Comparisons of imporper integrals

[ada0713]

Homework Statement

Find the upper bound for

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

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

 

[Rainbow Child]

If for two functions f,\,g you have f(x)&lt;g(x),\, \forall x&gt;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?

 

[ada0713]

That \int_{x_0}^\alpha f(x)\,d\,x,\ &lt; \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?<br /> <br /> But how do I find upper bound knowing this?

 

[ada0713]

I meant "integral of g(x)" for the first one and "integral for f(X)" for the second one. Sorry about the mess

 

[Rainbow Child]

Let f(x)=e^{-x^2},\,g(x)=e^{-3\,x}. Can you calculate
\int_3^\infty g(x)\,d\,x?

 

[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) ?

 

[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&lt;\frac{1}{3}\,e^{-9}

 

[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!

 

[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

 

[ada0713]

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