4. determine if the following improper integrals are convergent or divergent. justify your answers. (a) (10…

Explanation:

Step1: Use comparison - test

We know that $- 1\leqslant\sin(x)\leqslant1$. So, $3\leqslant4 + \sin(x)\leqslant5$. Then, $\frac{4+\sin(x)}{e^{3x}}\leqslant\frac{5}{e^{3x}}$.

Step2: Evaluate $\int_{0}^{\infty}\frac{5}{e^{3x}}dx$

We rewrite $\frac{5}{e^{3x}}$ as $5e^{-3x}$. Then, $\int_{0}^{\infty}5e^{-3x}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}5e^{-3x}dx$. Using the integral formula $\int e^{ax}dx=\frac{1}{a}e^{ax}+C$ ($a\neq0$), we have $\lim_{b\rightarrow\infty}\left[-\frac{5}{3}e^{-3x}\right]_{0}^{b}$.

Step3: Calculate the limit

$\lim_{b\rightarrow\infty}\left(-\frac{5}{3}e^{-3b}+\frac{5}{3}e^{0}\right)$. Since $\lim_{b\rightarrow\infty}e^{-3b}=0$, the value of $\lim_{b\rightarrow\infty}\left(-\frac{5}{3}e^{-3b}+\frac{5}{3}e^{0}\right)=\frac{5}{3}$.

Step4: Apply comparison - test conclusion

Since $\int_{0}^{\infty}\frac{5}{e^{3x}}dx$ converges and $\frac{4+\sin(x)}{e^{3x}}\leqslant\frac{5}{e^{3x}}$ for all $x\geqslant0$, by the comparison - test for improper integrals, $\int_{0}^{\infty}\frac{4 + \sin(x)}{e^{3x}}dx$ converges.

Answer:

The improper integral $\int_{0}^{\infty}\frac{4+\sin(x)}{e^{3x}}dx$ is convergent.