Areas under normal distribution curves can be used calculate probabilities of many random variables in the natural and social sciences.
The graph of the Standard Normal Curve, with a mean of 0 and a standard deviation of 1 is given below, and its function is given by
$\displaystyle {f{{\left({x}\right)}}}=\frac{{1}}{\sqrt{{{2}\pi}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}$.

The integral of $\displaystyle \frac{{1}}{\sqrt{{{2}\pi}}}{\int_{{-{{1.5}}}}^{{1.5}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}{\left.{d}{x}\right.}$ would be used to calculate the probability of an event that occurs between -1.5 to 1.5 standard deviations from the mean.

Unfortunately, $\displaystyle \frac{{1}}{\sqrt{{{2}\pi}}}{\int_{{-{{1.5}}}}^{{1.5}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}{\left.{d}{x}\right.}$ has no proper closed form antiderivative, so we must use numerical methods to approximate the values.



Using the Midpoint Rule with three rectangles of equal width, $\displaystyle \frac{{1}}{\sqrt{{{2}\pi}}}{\int_{{-{{1.5}}}}^{{1.5}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}{\left.{d}{x}\right.}\approx$

Using the Trapezoidal Rule with six trapezoids of equal width, $\displaystyle \frac{{1}}{\sqrt{{{2}\pi}}}{\int_{{-{{1.5}}}}^{{1.5}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}{\left.{d}{x}\right.}\approx$

Using Simpson's Rule, $\displaystyle \frac{{1}}{\sqrt{{{2}\pi}}}{\int_{{-{{1.5}}}}^{{1.5}}}{e}^{{-\frac{{x}^{{2}}}{{2}}}}{\left.{d}{x}\right.}\approx$

Round answers to 4 decimal places when appropriate.



(Note: The actual value to 4 places is 0.8664.)