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The Trapezoid Geometric Method of Finding Area

By Maurice Ticas

Integrals

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.

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