math 1b\nmidterm 1 (practice 1), page 4 of 5\n4. determine if the following improper integrals are…

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 first rewrite $\frac{5}{e^{3x}}$ as $5e^{-3x}$. Then $\int_{0}^{\infty}5e^{-3x}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}5e^{-3x}dx$. Let $u=-3x$, $du=-3dx$. When $x = 0$, $u = 0$; when $x = b$, $u=-3b$. So $\int5e^{-3x}dx=-\frac{5}{3}e^{-3x}+C$. Then $\lim_{b\rightarrow\infty}\int_{0}^{b}5e^{-3x}dx=\lim_{b\rightarrow\infty}\left(-\frac{5}{3}e^{-3x}\big|{0}^{b}\right)=\lim{b\rightarrow\infty}\left(-\frac{5}{3}e^{-3b}+\frac{5}{3}e^{0}\right)=\frac{5}{3}$.

Step3: Apply comparison - test result

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.