Integral Calculus, Area Under the Bell Curve

Area Under the Bell Curve

Let g(x) = exp(-x²). This is an important function in probability and statistics. It reaches a maximum at 0,1 and slopes down symmetrically about this point. It looks a bit like a bell, and is sometimes called the bell curve. The "tails" flatten out and approach 0 as x approaches ±∞. In fact the exponential curve drops to zero in a hurry, hence g is probably integrable over the entire x axis. How can we compute the area under g?

Let q be the area under this curve. We don't know what it is yet, but q is the integral of g as x runs from -∞ to +∞.

Let h(x,y) = g(x)×g(y). This looks like a bell surface in 3 dimensions. When y = 0 we have the original bell curve. Every other cross section y = c gives another bell curve multiplied by g(c).

The volume under the bell surface is a 2 dimensional integral, which becomes a nested integral. For each value of y, the integral with respect to x gives q×g(y). Integrate this with respect to y and get q². Thus the double integral of h = q². Compute this double integral and take the square root to find q.

What is the double integral of exp(-x²)×exp(-y²)?

In polar coordinates, x²+y² = r², hence the integrand becomes r×exp(-r²). The extra factor of r is introduced when we switch to polar coordinates, as described in the previous theorem.

The integral becomes -½exp(-r²). Evaluate as r runs from 0 to ∞ and get ½. Then let θ run from 0 to 2π, and the volume under h = π. Therefore the integral of exp(-x²) from -∞ to +∞ is sqrt(π), approximately 1.7724.