◀ ▲ ▶Branches / Analysis / Lemma: Trapezoid Rule

The idea of the following lemma, known as the trapezoid rule is illustrated in the following figure.

For a given, "sufficiently smooth" function $f:[a,b]\to\mathbb R$ we want to calculate the area under the curve of this function, which equals the integral $\int_{a}^b f(x)dx.$ If for whatever reasons, the exact calculation of the integral is too complicated, we can approximate this area by calculating the area of the trapezoid in the points $a$, $b$, $f(b)$ and $f(a)$ instead, which is easily found using the formula $$(b-a)\frac{f(a)+f(b)}{2}.$$

When doing so, we will make an error $R.$ We can expect that this error gets smaller and smaller if we replace the original trapezoid by the sum of areas of $n\ge 2$ smaller trapezoids with equal bases built as follows: * We choose the number of trapezoids $n\ge 2.$ * We choose the length of the basis of all trapezoids to be $h:=\frac{b-a}n,$ such that we have $n+1$ points $x_\nu:=a+\nu h.$ By this choice, we will have $x_0=a$ and $x_n=b.$ * Since the area of a single $\nu$-th smaller trapezoid (for $\nu=1,\ldots,n$) is given by $$h\frac{f(x_\nu)+f(x_{\nu-1})}{2},$$ the sum of all these $n$ trapezoids is given by $$h\left(\frac{f(a)}{2}+\sum_{\nu=1}^{n-1}f(x_\nu)+\frac{f(b)}{2}\right),$$ which is an approximation of the actual integral.

The following lemma specifies what exactly is meant by "sufficiently smooth" and how good we can expect this approximation to be.

Lemma: Trapezoid Rule

Let the function $f:\mathbb R\to\mathbb R$ be a twice continuously differentiable. Let $n\ge 1$ be a natural number. For the closed real interval $[a,b]$, the Riemann integral $\int_a^b f(x)dx$ can be approximated using the function values of $f$ via $$\int_a^b f(x)dx=\left(\frac{f(a)}{2}+\sum_{\nu=1}^{n-1}f(a+\nu h)+\frac{f(b)}{2}\right)h+R,$$ where $h:=\frac{b-a}n$ and the absolute error term $|R|$ has an upper bound using the supremum of the second derivative of $f$ on the interval $[a,b]$ defined as follows:

$$|R|\le \frac{(b-a)h^2}{12}\cdot \sup\{|f^{\prime\prime}(x)|\mid\text{for all } x\in[a,b]\}.$$

Notes

• Note that $\sup\{|f^{\prime\prime}(x)|\mid\text{for all } x\in[a,b]\}\cdot \frac{(b-a)}{12}$ is a constant. No matter how big this constant is because $h^2$ tends to $0$ as $n$ tends to $\infty$, we can make the error term $R$ arbitrarily small, and it will tend to $0$ at a "quadratic rate". Thus, the trapezoid rule is indeed suitable to approximate the value of the integral, as expected.
• The fact that we have an upper bound for the error term gives us even more information about how good this estimation is.
• There is also a factor of $h$ in the main term, not only in the error term. However, the main term does not tend to $0$ since it is not constant. It rather grows by the number of terms with a growing $n$ at the same rate at which $h$ gets smaller. Therefore, the factor $h$ has only a "norming" impact on the main term.

Table of Contents

Proofs: 1

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983