I have this integral that I am trying to evaluate by hand, but I am encountering some difficulties. According to Wolfram Alpha, the answer seems to be:
However, I do not understand how they got the indefinite integral. What integration technique should be used to approach this problem? Could anyone explain how to find the definite integral with work? All help would be greatly appreciated.
If we complete the square in the exponent, we have
$$\frac 1 2(2x - x^2) = \frac 1 2 \Big(-(x - 1)^2 + 1\Big) = -\left(\frac{x - 1}{\sqrt 2}\right)^2 + \frac 1 2$$
Hence,
\begin{align*} \int e^{\frac 1 2(2x - x^2)} dx &= \sqrt e \int e^{-\left(\frac{x - 1}{\sqrt 2}\right)^2} dx \end{align*}
Make the change of variables
$$u = \frac{x - 1}{\sqrt 2},\quad \sqrt 2 \, du = dx$$
and we'll find that our integral is
$$\sqrt{2e} \int e^{-u^2} du$$
Unfortunately, there is no closed form for this integral, but it can then be expressed in terms of the error function, which is defined to be
$$\operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_0^t e^{-u^2} du$$
Alternatively, since you do have bounds on your integral, you should look at the Gaussian integral, which gives that
$$\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$$