- #1
graphic7
Gold Member
- 451
- 2
I'm attempting to find the volume of the solid obtained by rotating the region under the curve:
[tex]e^{-x^2}[/tex] Bounded by y = 0, x = 0, and x = 1.
I've done quite a few of these problems before, however, none of them have involved an exponential. If I recall correctly [tex]e^{x^2}[/tex] cannot be integrated, at least symbolically. So, therein lies the problem.
So far this is what I've done:
[tex]\int_0^1 \pi\left( e^{-x^2}\right)^2 dx[/tex]
[tex]\pi\int_0^1 e^{-2x^2} dx[/tex]
I've integrated that integral with Mathematica, and it returns a function that uses the error function, which I doubt is anything that I'm expected to come up with in this class.
Therefore, the only way I believe this problem can be completed is by finding the area under [tex]e^{2x^2}[/tex] on the interval [tex]\left[0 ,1\right][/tex] using a Riemann Sum.
Any thoughts?
Edit: Fixed a typo involving a constant in the integral.
[tex]e^{-x^2}[/tex] Bounded by y = 0, x = 0, and x = 1.
I've done quite a few of these problems before, however, none of them have involved an exponential. If I recall correctly [tex]e^{x^2}[/tex] cannot be integrated, at least symbolically. So, therein lies the problem.
So far this is what I've done:
[tex]\int_0^1 \pi\left( e^{-x^2}\right)^2 dx[/tex]
[tex]\pi\int_0^1 e^{-2x^2} dx[/tex]
I've integrated that integral with Mathematica, and it returns a function that uses the error function, which I doubt is anything that I'm expected to come up with in this class.
Therefore, the only way I believe this problem can be completed is by finding the area under [tex]e^{2x^2}[/tex] on the interval [tex]\left[0 ,1\right][/tex] using a Riemann Sum.
Any thoughts?
Edit: Fixed a typo involving a constant in the integral.
Last edited: