If the value of the integral (I) is 4, the value of the constant b is ___________ (give answer up to 2 decimal places). $I = \int^\infin_{-\infin} e^{\frac{-x^2}{b}}dx$
The standard form of the Gaussian integral is:
$ \int_{-\infin}^{\infin} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}} $
This formula requires that $a > 0$. The integral $I$ given is:
$ I = \int_{-\infin}^{\infin} e^{\frac{-x^2}{b}} dx $
Comparing the given integral $I$ with the standard form, we can see that:
$ a = \frac{1}{b} $
For the integral to converge, we must have $a > 0$, which implies $\frac{1}{b} > 0$, meaning $b$ must be positive ($b > 0$).
Substituting $a = \frac{1}{b}$ into the standard formula, the value of the given integral $I$ is:
$ I = \sqrt{\frac{\pi}{(1/b)}} = \sqrt{\pi b} $
We are given that the value of the integral $I$ is 4. So, we set the expression for $I$ equal to 4:
$ \sqrt{\pi b} = 4 $
To find $b$, we first square both sides of the equation:
$ (\sqrt{\pi b})^2 = 4^2 $
$ \pi b = 16 $
Now, isolate $b$ by dividing both sides by $\pi$:
$ b = \frac{16}{\pi} $
Using the approximate value of $\pi \approx 3.14159265...$:
$ b \approx \frac{16}{3.14159265} $
$ b \approx 5.092958... $
Rounding the value of $b$ to two decimal places gives:
$ b \approx 5.09 $
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