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").

Let's check it out by using three trapezoids to approximate the area under the function $f(x) = 3 \ln(x)$ on the interval $[2, 8]$.

Here's how that looks in a diagram when we call the first trapezoid $T_1$, the second trapezoid $T_2$, and the third trapezoid $T_3$:

Recall that the area of a trapezoid is $h \left(\dfrac{b_1 + b_2}{2}\right)$ where $h$ is the height and $b_1$ and $b_2$ are the bases.

We need to think about the trapezoid as if it's lying sideways.

The height $h$ is the $2$ at the bottom of $T_1$ that spans $x = \greenD 2$ to $x = \maroonD 4$.

The first base $b_1$ is the value of $3 \ln(x)$ at $x = \greenD 2$, which is $3 \ln (\greenD 2)$.

The second base $b_2$ is the value of $3 \ln(x)$ at $x = \maroonD 4$, which is $3 \ln (\maroonD 4)$.

Here's how all of this looks visually:

Let's put this all together to find the area of $T_1$:

Simplify:

Let's find the height and both of the bases:

Plug in and simplify:

We find the total area by adding up the area of each of the three trapezoids:

Here's the final simplified answer: