Question

If, for some 𝛼∈(0, ∞),
\(∫^∞_02^{-x^2}dx=a\sqrt\pi,\) 
then 𝛼 equals (round off to two decimal places)

Answer 1

Correct Answer: 0.59

To find the value of 𝛼, consider the given integral: \(∫^∞_02^{-x^2}dx=a\sqrt\pi\) We need to evaluate this integral using a standard form.

The Gaussian integral is known for \(∫^∞_{−∞}e^{-x^2}dx=\sqrt\pi\) However, our integral is from 0 to ∞ and is in the form \(e^{-x^2\ln2}\) since \(2^{-x^2}=e^{-x^2\ln2}\)

Notice that:

\(∫^∞_02^{-x^2}dx=\frac{1}{2}∫^∞_{−∞}e^{-x^2\ln2}dx.\) By substituting \(x\sqrt{\ln2} for \ y\) we have:

\(dx=\frac{dy}{\sqrt{\ln2}}\) yielding:

\(∫e^{-y^2}\frac{dy}{\sqrt{\ln2}}=\frac{\sqrt\pi}{\sqrt{\ln2}}.\)

Thus, \(∫^∞_02^{-x^2}dx=\frac{\sqrt\pi}{2\sqrt{\ln2}}.\) Comparing with the given integral:

\(a\sqrt\pi=\frac{\sqrt\pi}{2\sqrt{\ln2}},\) therefore \(a=\frac{1}{2\sqrt{\ln2}}.\)

Calculating the numerical value:

\(a = \frac{1}{2\sqrt{\ln2}} ≈ 0.59.\)

Thus, the value of 𝛼 is approximately 0.59, which falls within the given range (0.59, 0.59).