# Difficulty with integration of exponential

• Jan 24th 2008, 01:22 PM
emttim84
Difficulty with integration of exponential
Ok, so this problem appears very simple, so I'm actually surprised that I'm stumped on it, but I'm very suspicious that the answer is in poor algebra skills since I'm thinking I may be forgetting my exponents rule here or applying incorrectly.

Anyway, the problem is to find the volume of a solid of revolution for the function e raised to the -x when $\displaystyle y=0$ and x is equal to or greater than 0, so the definite integral is from 0 to infinite. I have V = pie * the integral from 0 to infinite of e raised to the -x and then squared and I think this is where I'm running into trouble. I'm solving e raised to the -x and then squared as e raised to the $\displaystyle x^2$ and I don't see how you could solve the integral of e raised to the $\displaystyle x^2$ power because you can't do a U sub since du would be 2x and there's no x, so you can't factor out a coefficient...parts doesn't seem applicable, nor does tab, and can't integrate it as is or use algebra to integrate.

So my question is, would e raised to the -x power then squared be e raised to the -2x power instead? Because then that would be simple enough to integrate, then take the limit and solve the problem. I apologize in advance for not having everything in the math notation on here, but for some reason, I can't figure out how to depict e to a power such as -x with the math symbols.
• Jan 24th 2008, 01:35 PM
Jhevon
Quote:

Originally Posted by emttim84
Ok, so this problem appears very simple, so I'm actually surprised that I'm stumped on it, but I'm very suspicious that the answer is in poor algebra skills since I'm thinking I may be forgetting my exponents rule here or applying incorrectly.

Anyway, the problem is to find the volume of a solid of revolution for the function e raised to the -x when $\displaystyle y=0$ and x is equal to or greater than 0, so the definite integral is from 0 to infinite. I have V = pie * the integral from 0 to infinite of e raised to the -x and then squared and I think this is where I'm running into trouble. I'm solving e raised to the -x and then squared as e raised to the $\displaystyle x^2$ and I don't see how you could solve the integral of e raised to the $\displaystyle x^2$ power because you can't do a U sub since du would be 2x and there's no x, so you can't factor out a coefficient...parts doesn't seem applicable, nor does tab, and can't integrate it as is or use algebra to integrate.

So my question is, would e raised to the -x power then squared be e raised to the -2x power instead? Because then that would be simple enough to integrate, then take the limit and solve the problem. I apologize in advance for not having everything in the math notation on here, but for some reason, I can't figure out how to depict e to a power such as -x with the math symbols.

when you raise something to a power to another power, you multiply the powers, that is, $\displaystyle (x^a)^b = x^{ab}$

so, $\displaystyle V = \pi \int_0^{\infty} (e^{-x})^2~dx = \pi \int_0^{\infty} e^{-2x}~dx$

you could also think of $\displaystyle e^{-x}$ squared as $\displaystyle e^{-x} \cdot e^{-x}$, in which case you know that you add the powers to get $\displaystyle e^{-2x}$

i assume you are rotating about the x-axis, since that's what your formula says you're doing

nice signature
• Jan 24th 2008, 01:49 PM
emttim84
Quote:

Originally Posted by Jhevon
when you raise something to a power to another power, you multiply the powers, that is, $\displaystyle (x^a)^b = x^{ab}$

so, $\displaystyle V = \pi \int_0^{\infty} (e^{-x})^2~dx = \pi \int_0^{\infty} e^{-2x}~dx$

you could also think of $\displaystyle e^{-x}$ squared as $\displaystyle e^{-x} \cdot e^{-x}$, in which case you know that you add the powers to get $\displaystyle e^{-2x}$

i assume you are rotating about the x-axis, since that's what your formula says you're doing

nice signature

Ahh ok, so I /was/ making an exponents rule error...ok, well at least that explains why I couldn't integrate it. :p and yes, rotating it around the X axis. Thanks, I love that quote from Scrubs.

Now the integral should come out nice and neat, as should the limit. Thanks!
• Mar 17th 2008, 05:31 PM
AlexMack
Quote:

Originally Posted by emttim84
Ok, so this problem appears very simple, so I'm actually surprised that I'm stumped on it, but I'm very suspicious that the answer is in poor algebra skills since I'm thinking I may be forgetting my exponents rule here or applying incorrectly.

Anyway, the problem is to find the volume of a solid of revolution for the function e raised to the -x when $\displaystyle y=0$ and x is equal to or greater than 0, so the definite integral is from 0 to infinite. I have V = pie * the integral from 0 to infinite of e raised to the -x and then squared and I think this is where I'm running into trouble. I'm solving e raised to the -x and then squared as e raised to the $\displaystyle x^2$ and I don't see how you could solve the integral of e raised to the $\displaystyle x^2$ power because you can't do a U sub since du would be 2x and there's no x, so you can't factor out a coefficient...parts doesn't seem applicable, nor does tab, and can't integrate it as is or use algebra to integrate.

So my question is, would e raised to the -x power then squared be e raised to the -2x power instead? Because then that would be simple enough to integrate, then take the limit and solve the problem. I apologize in advance for not having everything in the math notation on here, but for some reason, I can't figure out how to depict e to a power such as -x with the math symbols.

Indeed, it has been proven that
$\displaystyle \int e^{x^{2}}\,dx$
cannot be evaluated in terms of elementary functions.