Assignment 6

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If the graph is decreasing then we know that the left end of an interval gives us a higher estimate than the right.

The trapezoidal estimate is the mean of the left and right estimates, so it will lie between the two. The trapezoidal estimate also corresponds to a straight-line estimate between the graph points at the left and right endpoing of each interval.

Since the graph is concave down, any straight line between two graph points will lie below the graph. In particular the value of the function at the midpoint will lie above the straight line. The trapezoidal estimate corresponds to the straight-line, estimate so in this case the midpoint estimate exceeds the trapezoidal estimate.

The straight-line trapezoidal area is also clearly less than the area beneath the curve, so the trapezoidal estimate is lower than the actual value of the integral.

The midpoint rule uses the function value at the midpoint for the altitude of the rectangle. Thus we imagine a rectangle whose top is a horizontal line segment through the midpoint value of the graph. You should sketch this.

If the function is concave downward, the part of the actual graph that lies between the endpoints and above the midpoint-value rectangle (this segment lies to the left of the midpoint) represents graph area under the actual graph which is left out of the midpoint approximation. The area under the horizontal segment at the midpoint which lies above the graph (this are lies to the right of the midpoint) represents area included in the midpoint approximation but which is not part of the area under the curve.

Since the function is concave down, the average altitude of the ‘left-out’ area to the right is greater than that of the ‘wrongly-excluded’ area to the left. Since the widths of the two regions are equal, the wrongly excluded area must be less than the wrongly-included area and the midpoint estimate must therefore be high. We conclude that the actual area is less than the midpoint area.

Since the trapezoidal approximation is less than the actual area, we have our final ordering:

LEFT < TRAP < exact < MID < RIGHT.

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GEOMETRIC SOLUTION:

First suppose n = 1. You have only one trapezoid. In this case left(n) represents the altitude of a rectangle constructed on the left-hand side of the trapezoid, so left(n) * `dx represents the area of this rectangle.

| f(b) - f(a) | is the altitude of the triangle formed by the top part of the trapezoid. Assuming b > a then if the slope of the top of the trapezoid is positive the area of the triangle is added to that of the rectangle to get the area of the trapezoid. If the slope is negative the area is subtracted.

The area of the triangle is 1/2 * | f(b) - f(a) | * `dx.

1/2 ( f(b) - f(a) ) * `dx is equal to the area if the slope is positive so that f(b) > f(a), and to the negative of the area if the slope is negative so that f(b) < f(a).

It follows that left(n) * `dx + 1/2 ( f(b) - f(a) ) * `dx = area of the trapezoid.

If you have n trapezoids then if a = x0, x1, x2, ..., xn = b represents the partition of the interval [ a, b ] then the contributions of the small triangles at the tops of the trapezoids are 1/2 ( f(x1) - f(x0) ), 1/2 ( f(x2) - f(x1) ), ..., 1/2 ( f(xn) - f(x(n-1)) ). These contributions add up to

1/2 ( (f (x1) - f(x0)) + (f (x2) - f(x1)) + ... + (f (xn) - f(x(n-1))) ). Rearranging we find that all terms except f(x0) and f(xn) cancel out:

1/2 ( -f(x0) + f(x1) - f(x1) + f(x2) - f(x2) + ... + f(x(n-1)) - f(x(n-1) + f(xn) ) =

1/2 (f(xn) - f(x0) ) = 1/2 ( f(b) - f(a) ).

Left(n) is the sum of all the areas of the left-hand rectangles, so left(n) + 1/2 ( f(b) - f(a) ) * `dx represents the areas of all the rectangles plus the contributions of all the small triangles, which gives us the trapezoidal approximation.**

SYMBOLIC SOLUTION: Assuming a < b and partition a = x0, x1, x2, ..., xn = b we have

left(n) = [ f(x0) + f(x1) + ... + f(x(n-1) ] * `dx

and

trap(n) = 1/2 [ f(x0) + 2 f(x1) + 2 f(x2) + ... + 2 f(x(n-1)) + f(xn) ] * `dx.

So

trap(n) - left(n) = [-1/2 f(x0) + 1/2 f(xn) ] * `dx. Since x0 = a and xn = b this gives us

trap(n) = left(n) + 1/2 ( f(b) - f(a) ) * `dx.

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See my notes and let me know if you have questions.