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.