# Area under Wien's displacement graph Why does the area under Wien's displacement graph give Stefan

Area under Wien's displacement graph
Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?
I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).
Is there a mathematical way to prove this?
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Since the question is a little terse, it is difficult to interpret. I think what must be happening is that the phrase 'Wien's displacement graph' is being used to mean the graph of $\rho \left(\omega \right)$ as a function of frequency $\omega$, where $\rho \left(\omega \right)$ is the energy density per unit frequency range in thermal or black body radiation. This graph implies Wien's displacement law if one studies it as a function of temperature. And the area under this graph is the total energy density in the radiation, which obeys the Stefan-Boltzmann law, as follows:
$\rho \left(\omega \right)=\frac{\hslash }{{\pi }^{2}{c}^{2}}\frac{{\omega }^{3}}{{e}^{\beta \hslash \omega }-1},$
$u={\int }_{0}^{\mathrm{\infty }}\rho \left(\omega \right)d\omega =\frac{{k}_{B}^{4}{T}^{4}}{{\pi }^{2}{c}^{3}{\hslash }^{3}}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{3}}{{e}^{x}-1}dx=\frac{{\pi }^{2}{k}_{B}^{4}{T}^{4}}{15{c}^{3}{\hslash }^{3}}=\frac{4\sigma }{c}{T}^{4}.$
The power per unit area emitted by the surface of a black body is related to this by
$I=\frac{1}{4}uc=\sigma {T}^{4}.$
Yasmine Larson