Am I doing this question right using CT? Can anyone please help me out?
Determine whether this integral is convergent or divergent.
$\int_{1}^{\infty}\frac{3+sin(x)}{e^x}$
$3\ge sin(x)$ for all x $\in[1,\infty)$
$\frac{3}{e^x}\ge \frac{sin(x)}{e^x}$
$0\le \frac{3}{e^x}\le \frac{sin(x)}{e^x}$
Let f(x) = $\frac{3}{e^x}$ Let g(x) = $\frac{sin(x)}{e^x}$
Since $\int_{1}^{\infty} \frac{3}{e^x}$ converges, $\int_{1}^{\infty} \frac{sin(x)}{e^x}$ also converges, so therefore, $\int_{1}^{\infty}\frac{3+sin(x)}{e^x}$ converges
That is not true because $\sin x$ could be negative for certain $x\in[1,\infty)$.
Rather, $\left|\dfrac{3+\sin x}{e^{x}}\right|\leq\dfrac{3+|\sin x|}{e^{x}}\leq\dfrac{4}{e^{x}}$ and we have $\displaystyle\int_{1}^{\infty}\dfrac{4}{e^{x}}dx=4e^{-1}<\infty$, so $\displaystyle\int_{1}^{\infty}\dfrac{3+\sin x}{e^{x}}dx$ converges absolutely.
In fact, $3+\sin x\geq 3-1=2>0$, so $\dfrac{3+\sin x}{e^{x}}>0$ and the absolute convergence is simply the usual convergence.