Evaluate the integral over the whole real line of e^{-2x^2 + 4x} dx in closed form, and give the numerical value to three decimal places.

Evaluate the integral over the whole real line of e^{-2x^2 + 4x} dx in closed form, and give the numerical value to three decimal places.

Approach: Complete the square in the exponent, shift the variable, and reduce the result to the standard Gaussian integral whose value comes from the polar coordinate argument.

e^2 * sqrt(pi/2). Completing the square gives -2x^2 + 4x = -2(x^2 - 2x) = -2(x-1)^2 + 2, so the integral is e^2 times the integral of e^{-2(x-1)^2} dx. Shifting by the change of variable u = x - 1 and then scaling with t = u sqrt(2) leaves e^2 * (1/sqrt(2)) times the integral of e^{-t^2} dt, which equals sqrt(pi). So the value is e^2 sqrt(pi/2) = 7.389 * 1.2533 = 9.261. The standard integral itself follows from squaring it: the product of two copies is a double integral of e^{-(x^2+y^2)} over the plane, and in polar coordinates that is the integral of e^{-r^2} r dr d(theta) = 2 pi * (1/2) = pi, so the single integral is sqrt(pi).

Follow-up: What is the same integral restricted to x >= 0, and how does the answer relate to the normal distribution function?

Key concepts: gaussian integral, completing the square, polar coordinates, change of variable.