In calculus, the trapezoidal rule (also known as the trapezoid rule or trapezium rule)^[a] is a technique for numerical integration, i.e., approximating the definite integral:

$${\displaystyle \int _{a}^{b}f(x)\,dx.}$$

The trapezoidal rule works by approximating the region under the graph of the function ${\displaystyle f(x)}$ as a trapezoid and calculating its area. It follows that

$${\displaystyle \int _{a}^{b}f(x)\,dx\approx (b-a)\cdot {\tfrac {1}{2))(f(a)+f(b)).}$$

The trapezoidal rule may be viewed as the result obtained by averaging the left and right Riemann sums, and is sometimes defined this way. The integral can be even better approximated by partitioning the integration interval, applying the trapezoidal rule to each subinterval, and summing the results. In practice, this "chained" (or "composite") trapezoidal rule is usually what is meant by "integrating with the trapezoidal rule". Let ${\displaystyle \{x_{k}\))$ be a partition of ${\displaystyle [a,b]}$ such that ${\displaystyle a=x_{0}<x_{1}<\cdots <x_{N-1}<x_{N}=b}$ and ${\displaystyle \Delta x_{k))$ be the length of the ${\displaystyle k}$-th subinterval (that is, ${\displaystyle \Delta x_{k}=x_{k}-x_{k-1))$), then

$${\displaystyle \int _{a}^{b}f(x)\,dx\approx \sum _{k=1}^{N}{\frac {f(x_{k-1})+f(x_{k})}{2))\Delta x_{k}.}$$

${\displaystyle \Delta x_{k))$

${\displaystyle \Delta x,}$

${\displaystyle \Delta x}$

$${\displaystyle \int _{a}^{b}f(x)\,dx\approx {\frac {\Delta x}{2))\left(f(x_{0})+2f(x_{1})+2f(x_{2})+2f(x_{3})+2f(x_{4})+\cdots +2f(x_{N-1})+f(x_{N})\right).}$$

The approximation becomes more accurate as the resolution of the partition increases (that is, for larger ${\displaystyle N}$, all ${\displaystyle \Delta x_{k))$ decrease).

As discussed below, it is also possible to place error bounds on the accuracy of the value of a definite integral estimated using a trapezoidal rule.

History

A 2016 Science paper reports that the trapezoid rule was in use in Babylon before 50 BCE for integrating the velocity of Jupiter along the ecliptic.^[1]

Numerical implementation

Non-uniform grid

When the grid spacing is non-uniform, one can use the formula

$${\displaystyle \int _{a}^{b}f(x)\,dx\approx \sum _{k=1}^{N}{\frac {f(x_{k-1})+f(x_{k})}{2))\Delta x_{k},}$$

${\displaystyle \Delta x_{k}=x_{k}-x_{k-1}.}$

Uniform grid

For a domain discretized into ${\displaystyle N}$ equally spaced panels, considerable simplification may occur. Let

$${\displaystyle \Delta x_{k}=\Delta x={\frac {b-a}{N))}$$

$${\displaystyle {\begin{aligned}\int _{a}^{b}f(x)\,dx&\approx {\frac {\Delta x}{2))\sum _{k=1}^{N}\left(f(x_{k-1})+f(x_{k})\right)\\[1ex]&={\frac {\Delta x}{2)){\Biggl (}f(x_{0})+2f(x_{1})+2f(x_{2})+2f(x_{3})+\dotsb +2f(x_{N-1})+f(x_{N}){\Biggr )}\\[1ex]&=\Delta x\left({\frac {f(x_{N})+f(x_{0})}{2))+\sum _{k=1}^{N-1}f(x_{k})\right).\end{aligned))}$$

Error analysis

The error of the composite trapezoidal rule is the difference between the value of the integral and the numerical result:

$${\displaystyle {\text{E))=\int _{a}^{b}f(x)\,dx-{\frac {b-a}{N))\left[{f(a)+f(b) \over 2}+\sum _{k=1}^{N-1}f\left(a+k{\frac {b-a}{N))\right)\right]}$$

There exists a number ξ between a and b, such that^[2]

$${\displaystyle {\text{E))=-{\frac {(b-a)^{3)){12N^{2))}f''(\xi )}$$

It follows that if the integrand is concave up (and thus has a positive second derivative), then the error is negative and the trapezoidal rule overestimates the true value. This can also be seen from the geometric picture: the trapezoids include all of the area under the curve and extend over it. Similarly, a concave-down function yields an underestimate because area is unaccounted for under the curve, but none is counted above. If the interval of the integral being approximated includes an inflection point, the error is harder to identify.

An asymptotic error estimate for N → ∞ is given by

$${\displaystyle {\text{E))=-{\frac {(b-a)^{2)){12N^{2))}{\big [}f'(b)-f'(a){\big ]}+O(N^{-3}).}$$

Several techniques can be used to analyze the error, including:^[3]

1. Fourier series
2. Residue calculus
3. Euler–Maclaurin summation formula^[4][5]
4. Polynomial interpolation^[6]

It is argued that the speed of convergence of the trapezoidal rule reflects and can be used as a definition of classes of smoothness of the functions.^[7]

Proof

First suppose that ${\displaystyle h={\frac {b-a}{N))}$ and ${\displaystyle a_{k}=a+(k-1)h}$. Let

$${\displaystyle g_{k}(t)={\frac {1}{2))t[f(a_{k})+f(a_{k}+t)]-\int _{a_{k))^{a_{k}+t}f(x)\,dx}$$

${\displaystyle |g_{k}(h)|}$

${\displaystyle [a_{k},a_{k}+h]}$

$${\displaystyle {dg_{k} \over dt}={1 \over 2}[f(a_{k})+f(a_{k}+t)]+{1 \over 2}t\cdot f'(a_{k}+t)-f(a_{k}+t),}$$

$${\displaystyle {d^{2}g_{k} \over dt^{2))={1 \over 2}t\cdot f''(a_{k}+t).}$$

Now suppose that ${\displaystyle \left|f''(x)\right|\leq \left|f''(\xi )\right|,}$ which holds if ${\displaystyle f}$ is sufficiently smooth. It then follows that

$${\displaystyle \left|f''(a_{k}+t)\right|\leq f''(\xi )}$$

${\displaystyle -f''(\xi )\leq f''(a_{k}+t)\leq f''(\xi )}$

${\displaystyle -{\frac {f''(\xi )t}{2))\leq g_{k}''(t)\leq {\frac {f''(\xi )t}{2)).}$

Since ${\displaystyle g_{k}'(0)=0}$ and ${\displaystyle g_{k}(0)=0}$,

$${\displaystyle \int _{0}^{t}g_{k}''(x)dx=g_{k}'(t)}$$

$${\displaystyle \int _{0}^{t}g_{k}'(x)dx=g_{k}(t).}$$

Using these results, we find

$${\displaystyle -{\frac {f''(\xi )t^{2)){4))\leq g_{k}'(t)\leq {\frac {f''(\xi )t^{2)){4))}$$

$${\displaystyle -{\frac {f''(\xi )t^{3)){12))\leq g_{k}(t)\leq {\frac {f''(\xi )t^{3)){12))}$$

Letting ${\displaystyle t=h}$ we find

$${\displaystyle -{\frac {f''(\xi )h^{3)){12))\leq g_{k}(h)\leq {\frac {f''(\xi )h^{3)){12)).}$$

Summing all of the local error terms we find

$${\displaystyle \sum _{k=1}^{N}g_{k}(h)={\frac {b-a}{N))\left[{f(a)+f(b) \over 2}+\sum _{k=1}^{N-1}f\left(a+k{\frac {b-a}{N))\right)\right]-\int _{a}^{b}f(x)dx.}$$

But we also have

$${\displaystyle -\sum _{k=1}^{N}{\frac {f''(\xi )h^{3)){12))\leq \sum _{k=1}^{N}g_{k}(h)\leq \sum _{k=1}^{N}{\frac {f''(\xi )h^{3)){12))}$$

$${\displaystyle \sum _{k=1}^{N}{\frac {f''(\xi )h^{3)){12))={\frac {f''(\xi )h^{3}N}{12)),}$$

so that

$${\displaystyle -{\frac {f''(\xi )h^{3}N}{12))\leq {\frac {b-a}{N))\left[{f(a)+f(b) \over 2}+\sum _{k=1}^{N-1}f\left(a+k{\frac {b-a}{N))\right)\right]-\int _{a}^{b}f(x)dx\leq {\frac {f''(\xi )h^{3}N}{12)).}$$

Therefore the total error is bounded by

$${\displaystyle {\text{error))=\int _{a}^{b}f(x)\,dx-{\frac {b-a}{N))\left[{f(a)+f(b) \over 2}+\sum _{k=1}^{N-1}f\left(a+k{\frac {b-a}{N))\right)\right]={\frac {f''(\xi )h^{3}N}{12))={\frac {f''(\xi )(b-a)^{3)){12N^{2))}.}$$

Periodic and peak functions

The trapezoidal rule converges rapidly for periodic functions. This is an easy consequence of the Euler-Maclaurin summation formula, which says that if ${\displaystyle f}$ is ${\displaystyle p}$ times continuously differentiable with period ${\displaystyle T}$

$${\displaystyle \sum _{k=0}^{N-1}f(kh)h=\int _{0}^{T}f(x)\,dx+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k)){(2k)!))(f^{(2k-1)}(T)-f^{(2k-1)}(0))-(-1)^{p}h^{p}\int _{0}^{T}{\tilde {B))_{p}(x/T)f^{(p)}(x)\,dx}$$

${\displaystyle h:=T/N}$

${\displaystyle {\tilde {B))_{p))$

${\displaystyle p}$

${\displaystyle O(h^{p})}$

A similar effect is available for peak-like functions, such as Gaussian, Exponentially modified Gaussian and other functions with derivatives at integration limits that can be neglected.^[9] The evaluation of the full integral of a Gaussian function by trapezoidal rule with 1% accuracy can be made using just 4 points.^[10] Simpson's rule requires 1.8 times more points to achieve the same accuracy.^[10][11]

Although some effort has been made to extend the Euler-Maclaurin summation formula to higher dimensions,^[12] the most straightforward proof of the rapid convergence of the trapezoidal rule in higher dimensions is to reduce the problem to that of convergence of Fourier series. This line of reasoning shows that if ${\displaystyle f}$ is periodic on a ${\displaystyle n}$-dimensional space with ${\displaystyle p}$ continuous derivatives, the speed of convergence is ${\displaystyle O(h^{p/d})}$. For very large dimension, the shows that Monte-Carlo integration is most likely a better choice, but for 2 and 3 dimensions, equispaced sampling is efficient. This is exploited in computational solid state physics where equispaced sampling over primitive cells in the reciprocal lattice is known as Monkhorst-Pack integration.^[13]

"Rough" functions

For functions that are not in C², the error bound given above is not applicable. Still, error bounds for such rough functions can be derived, which typically show a slower convergence with the number of function evaluations ${\displaystyle N}$ than the ${\displaystyle O(N^{-2})}$ behaviour given above. Interestingly, in this case the trapezoidal rule often has sharper bounds than Simpson's rule for the same number of function evaluations.^[14]

Applicability and alternatives

The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. Simpson's rule is another member of the same family, and in general has faster convergence than the trapezoidal rule for functions which are twice continuously differentiable, though not in all specific cases. However, for various classes of rougher functions (ones with weaker smoothness conditions), the trapezoidal rule has faster convergence in general than Simpson's rule.^[14]

Moreover, the trapezoidal rule tends to become extremely accurate when periodic functions are integrated over their periods, which can be analyzed in various ways.^[7][11] A similar effect is available for peak functions.^[10][11]

For non-periodic functions, however, methods with unequally spaced points such as Gaussian quadrature and Clenshaw–Curtis quadrature are generally far more accurate; Clenshaw–Curtis quadrature can be viewed as a change of variables to express arbitrary integrals in terms of periodic integrals, at which point the trapezoidal rule can be applied accurately.

Example

The following integral is given:

$${\displaystyle \int _{0.1}^{1.3}{5xe^{-2x}{dx))}$$

Solution