I have to find the volume of the solid which is obtained by rotating the following region about the $x$-axis: $$y = e^{-x^2},\quad y=0,\quad x=-1,\quad x=1.$$
I am struggling to calculate the volume and I have come this far:$$V=\int_{-1}^{1}A(x)\text{d}x=\pi\int_{-1}^{1}\bigl(e^{-x^2}\bigr)^2\mathrm{d}x=\pi\int_{-1}^{1}e^{-2x^2}\mathrm{d}x$$
Could anyone help me in the right direction? Thanks!
Since the integral is that bounded, you can make use of Taylor series for the integrand function:
$$e^{-2x^2} = \sum_{k = 0}^{+\infty} \frac{(-2x^2)^k}{k!}$$
thence substituting and taking the sum out of the integral sign, we have:
$$\sum_{k = 0}^{+\infty}\frac{(-2)^k}{k!}\int_{-1}^1 x^{2k}\ \text{d}x$$
Which is trivial.
After integrated you remain with:
$$\sum_{k = 0}^{+\infty}\frac{(-2)^k}{k!} \frac{(1 + (-1)^{2k})}{2k+1}$$
This series can be summed, and its sum is
$$\boxed{\sqrt{\frac{\pi}{2}}\text{erf}(\sqrt{2})}$$
Where erf is a Special Function called the Error Function which is exactly what you would have found for that integral, being that integral a non trivial integral.
More on Error Function
https://en.wikipedia.org/wiki/Error_function