# Homework Help: Comparisons of imporper integrals

1. Feb 15, 2008

### ada0713

1. The problem statement, all variables and given/known data
Find the upper bound for

$$\int$$ (from 3 to infinity) e^(-x^2)

[hint: e^(-x^2) < e^(-3x) for x>3]

3. The attempt at a solution

I don't understand what they mean by "upper bound"
Are they asking for the upper bound for the area below e^(-x^2)
for 3<x<infinity ?

I knew that the function converges by graphing the function
y= e^(-x^2) and y=e^(-x) and comparing between those two,
but I'm not sure what the problem is asking for.
Pleas help me witht the start!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 15, 2008

### Rainbow Child

If for two functions $f,\,g$ you have $f(x)<g(x),\, \forall x>x_0$ what can you say about the integrals $\int_{x_0}^\alpha f(x)\,d\,x,\,\int_{x_0}^\alpha g(x)\,d\,x$?

3. Feb 15, 2008

### ada0713

That $\int_{x_0}^\alpha f(x)\,d\,x,\ < \int_{x_0}^\alpha g(x)\,d\,x$ and that if
\int_{x_0}^\alpha g(x)\,d\,x[/itex] converges, then $\int_{x_0}^\alpha f(x)\,d\,x,\ converges as well? But how do I find upper bound knowing this? 4. Feb 15, 2008 ### ada0713 I meant "integral of g(x)" for the first one and "integral for f(X)" for the second one. Sorry about the mess 5. Feb 15, 2008 ### Rainbow Child Let [itex]f(x)=e^{-x^2},\,g(x)=e^{-3\,x}$. Can you calculate
$$\int_3^\infty g(x)\,d\,x$$?

6. Feb 15, 2008

### ada0713

Yes, I calculaed it and
lim[b to infinity] (-1/3 e^(-5b)) (1/3)
so the first part goes to zero when you plug in infinity for b
therefore it convererges to 1/3

.. so.. is it right that since f(x) is smaller than g(x) it converges as well
and therefore the possible upper bound should be 1/3 (or smaller than 1/3) ?

7. Feb 15, 2008

### Rainbow Child

I think the result is
$$\int_3^\infty e^{-3\,x}\,d\,x=\frac{1}{3}\,e^{-9}$$
And the answer is that one upper bound is $\frac{1}{3}\,e^{-9}$, i.e.

$$\int_3^\infty e^{-x^2}\,d\,x<\frac{1}{3}\,e^{-9}$$

Last edited: Feb 15, 2008
8. Feb 15, 2008

### ada0713

ooops,, I guess I made a calculation mistake somewhere in the middle, I'll figure it out what I did wrong. and THANK YOU for your help!!

9. Feb 15, 2008

### Rainbow Child

Your are faster in typing!
The result I posted it's the correct one, and one upper bound is $\frac{1}{3}\,e^{-9}$, or every other number bigger than that, not smaller!

Glad I helped

10. Feb 15, 2008

### ada0713

I noticed that you meant 1/3 e^(-9)
:) Thanks again!

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