Walk through an example using the trapezoid rule, then try a couple of practice problems on your own.
By now you know that we can use Riemann sums to approximate the area under a function. Riemann sums use rectangles, which make for some pretty sloppy approximations. But what if we used trapezoids to approximate the area under a function instead?
Key idea: By using trapezoids (aka the "trapezoid rule") we can get more accurate approximations than by using rectangles (aka "Riemann sums").
An example of the trapezoid rule
Let's check it out by using three trapezoids to approximate the area under the function f(x)=3ln(x) on the interval [2,8].
Here's how that looks in a diagram when we call the first trapezoid T1, the second trapezoid T2, and the third trapezoid T3:
Recall that the area of a trapezoid is h(2b1+b2) where h is the height and b1 and b2 are the bases.
Finding the area of T1
We need to think about the trapezoid as if it's lying sideways.
The height h is the 2 at the bottom of T1 that spans x=2 to x=4.
The first base b1 is the value of 3ln(x) at x=2, which is 3ln(2).
The second base b2 is the value of 3ln(x) at x=4, which is 3ln(4).
Here's how all of this looks visually:
Let's put this all together to find the area of T1: