Consder a real-valued continuous function from points a, b of its domain where a < b. We will use the formula for the area of a trapezoid to find an approximation of the area \(\int_{a}^{b} f(x) \, dx\). Let's begin.

The formula for the area of a trapezoid is \[(c + d)/2 \times h \] where \(c, d\) are the lengths of the trapezoid basis, respectively, and \(h\) is the trapezoid height. After dividing our closed interval \([a,b]\) into n smaller subintervals of width \(\frac{b-a}{n}\), we then have that \(\int_{a}^{b} f(x) \, dx \) is approximately equal to \[ \frac{f(x_0) + f(x_1)}{2} \times \frac{b-a}{n} + \frac{f(x_1) + f(x_2)}{2} \times \frac{b-a}{n} + \dots + \frac{f(x_n-1) + f(x_n)}{2} \times \frac{b-a}{n} \\ = \frac{b-a}{2n} ( f(x_0) + 2 \, f(x_1) + \cdots + 2 \, f(x_{n-1}) + f(x_n)) \]

Our geometric intuition would allow us to be convinced that when \(f\) is concave up on \([a, b]\), then the trapezoid method is an over-approximation. When \(f\) is concave down on \([a, b]\), the method is an under-approximation.

Now consider applying the method for \(\int_{-\pi}^{\pi} \sin(x) \, dx\) for when \(n = 3\) and allow your imagination to give you the geometric intuition to see that the approximation gives the correct answer of \(0\). In this case, the trapezoids become triangles and what your are left with are two pairs of triangles with areas that cancel each other out to zero.