Originally Posted by

**ticbol** The trapezoidal rule was derived fro the area of a trapezoid whose two bases, b1 and b2, are parallel, and are separated by a perpendicular distace or altitude, h:

A = (b1 +b2)/2]*h

In your given table, the bases are the y-values, and the h is 0.5, which is the constant interval in the x-values.

So in the 1st trapezoid, b1 = 5 and b2 = 1.

Hence A1 = (5+1)/2]*0.5 = 1.5

In the 2nd trapezoid, b1 = 1 and b2 = -2

so, A2 = (1 +(-2))/2]*0.5 = -0.25

In the 3rd trapezoid, b1 = -2, and b2 = 3, so,

A3 = (-2 +3)/2]*0.5 = 0.25

A4 = [(3 +7)/2]*0.5 = 2.5

Therefore, total area A = A1 +A2 +A3 +A4 = 1.5 -0.25 +0.25 +2.5 = 4 sq.units.

Using Simpson's Trapezoidal Rule,

A = (d/2)[h1 +2h2 +2h3 +2h4 +h5]

A = (0.5 /2)[5 +2(1) +2(-2) +2(3) +7]

A = 0.25[16]

A = 4 sq.units ---------------the same as above.