determine whether the following integral converges or diverges.\n∫0,∞ dx/(e^x + 3)\nas x approaches ∞, which…

Explanation:

Step1: Analyze the integrand behavior

For positive - valued functions, when comparing $\frac{1}{e^{x}+3}$ and $\frac{1}{e^{x}}$, since $e^{x}+3>e^{x}$ for all $x\geq0$, then $\frac{1}{e^{x}+3}<\frac{1}{e^{x}}$.

Step2: Recall the convergence of $\int_{0}^{\infty}\frac{1}{e^{x}}dx$

We know that $\int_{0}^{\infty}\frac{1}{e^{x}}dx=\int_{0}^{\infty}e^{-x}dx$. Using the integral formula $\int e^{-x}dx=-e^{-x}+C$. Then $\lim_{b\rightarrow\infty}\int_{0}^{b}e^{-x}dx=\lim_{b\rightarrow\infty}(-e^{-b}+e^{0}) = 1$, so $\int_{0}^{\infty}\frac{1}{e^{x}}dx$ converges.

Step3: Apply the Comparison Test for improper integrals

The Comparison Test states that if $0\leq f(x)\leq g(x)$ for all $x\geq a$ and $\int_{a}^{\infty}g(x)dx$ converges, then $\int_{a}^{\infty}f(x)dx$ converges. Here, $f(x)=\frac{1}{e^{x}+3}$, $g(x)=\frac{1}{e^{x}}$, $a = 0$. Since $\int_{0}^{\infty}\frac{1}{e^{x}}dx$ converges and $\frac{1}{e^{x}+3}<\frac{1}{e^{x}}$ for $x\geq0$, $\int_{0}^{\infty}\frac{1}{e^{x}+3}dx$ converges.

Answer:

1. For the first - part question: C. $\frac{1}{e^{x}+3}<\frac{1}{e^{x}}$
2. For the second - part question: A. $\int_{0}^{\infty}\frac{dx}{e^{x}+3}$ converges.