Trapezoidal sums (College Board AP® Calculus BC): Revision Note

Trapezoidal sums

What is a trapezoidal sum?

• A trapezoidal sum is another method for approximating the exact value of an accumulation of change

  • Equivalently, it is a method for approximating the exact value of a definite integral

  • Or the exact area between a curve and the x-axis

• The approximation is made by adding up the areas of a number of trapezoids

How do I calculate a trapezoidal sum?

• To calculate the trapezoidal sum of a function $f$ between $x=a$ and $x=b$ (where $a<b$):

  • Divide the interval into $n$ subintervals by choosing values $x_0,x_1,...,x_n$ such that $a=x_0<x_1<\cdots <x_n=b$

    • The intervals do not need to be the same size

  • Let this define $n$ trapezoids

    • The $i$th trapezoid has a width of $\left(x_i-x_{i-1}\right)$

      • This is the distance from the left-hand side of the trapezoid to the right-hand side

    • The parallel sides of the $i$th trapezoid have heights of $f\left(x_{i-1}\right)$ and $f\left(x_i\right)$

      • These are the values of the function at the left-hand side and right-hand side of the trapezoid

    • The area of the $i$th trapezoid is $\left(x_i-x_{i-1}\right)·\frac{f\left(x_{i-1}\right)+f\left(x_i\right)}{2}$

  • The trapezoidal sum is the sum of the areas of these $n$ trapezoids

    • $\left(x_1-x_0\right)·\frac{f\left(x_0\right)+f\left(x_1\right)}{2}+\left(x_2-x_1\right)·\frac{f\left(x_1\right)+f\left(x_2\right)}{2}+\dots +\left(x_n-x_{n-1}\right)·\frac{f\left(x_{n-1}\right)+f\left(x_n\right)}{2}$

• In general, increasing the number of trapezoids, $n$, gives a more accurate approximation

• On the exam you may just be given values of the function in a table, rather than being given the function explicitly

  • See the Worked Example

What if all the intervals in a trapezoidal sum have the same size?

• If the intervals used for a trapezoidal sum all have the same size then the formula can be simplified slightly

• The width of each trapezoid will be $\frac{b-a}{n}$

• So the trapezoidal sum will become

  • $\frac{b-a}{n}·\frac{f\left(x_0\right)+f\left(x_1\right)}{2}+\frac{b-a}{n}·\frac{f\left(x_1\right)+f\left(x_2\right)}{2}+\dots +\frac{b-a}{n}·\frac{f\left(x_{n-1}\right)+f\left(x_n\right)}{2}$

• By collecting terms and simplifying, this can be rearranged as

  • $\frac{b-a}{n}·\left[\frac{f\left(x_0\right)+f\left(x_n\right)}{2}+f\left(x_1\right)+\dots +f\left(x_{n-1}\right)\right]$

  • or $\frac{b-a}{2n}·\left[f\left(x_0\right)+f\left(x_n\right)+2\left(f\left(x_1\right)+\dots +f\left(x_{n-1}\right)\right)\right]$

• It may be easier for you to understand how the trapezoidal sum is calculated

  • than to try and remember that formula

How can I tell if a trapezoidal sum is an underestimate or an overestimate?

• If a function is concave up over the interval for which a trapezoidal sum is being calculated

  • then the trapezoidal sum will be an overestimate

• If a function is concave down over the interval for which a trapezoidal sum is being calculated

  • then the trapezoidal sum will be an underestimate

• If a function has portions that are both concave up and concave down, then it is not immediately obvious whether a trapezoidal sum will be an underestimate or an overestimate