All of these summations are starting to feel like Rube Goldberg Machines. Granted, Rube Goldberg Machines are awesome, but do we seriously need this many methods to sum up intervals? Trust us, they *are* all useful in their own way. Just one big one to go, call it the grand finale.

A trapezoid sum is different from a left-hand sum, right-hand sum, or midpoint sum. Instead of drawing a rectangle on each sub-interval, we draw a trapezoid on each sub-interval. We do this by connecting the points on the function at the endpoints of the sub-interval.

First, a note on the area of trapezoids. A trapezoid that looks like this:

The area of the trapezoid is the average of the areas of two rectangles.

Thanks to the distributive property, this can be rewritten as

## Practice:

Let *f* (*x*) = *x*^{2} + 1 and let *R* be the region between the graph of *f* and the *x*-axis on [0,4]. Use a trapezoid sum with 2 sub-intervals to estimate the area of *R*. Is this an over-estimate or an under-estimate? | |

First we divide the interval [0,4] into two sub-intervals, as usual. On [0,2] we find the value of *f* at each endpoint of the sub-interval and plot the appropriate points. The area of this trapezoid is On [2,4] we find the value of *f* at each endpoint of the sub-interval and plot the appropriate points. The two heights of the trapezoid are *f (2) = 5 and *f (4) = 17, so the area of this trapezoid is We add the areas of the trapezoids and estimate that the area of *R* is 6 + 44 = 50. Since the trapezoids cover *R* and then some, this is an overestimate. Graphed, the trapezoids look like this:
| |

Let *f* (*x*) = *x*^{2} + 1 and let *R* be the region between the graph of *f* and the *x*-axis on [0,4]. Use a trapezoid sum with 4 sub-intervals to estimate the area of *R*. Is this an over-estimate or an under-estimate? | |

We divide [0,4] into four sub-intervals of width 1. The first trapezoid looks like this, with heights *f* (0) = 1 and *f* (1) = 2. It has area The second trapezoid looks like this, with heights 2 (we know this from the previous trapezoid) and *f** *(2) = 5. It has area The third trapezoid has heights 5 (known from previous trapezoid) and *f* (3) = 10. It has area . The third trapezoid has heights 10 (known from previous trapezoid) and *f* (4) = 17. It has area Adding up the areas of the trapezoids, we estimate the area of *R* to be 1.5 + 3.5 + 7.5 + 13.5 = 26. This is an overestimate, since again the trapezoids cover all of *R* and then some. | |

Let . Use a trapezoid sum with 3 sub-intervals to estimate the area between *f* and the *x*-axis on [0,9]. Is this an overestimate or underestimate?

Answer

We divide [0,9] into 3 sub-intervals of width 3.The first trapezoid has

The second trapezoid has

The third trapezoid has

Adding up the areas of the trapezoids, we get

71.625 + 58.125 + 31.135 = 160.875.

This is an underestimate, since the trapezoids don't cover the whole area between *f* and the *x*-axis on [0,9].

The table below shows partial values of the function *f* (*x*). Use a trapezoid sum with 4 sub-intervals to estimate the area between the graph of *f* and the *x*-axis on [0,1].

Answer

Each sub-interval has width 0.25. First trapezoid, on [0,0.25]:

Second trapezoid, on [.25,.5]:

Third trapezoid, on [.5,.75]:

Fourth trapezoid, on [0.75,1]:

We add the areas of the trapezoids and get

0.75 + 1.25 + 1.875 + 2.5 = 6.375.