How can I solve this integral with the comparison theorem?

Question

I have an integral that I am not sure how to solve with the comparison theorem to see if it is divergent or convergent.

$$\int_1^\infty\frac{e^{-2x}}{\sqrt{x+16}}\;dx$$

How can I solve this with the comparison theorem to see if it is divergent or convergent?

Answer 1

Note that $0 \leq \frac{e^{-2x}}{\sqrt{x+16}} \leq \frac{e^{-2x}}{4}$ for $x \geq 0$

now compute

$$\int_1^\infty \frac{e^{-2x}}{\sqrt{x+16}} \leq \int_1^\infty\frac{e^{-2x}}{4}< \infty$$

Answer 2

Consider by itself

$$ \frac{1}{\sqrt{x + 16}}$$

It is clear that for $1 \le x $

$$ 0 \le \frac{1}{\sqrt{x + 16}} \le \frac{1}{\sqrt{17}}$$

Thus we conclude that

$$ \int_1^{\infty} \frac{e^{-2x}}{\sqrt{x+17}} dx \le \frac{1}{\sqrt{17}} \int_0^\infty e^{-2x} dx = \frac{1}{\sqrt{17}} \frac{1}{2} $$

Furthermore since $$ \frac{e^{-2x}}{\sqrt{x+16}} > 0 \forall x$$

Then it follows that

$$ 0 \le \int_1^{\infty} \frac{e^{-2x}}{\sqrt{x+17}} dx \le \frac{1}{\sqrt{17}} \frac{1}{2}$$

Thus it is bounded and therefore converges.