Comparison Theorem: Convergence of Integral from 0-->1

[fk378]

Homework Statement

Use the Comparison Theorem to determine whether the integral below is convergent or divergent:

e^-x / sqrt x dx integrated from 0-->1

The Attempt at a Solution

I think it is convergent because 1/e^x is convergent. I set the original integral less than or equal to the integral of 1/(e^x) dx

When I solved for it, I got -1/e + 1, therefore it is convergent. However, my professor marked my paper as saying it's not true. He set the original integral less than or equal to 1/sqrt x, and solving for that, got 2. Why can't my comparison hold true?

 

[quasar987]

Because it's not true in (0,1] that e^-x / sqrt x <e^-x

 

[Kreizhn]

Your comparison doesn't hold true because on (0,1), we have that

$$\frac{e^{-x}}{\sqrt{x}} > e^{-x}$$

This follows from the fact that on (0,1), $\sqrt{x}< 1 \Rightarrow \frac{1}{\sqrt{x}} > 1}$

 

[Kreizhn]

Haha, barely beaten to it. I knew I shouldn't have wasted my time previewing the post :tongue: