What is the integral of 2^(-x)^2?

This is a pretty naughty integral. I can't say in good confidence that I can give you the correct answer. However, if we let the base of the integrating power function which is 2 be replaced with e, Euler's number, which is about 2.178281828, then treating the integral as a Gaussian Integral, the result of the integral, over the bounds bounds as x ranges from negative infinity to positive infinity over the whole real number line, then the integral:

(from negative infinity to positive infinity) ∫ e^(-x^2)dx = √(π)

Integrating this requires several steps like squaring the integral, using polar coordinates, introducing a Jacobean factor, evaluating a double integral, and then taking the square root of the integrated Gaussian function to cancel the initial squaring. Anyway, I found this to be a pretty good video on how to evaluate a Gaussian integral.

https://www.youtube.com/watch?v=nqNzKeVCYBU

So anyway, your integral, should be in that ballpark range of √(π) if you use the similar techniques shown in the video.

However, if you wanted to evaluate it as any definite integral:

(from a to b) ∫2^(-x^2)dx,

then you would have to expand 2^(-x^2)dx as a Maclaurin series. I would take it out to about the 4th or 5th derivative, and then you input the Maclaurin expansion into the integral and integrate each term of the series. You could get the Maclaurin series for a^(-x^2) treating a as a constant term to get a generalization or any value of a, whether it be 2,3,4, etc.

the integral of 2^(-x)^2 dx =the integral of 2^(-2x) dx

the integral of a^(bx) dx= ( a^(bx))/(b ln(a))+ C

int 2^(-x)^2 = int 2^x^2 dx = int sum_(n=0)^infinity (x^(2n))*ln^n(2)/(n!) dx

int 2^x^2 dx = sum_(n=0)^infinity [x^(2n + 1)]*[ln^n(2)] /[(2n + 1)*(n!)]