(1 point)\ndetermine if the improper integral converges and\n int_{3}^{infty} \frac{e^{sqrt{x}}}{sqrt{x}}dx…

Explanation:

Step1: Use substitution

Let $u = \sqrt{x}$, then $du=\frac{1}{2\sqrt{x}}dx$ and $dx = 2\sqrt{x}du$. When $x = 3$, $u=\sqrt{3}$; as $x\rightarrow\infty$, $u\rightarrow\infty$. The integral $\int_{3}^{\infty}\frac{e^{\sqrt{x}}}{\sqrt{x}}dx$ becomes $2\int_{\sqrt{3}}^{\infty}e^{u}du$.

Step2: Evaluate the new - integral

The antiderivative of $e^{u}$ is $e^{u}$. So, $2\int_{\sqrt{3}}^{\infty}e^{u}du=2\lim_{b\rightarrow\infty}\int_{\sqrt{3}}^{b}e^{u}du$. [ \begin{align*} 2\lim_{b\rightarrow\infty}\int_{\sqrt{3}}^{b}e^{u}du&=2\lim_{b\rightarrow\infty}(e^{u}\big|{\sqrt{3}}^{b})\ &=2\lim{b\rightarrow\infty}(e^{b}-e^{\sqrt{3}}) \end{align*} ] Since $\lim_{b\rightarrow\infty}e^{b}=\infty$, the value of $2\lim_{b\rightarrow\infty}(e^{b}-e^{\sqrt{3}})=\infty$.

Answer:

F