Comparison test

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1. Oct 14, 2016

Jeanclaud

Hi, so the question is I have to tell if this integral diverges or converges.(without solving it)
integral(1/(e^x sqrt(x)))dx from 1 to +inf

2. Relevant equations
integration techniques.

3. The attempt at a solution
my answer: let 1/e^x >1/(e^x sqrt(x))
then I solved the definite integral(1/e^x)from 1 to +inf and got 1/e which means it converges.
so the given integral has to converge also since it is smaller than the 1/e^x.
that was my answer in the exam but they considered it all wrong so please can anybody tell me the reason. Thank you.

2. Oct 14, 2016

Simon Phoenix

Looking at the form of this integral $$\int_1^\infty \frac 1 { \sqrt x e^x} \, dx$$

it seems fairly clear we're expecting convergence here - so for the comparison test we are looking for a larger integral on the domain that converges. We have
$$\frac 1 { \sqrt x e^x } \leq \frac 1 { e^x }$$ on this domain. Everything's all positive so we're good to go.

The integral $$\int_1^\infty \frac 1 { e^x } \, dx$$ is clearly convergent, so our integral of interest is also convergent.

I can't see why your answer was considered to be incorrect either

3. Oct 14, 2016

Staff: Mentor

4. Oct 14, 2016

Staff: Mentor

5. Oct 14, 2016

Jeanclaud

nope.

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6. Oct 14, 2016

Simon Phoenix

Well, I can't see why your answer was considered to be wrong. Just a silly thought; you did actually copy the integral down from the exam paper correctly?

That's the only thing I can think of because your answer looks OK to me.

You could probably structure your answer a bit better - does your answer look anything like the worked examples in your textbooks in terms of how it's laid out? Try to copy this 'style' and set out your answers in a clear step-by-step fashion and explain what you're doing (only takes a few words here and there). It's a bit of a pain to do this initially but it becomes easier with practice and eventually becomes second nature - and you (should) find your understanding and maybe even your marks improve the more you do this

7. Oct 14, 2016

Ray Vickson

I second the remarks of Simon Phoenix regarding your presentation style. I would add that some of the things you wrote are technically wrong without further qualification. In particular, the inequality $e^{-x} > e^{-x}/\sqrt{x}$ is false when $x < 1$, but true when $x > 1$. You could say something like "since we want x > 1, ..." and then what you wrote would be correct. Just a few words of explanation is all you need; it would take < 5 seconds to write them.

8. Oct 14, 2016

Jeanclaud

thanks you.

9. Oct 14, 2016

Staff: Mentor

But since the interval of integration is $[1, \infty)$, I don't think it's absolutely necessary to stipulate that x > 1.

Last edited: Oct 14, 2016
10. Oct 14, 2016

Ray Vickson

We are all just trying to figure out why he lost marks on correct work.