1. The problem involves understanding the formula for luminance $Df=K\int S(\lambda)R(\lambda)V(\lambda)d\lambda$ and the related color space luminance equation $Y=0.299R+0.587G+0.144B$. 2. The integral $Df=K\int S(\lambda)R(\lambda)V(\lambda)d\lambda$ represents the calculation of luminance by integrating the product of spectral power distribution $S(\lambda)$, reflectance $R(\lambda)$, and the photopic luminous efficiency function $V(\lambda)$ over wavelength $\lambda$. 3. The constants in the $Y$ equation represent the contribution of red, green, and blue components to perceived luminance. 4. The graphs show $S(\lambda)$ and $R(\lambda)$ with peaks near 450 nm and 600 nm, and $V(\lambda)$ peaking near the middle of the visible spectrum, indicating how these functions overlap to influence luminance. 5. To compute luminance $Df$, multiply the values of $S(\lambda)$, $R(\lambda)$, and $V(\lambda)$ at each wavelength, then integrate over the visible range (400-700 nm), scaling by constant $K$. 6. The square root symbol and $\Delta L$ (change in luminance) relate to perceptual differences in luminance, often used in color difference formulas. Final answer: The luminance $Df$ is calculated by the integral formula given, and the perceived luminance $Y$ from RGB values is given by $Y=0.299R+0.587G+0.144B$.