Integral of e^(-x^2) ?

[fbellman]

Hi everyone! Sorry about the influx of posts lately (well, I guess two isn't really an influx...)

I have a problem where I have to find the area (and, eventually, the volume) of an area between two graphs. The two graphs are:

y=e^(-x^2) and
y=1-cos(x).

To find the area, I know you take the integral of f(x)-g(x). So I would have the integral of [e^(-x^2) - (1-cosx)]*dx on the interval of 0 to 1.

The problem is, I have no idea how to take the antiderivative of e^(-x^2). I know the integral of e^x is e^x. And the integral of e^(4x) is (e^4x)/4. But obviously the integral of e^(-x^2) can't be (-2x)e^(-x^2).... Help?

Thanks for any help you can offer!!

 

[stapel]

I couldn't come up with anything, so I checked a table of integrals, and this one wasn't included. So I checked "The Integrator" at the Wolfram site, and got:

. . . . .\(\displaystyle \large{\int{e^{-x^2}\,}dx\, =\,\frac{1}{2}\,\sqrt{x}\,\mbox{Erf[x]}}\)

In other words, this ain't a nice one. :shock:

Eliz.

 

[skeeter]

this problem is meant to be done using technology, i.e. a calculator.

using a TI series calculator ...

Y1 = e^-x²
Y2 = 1 - cos(x)

both functions are symmetrical to the y-axis ... x-values of intersection are +/- 0.942
I would store the unrounded value for the positive x-value of intersection someplace, say register B.

Now the area between the curves may be calculated using the following syntax ...

2fnint(Y1-Y2, x, 0, B)

you should get approx 1.182 for the area between the curves