The integral of e^(−x²) over all x equals √π ≈ 1.7725. This is the Gaussian integral. Its square root divided by √(2π) gives the standard normal distribution curve.
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Normal distribution formula
f(x) = (1/σ√(2π)) · e^(−(x−μ)²/2σ²)
σ = stdanard deviation, μ = mean
The 1/√(2π) normalisation factor comes directly from the Gaussian integral: ∫e^(−x²)dx = √π.
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The squaring trick: ∫e^(−x²)dx = √π
I² = ∫∫ e^(−x²−y²) dx dy = ∫₀^∞ e^(−r²) 2πr dr = π
Step 1: Square I – convert to double integral over bidang
Step 2: Switch to polar coordinates (r, θ) – the θ integral gives 2π
Step 3: Substitute u = r² – the r integral gives 1/2. Therefore I² = π, so I = √π.
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Bagaimana Feynman menggunakan varian integral Gauss dalam fisika?