This question is regarding White Noise Equivalent Bandwidth (NEB).I was going through this material from Renesas .
The definition of NEB is given as when running white noise through a first order low-pass filter with fH as its -3dB frequency. In this case, the noise behaves as if filtered by a brick-wall filter with a higher cutoff frequency of fc = 1.57fH.This increased noise bandwidth is known as the white noise equivalent bandwidth (NEB).
My question is why white noise is behaving like that.
This is a property of noise; when you measure just the RMS result of the noise, you throw away the information about its frequency. This is generally a useful way to measure noise for lots of applications where you want to consider, say, the signal to noise ratio, where all you care about is the power of the noise (its total RMS value) and the power of the signal (its RMS value).
This rule of thumb is generally used for noise densities; for example see the below snippet for the MCP6H01 op-amp, if you want to calculate the input noise density with a 1kHz bandwidth instead of evaluating the integral of \$H(f)\$ (below) you can just* multiply the noise density by 1.57 times 1kHz; \$\sqrt{1.57\cdot\text{1kHz}}\cdot 35\text{nV}=1.39\mu\text{v}\$. 
We can calculate The noise equivalent bandwidth (NEB) by integrating the response from a first-order filter from 0 to infinity. The filter's transfer function is given by:
$$H(f) = \frac{1}{1 + j\frac{f}{f_H}}$$
Where (\$f_H\$ ) is the -3dB cutoff frequency. The magnitude response is:
$$|H(f)|^2 = \frac{1}{1 + \left(\frac{f}{f_H}\right)^2}$$
The Noise Equivalent Bandwidth (NEB) is the area under this magnitude response curve, calculated as:
$$NEB = \int_0^{\infty} |H(f)|^2 df$$
Performing this integration gives: \$NEB \approx 1.57 f_H\$
This means that the white noise passing through a first-order low-pass filter behaves as if it has been filtered by an ideal (brick-wall) filter with a cutoff frequency of \$f_H\$.
Bonus NEB for orders 1-10:
*For the pedantic, the noise analysis is more complex; the corner frequency for this part is pretty high, plus the 0.1 to 10Hz noise and the noise gain of the config. This is just meant to be an easy-to-follow application.
From "Low-Noise Electronic Design", Motchenbacher, chapter 1-4: "The noise bandwidth, \$\Delta f\$, is the frequency span of a rectangularly shaped power gain curve equal in area to the area of the actual power gain versus frequency curve."
In equation form for voltages: $$ \Delta f = {1 \over {A_{vo}}^2}{\int_{0}^{\infty}[{{A_v}^2(f)}]\;df} $$
Where:
\${A_v}^2(f)\$ = voltage gain as a function of frequency
\$A_{vo}^2\$ = midband voltage gain
Basically, since the low-pass filter given in the example has noise energy past the corner frequency, this noise energy will contribute to the total noise energy.
Fun exercise:
The gain function for a simple R-C low-pass filter is: $$ h(\omega)= {\left| {1 \over {1+j\omega R C}} \right|} = {1 \over \sqrt{1+\omega^2 R^2 C^2}}$$ Set R = 1.0, C = 1.0 which gives a corner frequency of 1.0.
Integrate from 0 to say 1000 and see what answer you get.
