sketch the region bounded by the curves, and visually estimate the location of the centroid.\ny = e^{x}, y =…

Explanation:

Step1: Calculate the area (A)

The area (A) of the region bounded by (y = e^{x}), (y = 0), (x = 0), and (x = 3) is given by the integral: $$A=\int_{0}^{3}e^{x}dx$$ Using the integral formula (\int e^{x}dx=e^{x}+C), we have: $$A = e^{x}\big|_{0}^{3}=e^{3}-e^{0}=e^{3}-1$$

Step2: Calculate (\bar{x})

The formula for (\bar{x}) is (\bar{x}=\frac{1}{A}\int_{0}^{3}x\cdot e^{x}dx) Use integration by parts. Let (u = x), (dv=e^{x}dx), then (du = dx), (v = e^{x}) By integration - by - parts formula (\int u;dv=uv-\int v;du), we get: (\int_{0}^{3}x\cdot e^{x}dx=x e^{x}\big|{0}^{3}-\int{0}^{3}e^{x}dx) (=3e^{3}-e^{x}\big|_{0}^{3}=3e^{3}-(e^{3}-1)=2e^{3}+1) So (\bar{x}=\frac{2e^{3}+1}{e^{3}-1})

Step3: Calculate (\bar{y})

The formula for (\bar{y}) is (\bar{y}=\frac{1}{A}\int_{0}^{3}\frac{1}{2}(e^{x})^{2}dx=\frac{1}{2A}\int_{0}^{3}e^{2x}dx) Using the integral formula (\int e^{ax}dx=\frac{1}{a}e^{ax}+C) ((a = 2) here) (\int_{0}^{3}e^{2x}dx=\frac{1}{2}e^{2x}\big|_{0}^{3}=\frac{1}{2}(e^{6}-1)) So (\bar{y}=\frac{e^{6}-1}{4(e^{3}-1)}=\frac{e^{3}+1}{4})

Answer:

(\left(\frac{2e^{3}+1}{e^{3}-1},\frac{e^{3}+1}{4}\right))