1. ## Comparison Theorem Integrals

Use the comparison Theorem to determine whether the integral is convergent or divergent

integral from 0,1

[e^(-x)]/sqrt(x)

thanks guys

2. Originally Posted by Asuhuman18
Use the comparison Theorem to determine whether the integral is convergent or divergent

integral from 0,1

[e^(-x)]/sqrt(x)

thanks guys
$\frac{e^{-x}}{\sqrt{x}} \leq \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ \forall \,\ x \in (0,1]$.
and it is easy to show that the improper integral $\int_0^1 \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ dx$ converges.

T

3. or $\frac{e^{-x}}{\sqrt{x}}\le \frac{1}{\sqrt{x}}$ for $x>0$ and in particular for $x\in(0,1]$ holds and $\int_0^1\frac{dx}{\sqrt x}$ verifies the convergence of the original integral.

4. Similar to what Krizalid has shown, we can also find a lower bound.
For all $x\in[0,1]$, we have
$\frac{1}{\mathrm{e}}\leq\mathrm{e}^{-x}\leq1$, which yields by integrating after dividing by $\sqrt{x}$
$\frac{2}{\mathrm{e}}\leq\int_{0}^{1}\frac{\mathrm{ e}^{-x}}{\sqrt{x}}\mathrm{d}x\leq2.$
Note that above, we apply improper integration and that
$\int_{a}^{b}\frac{1}{\sqrt{x}}\mathrm{d}x=2\sqrt{x }\big|_{x=a}^{x=b}$ provided that $b>a>0$.

5. Originally Posted by bkarpuz
Similar to what Krizalid has shown, we can also find a lower bound.
For all $x\in[0,1]$, we have
$\frac{1}{\mathrm{e}\sqrt{x}}\leq\frac{\mathrm{e}^{-x}}{\sqrt{x}}\leq\frac{1}{\sqrt{x}}$, which yields by integrating
$\frac{2}{\mathrm{e}}\leq\int_{0}^{1}\frac{\mathrm{ e}^{-x}}{\sqrt{x}}\mathrm{d}x\leq2.$
Recall that
$\int_{a}^{b}\frac{1}{\sqrt{x}}\mathrm{d}x=2\sqrt{x }\big|_{x=a}^{x=b}.$
It should be ${\color{red} (}0,1]$.