frac{1}{2}). Using a calculator, (e\approx2.71828), (e^{2}\approx7.38906), (e^{4}\approx54.59815). (V=\frac{54.59815}{2}+2\times7.38906-\frac{1}{2}). (V = 27.299075+14.77812 - 0.5). (V\approx41.577).

Answer:

(41.577)