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== Numerical Integration ==

=== Introduction ===

There exists two cases, when the computation of the value of a definite integral by numerical methods is needed. One of them is the calculation of the area below the curve defined by a set of experimental data, and another is the calculation of the definite integral of a mathematical function, for which no known integral is known. The former is often the case of response functions in the experimental labors of science and engineering, while the latter is normally the case in the practical investigations of physics, mathematics, and engineering.

Independently of it, the development of numerical methods for integration purposes, a field that belongs to the department of applied mathematics, is based on the simple idea from which it stems, i.e., if <math>y(x) = f(x)</math> is a real-valued (the complex-valued case can be treated analogously, by separating it into its real and imaginary parts) continuous function of <math>x</math> defined in an interval <math>(a,b)</math>, its definite integral,

<math>
\int_{a}^{b} f(x) dx = \sum_{x = a}^{x = b} \lim_{\delta x \rightarrow 0} f(x) \delta x = \lim_{\delta x \rightarrow 0} \sum_{x = a}^{x = b} f(x) \delta x \approx \sum_{x = a}^{x = b} f(x) \Delta x
</math>,

can be calculated approximately as the finite sum of the product <math>f(x) \Delta x</math> evaluated at some given points in the interval <math>(a,b)</math>.

In the case of experimental data, the set of points at which the value of the function is measured is usually not regularly distributed (i.e., the points are not equispaced), so the value of the definite integral must be calculated in the form:

<math>
\int_{a}^{b} f(x) dx = \int_{{x_0} = a}^{{x_1}} f(x) dx + \int_{{x_1}}^{{x_2}} f(x) dx + ... + \int_{{x_{n-2}}}^{{x_{n-1}}} f(x) dx + \int_{{x_{n-1}}}^{{x_n = b}} f(x) dx
</math>,

which can be approximated either as:

<math>
\int_{a}^{b} f(x) dx \approx f({x_0} = a) ({x_1} - {x_0}) + f({x_1}) ({x_2} - {x_1}) + ... + f({x_{n-2}}) ({x_{n-1}} - {x_{n-2}}) + f({x_{n-1}}) ({x_n} - {x_{n-1}})
</math>,

or as:

<math>
\int_{a}^{b} f(x) dx \approx f({x_1}) ({x_1} - {x_0}) + f({x_2}) ({x_2} - {x_1}) + ... + f({x_{n-1}}) ({x_{n-1}} - {x_{n-2}}) + f({x_n} = b) ({x_n} - {x_{n-1}})
</math>.

In the first case, the value of the integral is underestimated (overestimated) in the case of monotonically ascending (descending) functions, since the value of <math>f(x)</math> taken in each evaluation is always the lowest (highest) in every subinterval, and hence constituting an absolute lower (upper) bound to the value of the integral, while in the second case, the value of the integral is overestimated (underestimated) in the case of monotonically descending (ascending) functions, since the value of <math>f(x)</math> taken in each evaluation is always the highest (lowest) in every subinterval, and hence constituting an absolute upper (lower) bound to the value of the integral.

According to the Mean-Value Theorem of Calculus, the value of a definite integral can also be calculated as:

<math>
\int_{a}^{b} f(x) dx = f(\gamma) (b - a)
</math>,

for some value <math>\gamma</math> in <math>(a,b)</math> for which <math>f(\gamma)</math> represents the mean value of <math>f(x)</math> in <math>(a,b)</math>, so it is then a better approximation to calculate the definite integral of a set of experimental data as:

<math>
\int_{a}^{b} f(x) dx = \int_{{x_0} = a}^{{x_1}} f(x) dx + \int_{{x_1}}^{{x_2}} f(x) dx + ... + \int_{{x_{n-2}}}^{{x_{n-1}}} f(x) dx + \int_{{x_{n-1}}}^{{x_n = b}} f(x) dx
</math>,

<math>
\int_{a}^{b} f(x) dx = f(\gamma_{(0,1)}) ({x_1} - {x_0}) + f(\gamma_{(1,2)}) ({x_2} - {x_1}) + ... + f(\gamma_{(n-2,n-1)}) ({x_{n-1}} - {x_{n-2}}) + f(\gamma_{(n-1,n)}) ({x_n} - {x_{n-1}})
</math>,

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_0}) + f({x_1})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) + f({x_2})}{2} ({x_2} - {x_1}) + ... + \frac{f({x_{n-2}}) + f({x_{n-1}})}{2} ({x_{n-1}} - {x_{n-2}}) + \frac{f({x_{n-1}}) + f({x_n})}{2} ({x_n} - {x_{n-1}})
</math>.

For reasons that we shall see later, this is equal to assume a piecewise linear interpolating function between the different points, and the value of the integral so calculated is exact for linear functions (i.e., functions for which its slope changes at constant rate), although it is underestimated for functions for which its slope grows at non-constant rate (i.e., its second derivative is strictly positive in the considered interval), and it is overestimated for functions for which its slope decreases at non-constant rate (i.e., its second derivative is strictly negative in the considered interval). The value so calculated constitutes a better approximation than the lower and upper bounds presented before, and in the case that the second derivative of the function (which can be calculated from the experimental data with the help of second or central differences) changes signs between subintervals, the value is expected to be close to the actual value due to the cancellation of the errors in the approximation of the mean values.

In the case of mathematical functions, there is more information about the function, since it is possible not only to calculate the value of the function at any given point, but also to compute first, second, and higher-order derivatives with any degree of accuracy. 

Let us elaborate some mathematical results, going from simple to more elaborate methods:

The main theme in the development of numerical methods, together with the study of the stability (i.e., if a method converges), is the rate of convergence of the method, which studies how many evaluations are needed and the error in the approximation, for non-iterative methods, or how many iterations are needed and how the error is minimized in each iteration, for iterative methods.

In the case of the study of the stability, and as we have seen before, the value of the definite integral of a function <math>f(x)</math> defined in an interval <math>(a,b)</math>,

<math>
\int_{a}^{b} f(x) dx = \sum_{x = a}^{x = b} \lim_{\delta x \rightarrow 0} f(x) \delta x = \lim_{\delta x \rightarrow 0} \sum_{x = a}^{x = b} f(x) \delta x \approx \sum_{x = a}^{x = b} f({x_k}) \Delta {x_k}
</math>,

can be calculated approximately as the finite sum of the product <math>f(x) \Delta x</math> evaluated at some given points in the interval <math>(a,b)</math>.

In the limit <math>\Delta x \rightarrow \delta x \rightarrow 0</math>, the finite sum tends to the infinite integral, and so convergence is assured.

In the case of the study of the rate of convergence, one is interested in increasing the accuracy of the approximation, while retaining the number of subintervals, with a minor increase in computational complexity.

The approach used consists normally in using a polynomial approximation for the evaluation of the function <math>f(x)</math> in each subinterval, using the information provided by the value of the function at several points in the subinterval.

Let us consider the case of equally-spaced points (although this restriction can be easily lifted):

According to the definition,

<math>
\int_{a}^{b} f(x) dx = \sum_{x = a}^{x = b} \lim_{\delta x \rightarrow 0} f(x) \delta x = \lim_{\delta x \rightarrow 0} \sum_{x = a}^{x = b} f(x) \delta x \approx \sum_{x = a}^{x = b} f({x_k}) \Delta {x_k} = \sum_{x = a}^{x = b} f({x_k}) ({x_{j+1}} - {x_{j}})
</math>,

with <math>k</math> being indicative of the subinterval, <math>{x_k}</math> being some arbitrary number in every subinterval <math>({x_{j}},{x_{j+1}})</math>, with <math>j = 0, 1, 2, ..., n-1</math>, and <math>{x_{0}} = a, {x_{n}} = b</math>, and <math>\Delta {x_{k}} = ({x_{j+1}} - {x_{j}})</math>, with <math>k = 1, 2, ..., n</math>, and <math>j = k - 1</math>.

The Mean-Value Theorem of Calculus tells us, that if

<math>
\int_{a}^{b} f(x) dx
</math>

is the definite integral of <math>f(x)</math> in <math>(a,b)</math>, there exists a value <math>\gamma</math> in <math>(a,b)</math>, such that

<math>
\int_{a}^{b} f(x) dx = f(\gamma) (b - a)
</math>.

Additionally, by definition, if

<math>
\int_{a}^{b} f(x) dx
</math>

is the definite integral of <math>f(x)</math> in <math>(a,b)</math>, this one can also be understood as being composed of the individual contributions

<math>
\int_{a}^{b} f(x) dx = \int_{{x_0} = a}^{{x_1}} f(x) dx + \int_{{x_1}}^{{x_2}} f(x) dx + ... + \int_{{x_{n-2}}}^{{x_{n-1}}} f(x) dx + \int_{{x_{n-1}}}^{{x_n = b}} f(x) dx
</math>

for arbitrary values <math>{x_{0}} = a, {x_{1}}, {x_{2}}, ..., {x_{n-2}}, {x_{n-1}}, {x_{n}} = b</math>.

Now, applying the Mean-Value Theorem to each individual contribution yields the result:

<math>
\int_{a}^{b} f(x) dx = f(\gamma_{({x_{0}},{x_{1}})}) ({x_{1}} - {x_{0}}) + f(\gamma_{({x_{1}},{x_{2}})}) ({x_{2}} - {x_{1}}) + ... + f(\gamma_{({x_{n-2}},{x_{n-1}})}) ({x_{n-1}} - {x_{n-2}}) + f(\gamma_{({x_{n-1}},{x_{n}})}) ({x_{n}} - {x_{n-1}})
</math>,

which is exact.

In the particular case that every subinterval is of equal size, i.e., <math>({x_{1}} - {x_{0}}) = ({x_{2}} - {x_{1}}) = ... = ({x_{n-1}} - {x_{n-2}}) = ({x_{n}} - {x_{n-1}}) = \Delta x</math>, then the expression reduces to

<math>
\int_{a}^{b} f(x) dx = \left(f(\gamma_{({x_{0}},{x_{1}})}) + f(\gamma_{({x_{1}},{x_{2}})}) + ... + f(\gamma_{({x_{n-2}},{x_{n-1}})}) + f(\gamma_{({x_{n-1}},{x_{n}})})\right) \Delta x
</math>.

In this way, the calculation of the initial definite integral reduces to the calculation of the mean values

<math>
f(\gamma_{({x_{0}},{x_{1}})}), f(\gamma_{({x_{1}},{x_{2}})}), ..., f(\gamma_{({x_{n-2}},{x_{n-1}})}), f(\gamma_{({x_{n-1}},{x_{n}})})
</math>.

In a first approximation, with only one point,

<math>
f(\gamma_{({x_{0}},{x_{1}})}) \approx f({x_{i}}_{({x_{0}},{x_{1}})}), f(\gamma_{({x_{1}},{x_{2}})}) \approx f({x_{i}}_{({x_{1}},{x_{2}})}), ..., f(\gamma_{({x_{n-2}},{x_{n-1}})}) \approx f({x_{i}}_{({x_{n-2}},{x_{n-1}})}), f(\gamma_{({x_{n-1}},{x_{n}})}) \approx f({x_{i}}_{({x_{n-1}},{x_{n}})})
</math>,

with <math>{x_{i}}</math> being the value of <math>x</math> in the middle of each subinterval, which leads to

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{i}}_{({x_{0}},{x_{1}})}) + f({x_{i}}_{({x_{1}},{x_{2}})}) + ... + f({x_{i}}_{({x_{n-2}},{x_{n-1}})}) + f({x_{i}}_{({x_{n-1}},{x_{n}})})\right) \Delta x
</math>.

In a second approximation, with only two points,

<math>
f(\gamma_{({x_{0}},{x_{1}})}) \approx \frac{f({x_{0}}) + f({x_{1}})}{2}, f(\gamma_{({x_{1}},{x_{2}})}) \approx \frac{f({x_{1}}) + f({x_{2}})}{2}, ..., f(\gamma_{({x_{n-2}},{x_{n-1}})}) \approx \frac{f({x_{n-2}}) + f({x_{n-1}})}{2}, f(\gamma_{({x_{n-1}},{x_{n}})}) \approx \frac{f({x_{n-1}}) + f({x_{n}})}{2}
</math>,

with <math>{x_{n-1}}, {x_{n}}</math> being the value of <math>x</math> at the beginning and at the end of each subinterval, which leads to

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + 2 f({x_{1}}) + ... + 2 f({x_{n-1}}) + f({x_{n}})\right) \Delta x / 2
</math>.

In a third approximation, with only three points,

<math>
f(\gamma_{({x_{0}},{x_{1}})}) \approx \frac{f({x_{0}}) + f({x_{i}}_{({x_{0}},{x_{1}})}) + f({x_{1}})}{3}, f(\gamma_{({x_{1}},{x_{2}})}) \approx \frac{f({x_{1}}) + f({x_{i}}_{({x_{1}},{x_{2}})}) + f({x_{2}})}{3}, ..., f(\gamma_{({x_{n-2}},{x_{n-1}})}) \approx \frac{f({x_{n-2}}) + f({x_{i}}_{({x_{n-2}},{x_{n-1}})}) + f({x_{n-1}})}{3}, f(\gamma_{({x_{n-1}},{x_{n}})}) \approx \frac{f({x_{n-1}}) + f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})}{3}
</math>,

with <math>{x_{n-1}}, {x_{i}}, {x_{n}}</math> being the value of <math>x</math> at the beginning, in the middle, and at the end of each subinterval, which leads to

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + f({x_{i}}_{({x_{0}},{x_{1}})}) + 2 f({x_{1}}) + f({x_{i}}_{({x_{1}},{x_{2}})}) + 2 f({x_{2}}) + ... + 2 f({x_{n-2}}) + f({x_{i}}_{({x_{n-2}},{x_{n-1}})}) + 2 f({x_{n-1}}) + f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})\right) \Delta x / 3
</math>.

In a fourth approximation, while still making use of the evaluation of the function <math>f(x)</math> at the beginning, in the middle, and at the end of each subinterval, one can use the result that if <math>f(\gamma^{-})</math> and <math>f(\gamma^{+})</math> are estimates for <math>f(\gamma)</math>, then its arithmetic mean <math>\left(f(\gamma^{-}) + f(\gamma^{+})\right) / 2</math> is also another estimate with at least the same accuracy, if not better.

So, adding the results for the first and second approximation, and dividing by two,

<math>
f(\gamma_{({x_{n-1}},{x_{n}})}) \approx \frac{f({x_{i}}_{({x_{n-1}},{x_{n}})}) + \frac{f({x_{n-1}}) + f({x_{n}})}{2}}{2} = \frac{f({x_{n-1}}) + 2 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})}{4}
</math>,

which leads to the result

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + 2 f({x_{i}}_{({x_{0}},{x_{1}})}) + 2 f({x_{1}}) + 2 f({x_{i}}_{({x_{1}},{x_{2}})}) + 2 f({x_{2}}) + ... + 2 f({x_{n-2}}) + 2 f({x_{i}}_{({x_{n-2}},{x_{n-1}})}) + 2 f({x_{n-1}}) + 2 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})\right) \Delta x / 4
</math>.

Adding the results for the first and third approximation, and dividing by two,

<math>
f(\gamma_{({x_{n-1}},{x_{n}})}) \approx \frac{f({x_{i}}_{({x_{n-1}},{x_{n}})}) + \frac{f({x_{n-1}}) + f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})}{3}}{2} = \frac{f({x_{n-1}}) + 4 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})}{6}
</math>,

which leads to the result

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + 4 f({x_{i}}_{({x_{0}},{x_{1}})}) + 2 f({x_{1}}) + 4 f({x_{i}}_{({x_{1}},{x_{2}})}) + 2 f({x_{2}}) + ... + 2 f({x_{n-2}}) + 4 f({x_{i}}_{({x_{n-2}},{x_{n-1}})}) + 2 f({x_{n-1}}) + 4 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})\right) \Delta x / 6
</math>.

In practice, one can do no better with only three evaluations in an interval, but the results obtained are simple and accurate enough, even in the case of one single interval.

Let us illustrate the case by means of an example:

Let us suppose, that we wish to calculate the definite integral of the function <math>f(x) = exp(x)</math> in the interval <math>(0,1)</math>, for which we know its exact value, <math>F(x)|_{0}^{1} = \int_{0}^{1} exp(x) dx = exp(x)|_{0}^{1} = exp(1) - exp(0) = 2.71828 - 1.00000 = 1.71828</math>.

Let us also keep the problem simple and do the calculations with a single interval, i.e., <math>{x_{0}} = a = 0.0</math>, <math>{x_{n}} = b = 1.0</math>, and <math>x_i = 0.5</math>.

So, we have:

First approximation:

<math>
\int_{a}^{b} f(x) dx \approx f(x_i) (b - a) = exp(0.5) (1.0 - 0.0) = 1.64872
</math>

This is equal to a relative error of

<math>
error = \frac{1.64872 - 1.71828}{1.71828} \times 100 = -4.04823%
</math>.

Second approximation:

<math>
\int_{a}^{b} f(x) dx \approx \frac{f(x_a) + f(x_b)}{2} (b - a) = \frac{exp(0.0) + exp(1.0)}{2} (1.0 - 0.0) = 1.85914
</math>

This is equal to a relative error of

<math>
error = \frac{1.85914 - 1.71828}{1.71828} \times 100 = +8.19774%
</math>.

Third approximation:

<math>
\int_{a}^{b} f(x) dx \approx \frac{f(x_a) + f(x_i) + f(x_b)}{3} (b - a) = \frac{exp(0.0) + exp(0.5) + exp(1.0)}{3} (1.0 - 0.0) = 1.78900
</math>

This is equal to a relative error of

<math>
error = \frac{1.85914 - 1.71828}{1.71828} \times 100 = +4.11575%
</math>.

Fourth approximation, first and second terms (linear expansion):

<math>
\int_{a}^{b} f(x) dx \approx \frac{f(x_a) + 2 f(x_i) + f(x_b)}{4} (b - a) = \frac{exp(0.0) + 2 \times exp(0.5) + exp(1.0)}{4} (1.0 - 0.0) = 1.75393
</math>

This is equal to a relative error of

<math>
error = \frac{1.75393 - 1.71828}{1.71828} \times 100 = +2.07475%
</math>.

Fourth approximation, first and third terms (quadratic expansion):

<math>
\int_{a}^{b} f(x) dx \approx \frac{f(x_a) + 4 f(x_i) + f(x_b)}{6} (b - a) = \frac{exp(0.0) + 4 \times exp(0.5) + exp(1.0)}{6} (1.0 - 0.0) = 1.71886
</math>

This is equal to a relative error of

<math>
error = \frac{1.71886 - 1.71828}{1.71828} \times 100 = +0.0337589%
</math>.

In case more accuracy is needed, one can analogously introduce higher-order terms in the expansion of <math>f(x)</math> around <math>x_i</math>, or half the integration step in one of the methods considered before.

Let us consider the case, when higher-order terms are considered in the expansion, i.e.,

<math>
f(\gamma_{({x_{n-1}},{x_{n}})}) \approx \frac{f({x_{i}}_{({x_{n-1}},{x_{n}})}) + \frac{f({x_{n-1}}) + f({x_{i}}_{1/4}) + f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{i}}_{3/4}) + f({x_{n}})}{5}}{2} = \frac{f({x_{n-1}}) + f({x_{i}}_{1/4}) + 6 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{i}}_{3/4}) + f({x_{n}})}{10}
</math>,

which leads to the result

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + f({x_{1/4}}_{({x_{0}},{x_{1}})}) + 6 f({x_{i}}_{({x_{1}},{x_{2}})}) + f({x_{3/4}}_{({x_{0}},{x_{1}})}) + 2 f({x_{1}}) + ... + 2 f({x_{n-1}}) + f({x_{1/4}}_{({x_{n-1}},{x_{n}})}) + 6 f({x_{i}}_{({x_{n-1}},{x_{n}})}) + f({x_{1/4}}_{({x_{n-1}},{x_{n}})}) + f({x_{n}})\right) \Delta x / 10
</math>.

Let us illustrate the case by means of an example:

Let us suppose, that we wish to calculate the definite integral of the function <math>f(x) = exp(x)</math> in the interval <math>(0,1)</math>, for which we know its exact value, <math>F(x)|_{0}^{1} = \int_{0}^{1} exp(x) dx = exp(x)|_{0}^{1} = exp(1) - exp(0) = 2.71828 - 1.00000 = 1.71828</math>.

Let us also keep the problem simple and do the calculations with two intervals, i.e., <math>{x_{0}} = a = 0.0</math>, <math>{x_{n}} = b = 1.0</math>, <math>{x_{1}} = 0.5</math>, <math>{x_{i}}_{({x_{0}},{x_{1}})} = 0.25</math> and <math>{x_{i}}_{({x_{1}},{x_{2}})} = 0.25</math>.

Applying second order expansion

<math>
\int_{a}^{b} f(x) dx \approx \left(f({x_{0}}) + 4 f({x_{i}}_{({x_{0}},{x_{1}})}) + 2 f({x_{1}}) + 4 f({x_{i}}_{({x_{1}},{x_{2}})}) + f({x_{2}})\right) \Delta x / 6
</math>,

which yields

<math>
\int_{a}^{b} f(x) dx \approx \left(exp(0.0) + 4 exp(0.25) + 2 exp(0.5) + 4 exp(0.75) + exp(1.0)\right) \times 0.5 / 6 = 1.71832
</math>,

which is equal to a relative error of

<math>
error = \frac{1.71832 - 1.71828}{1.71828} \times 100 = +0.00216456%
</math>.

Applying fourth order expansion

<math>
\int_{a}^{b} f(x) dx \approx \left(f(0.0) + f(0.25) + 6 f(0.50) + f(0.75) + f(1.0)\right) 1.0 / 10
</math>,

which yields

<math>
\int_{a}^{b} f(x) dx \approx \left(exp(0.0) + 7 exp(0.25) + 2 exp(0.5) + 7 exp(0.75) + exp(1.0)\right) \times 1.0 / 10
</math>.

We know from the Calculus of Differences, that any function can be expressed as the polynomial expansion

<math>
f(x) = f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_0})^2 + ...
</math>

in the case of forward differences, or

<math>
f(x) = f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_1})^2 + ...
</math>

in the case of backward differences, or

<math>
f(x) = f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_i})^2 + ...
</math>

in the case of central differences.

In a first approximation,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_0}) + h({x_0})(x - {x_0}) \right) dx \approx \int_{a}^{b} f({x_0}) dx = f({x_0})(b - a) = f({x_0})({x_1} - {x_0}) = f({x_0}) \Delta x
</math>

<math>
\int_{a}^{b} h({x_0})(x - {x_0}) dx = h({x_0})\frac{(x - {x_0})^2}{2}|_{a}^{b} = h({x_0})\frac{(b - {x_0})^2 - (a - {x_0})^2}{2} = h({x_0})\frac{({x_1} - {x_0})^2 - ({x_0} - {x_0})^2}{2} = h({x_0})\frac{({x_1} - {x_0})^2}{2} = h({x_0})\frac{\Delta x^2}{2} \simeq f'({x_0}) \frac{\Delta x^2}{2}
</math>

in the case of forward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_1}) + h({x_1})(x - {x_1}) \right) dx \approx \int_{a}^{b} f({x_1}) dx = f({x_1})(b - a) = f({x_1})({x_1} - {x_0}) = f({x_1}) \Delta x
</math>

<math>
\int_{a}^{b} h({x_1})(x - {x_1}) dx = h({x_1})\frac{(x - {x_1})^2}{2}|_{a}^{b} = h({x_1})\frac{(b - {x_1})^2 - (a - {x_1})^2}{2} = h({x_1})\frac{({x_1} - {x_1})^2 - ({x_0} - {x_1})^2}{2} = h({x_1})\frac{-({x_0} - {x_1})^2}{2} = - h({x_1})\frac{\Delta x^2}{2} \simeq -f'({x_1}) \frac{\Delta x^2}{2}
</math>

in the case of backward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_i}) + h({x_i})(x - {x_i}) \right) dx \approx \int_{a}^{b} f({x_i}) dx = f({x_i})(b - a) = f({x_i})({x_1} - {x_0}) = f({x_i}) \Delta x
</math>

<math>
\int_{a}^{b} h({x_i})(x - {x_i}) dx = h({x_i})\frac{(x - {x_i})^2}{2}|_{a}^{b} = h({x_i})\frac{(b - {x_i})^2 - (a - {x_i})^2}{2} = h({x_i})\frac{({x_1} - {x_i})^2 - ({x_0} - {x_i})^2}{2} = h({x_i})\frac{(\Delta x /2)^2 - (-\Delta x /2)^2}{2} = h({x_i})\frac{\Delta x^2 - \Delta x^2}{4} = 0
</math>

in the case of central differences.

So, in the case of evaluating the integral with one single point, the value of it is always underestimated (overestimated) with forward (backward) differences if the value of the slope of the function at the evaluation point is positive, and underestimated (overestimated) with forward (backward) differences if the value of the slope of the function at the evaluation point is negative, being the error in the estimation directly proportional to the slope of the function at the evaluation point, and to one half of the square of the integration step. In the case of evaluating the integral with central differences, the value is exact for linear functions.

In a second approximation,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + h({x_0})(x - {x_0})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) \right) dx = f({x_0})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_0})^2 - ({x_0} - {x_0})^2}{2}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_0})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_0})^2}{2} = f({x_0})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{2} ({x_1} - {x_0}) = \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) = \frac{f({x_1}) + f({x_0})}{2} \Delta x
</math>

<math>
\int_{a}^{b} h({x_0})(x - {x_0})^2 dx = h({x_0})\frac{(x - {x_0})^3}{3}|_{a}^{b} = h({x_0})\frac{({x_1} - {x_0})^3 - ({x_0} - {x_0})^3}{3} = h({x_0})\frac{({x_1} - {x_0})^3}{3} = h({x_0})\frac{\Delta x^3}{3} \simeq f''({x_0}) \frac{\Delta x^3}{3}
</math>

in the case of forward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + h({x_1})(x - {x_1})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) \right) dx = f({x_1})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_1})^2 - ({x_0} - {x_1})^2}{2}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_1})({x_1} - {x_0}) - \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_0})^2}{2} = f({x_1})({x_1} - {x_0}) - \frac{f({x_1}) - f({x_0})}{2} ({x_1} - {x_0}) = \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) = \frac{f({x_1}) + f({x_0})}{2} \Delta x
</math>

<math>
\int_{a}^{b} h({x_1})(x - {x_1})^2 dx = h({x_1})\frac{(x - {x_1})^3}{3}|_{a}^{b} = h({x_1})\frac{({x_1} - {x_1})^3 - ({x_0} - {x_1})^3}{3} = h({x_1})\frac{({x_1} - {x_0})^3}{3} = h({x_1})\frac{\Delta x^3}{3} \simeq f''({x_1}) \frac{\Delta x^3}{3}
</math>

in the case of backward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + h({x_i})(x - {x_i})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) \right) dx = f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_i})^2 - ({x_0} - {x_i})^2}{2}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_i})^2 - ({x_i} - {x_0})^2}{2} = f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} \frac{({x_1} - {x_0})^2 - ({x_1} - {x_0})^2}{8} = f({x_i})({x_1} - {x_0}) = f({x_i}) \Delta x
</math>

<math>
\int_{a}^{b} h({x_i})(x - {x_i})^2 dx = h({x_i})\frac{(x - {x_i})^3}{3}|_{a}^{b} = h({x_i})\frac{({x_1} - {x_i})^3 - ({x_0} - {x_i})^3}{3} = h({x_i})\frac{({x_1} - {x_i})^3 + ({x_i} - {x_0})^3}{3} = h({x_i})\frac{({x_1} - {x_0})^3}{12} = h({x_i})\frac{\Delta x^3}{12} \simeq f''({x_i}) \frac{\Delta x^3}{12}
</math>

in the case of central differences.

So, in the case of evaluating the integral with a linear function, there is no difference between forward and backward differences, and the value of the integral is equal to the product of the mean of the values of the function at the beginning and at the end of the interval times the integration step, being the value so calculated underestimated (overestimated) for monotonically increasing (decreasing) functions with a positive (negative) second derivative, the error in the approximation being directly proportional to the second derivative of the function and to the cube of the integration step (i.e., as one increases the number of subintervals, one reduces the deviation proportional to the cube of the number of subintervals). In the case of central differences, the value of the integral is equal to the product of the value of the function in the middle of the interval times the integration step, as in the previous case, the error in the approximation being reduced by a factor of four with relation to the case of forward and backward differences.

In a third approximation,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_0})^2 + h({x_0})(x - {x_0})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_0})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3 - ({x_0} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{3} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{3 f({x_1}) + 3 f({x_0}) + 2 f({x_1}) - 4 f({x_i}) + 2 f({x_0})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{5 f({x_0}) - 4 f({x_i}) + 5 f({x_1})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_0})(x - {x_0})^3 dx = h({x_0})\frac{(x - {x_0})^4}{4}|_{a}^{b} = h({x_0})\frac{({x_1} - {x_0})^4 - ({x_0} - {x_0})^4}{4} = h({x_0})\frac{({x_1} - {x_0})^4}{4} = h({x_0})\frac{\Delta x^4}{4} \simeq f'''({x_0}) \frac{\Delta x^4}{4}
</math>

in the case of forward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_1})^2 + h({x_1})(x - {x_1})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_1})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_1})^3 - ({x_0} - {x_1})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) - \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) - \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{3} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{3 f({x_1}) + 3 f({x_0}) - 2 f({x_1}) + 4 f({x_i}) - 2 f({x_0})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_0}) + 4 f({x_i}) + f({x_1})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_1})(x - {x_1})^3 dx = h({x_1})\frac{(x - {x_1})^4}{4}|_{a}^{b} = h({x_1})\frac{({x_1} - {x_1})^4 - ({x_0} - {x_1})^4}{4} = -h({x_1})\frac{({x_1} - {x_0})^4}{4} = -h({x_1})\frac{\Delta x^4}{4} \simeq -f'''({x_1}) \frac{\Delta x^4}{4}
</math>

in the case of backward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_i})^2 + h({x_i})(x - {x_i})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_i})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_i})^3 - ({x_0} - {x_i})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{12}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{12 f({x_i}) + f({x_1}) - 2 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + 10 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_i})(x - {x_i})^3 dx = h({x_i})\frac{(x - {x_i})^4}{4}|_{a}^{b} = h({x_i})\frac{({x_1} - {x_i})^4 - ({x_0} - {x_i})^4}{4} = h({x_i})\frac{({x_1} - {x_0})^4 - ({x_1} - {x_0})^4}{64} = 0
</math>

in the case of central differences.

In a fourth approximation,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_0})^2 + \frac{\frac{f({x_1}) - f({x_i})}{{x_1} - {x_i}} - \frac{f({x_i}) - f({x_0})}{{x_i} - {x_0}}}{({x_1} - {x_0})^2} ({x} - {x_0})^3 + h({x_0})(x - {x_0})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_0}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_0})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3 - ({x_0} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{3} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{3 f({x_1}) + 3 f({x_0}) + 2 f({x_1}) - 4 f({x_i}) + 2 f({x_0})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{5 f({x_0}) - 4 f({x_i}) + 5 f({x_1})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_0})(x - {x_0})^3 dx = h({x_0})\frac{(x - {x_0})^4}{4}|_{a}^{b} = h({x_0})\frac{({x_1} - {x_0})^4 - ({x_0} - {x_0})^4}{4} = h({x_0})\frac{({x_1} - {x_0})^4}{4} = h({x_0})\frac{\Delta x^4}{4} \simeq f'''({x_0}) \frac{\Delta x^4}{4}
</math>

in the case of forward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_1})^2 + h({x_1})(x - {x_1})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_1}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_1}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_1})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_1})^3 - ({x_0} - {x_1})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) - \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + f({x_0})}{2} ({x_1} - {x_0}) - \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{3} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{3 f({x_1}) + 3 f({x_0}) - 2 f({x_1}) + 4 f({x_i}) - 2 f({x_0})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_0}) + 4 f({x_i}) + f({x_1})}{6} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_1})(x - {x_1})^3 dx = h({x_1})\frac{(x - {x_1})^4}{4}|_{a}^{b} = h({x_1})\frac{({x_1} - {x_1})^4 - ({x_0} - {x_1})^4}{4} = -h({x_1})\frac{({x_1} - {x_0})^4}{4} = -h({x_1})\frac{\Delta x^4}{4} \simeq -f'''({x_1}) \frac{\Delta x^4}{4}
</math>

in the case of backward differences,

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_i})^2 + h({x_i})(x - {x_i})^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} \left( f({x_i}) + \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}} ({x} - {x_i}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} ({x} - {x_i})^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_i})^3 - ({x_0} - {x_i})^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{({x_1} - {x_0})^2} \frac{({x_1} - {x_0})^3}{12}
</math>

<math>
\int_{a}^{b} f(x) dx \approx f({x_i})({x_1} - {x_0}) + \frac{f({x_1}) - 2 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{12 f({x_i}) + f({x_1}) - 2 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} f(x) dx \approx \frac{f({x_1}) + 10 f({x_i}) + f({x_0})}{12} ({x_1} - {x_0})
</math>

<math>
\int_{a}^{b} h({x_i})(x - {x_i})^3 dx = h({x_i})\frac{(x - {x_i})^4}{4}|_{a}^{b} = h({x_i})\frac{({x_1} - {x_i})^4 - ({x_0} - {x_i})^4}{4} = h({x_i})\frac{({x_1} - {x_0})^4 - ({x_1} - {x_0})^4}{64} = 0
</math>

in the case of central differences.

The divided differences are:

<math>
\frac{df(x)}{dx} \approx \frac{f({x_1}) - f({x_0})}{{x_1} - {x_0}}
</math>

<math>
\frac{d^2f(x)}{dx^2} \approx \frac{\frac{f({x_1}) - f({x_i})}{{x_1} - {x_i}} - \frac{f({x_i}) - f({x_0})}{{x_i} - {x_0}}}{{x_1} - {x_0}} = \frac{2 (f({x_1}) - 2 f({x_i}) + f({x_0}))}{({x_1} - {x_0})^2}
</math>

<math>
\frac{d^3f(x)}{dx^3} \approx \frac{\frac{\frac{f({x_1}) - f(({x_1} + {x_i})/2)}{\frac{{x_1} - {x_0}}{4}} - \frac{f(({x_1} + {x_i})/2) - f({x_i})}{\frac{{x_1} - {x_0}}{4}}}{\frac{{x_1} - {x_0}}{2}} - \frac{\frac{f({x_i}) - f(({x_i} + {x_0})/2)}{\frac{{x_1} - {x_0}}{4}} - \frac{f(({x_i} + {x_0})/2) - f({x_0})}{\frac{{x_1} - {x_0}}{4}}}{\frac{{x_1} - {x_0}}{2}}}{{x_1} - {x_0}} = \frac{8 (f({x_1}) - 2 f(({x_1} + {x_i})/2) + 2 f(({x_i} + {x_0})/2) - f({x_0}))}{({x_1} - {x_0})^3}
</math>

<math>
\frac{d^4f(x)}{dx^4} \approx \frac{\frac{\frac{\frac{f({x_1}) - f(({x_1} + {x_i})/4)}{\frac{{x_1} - {x_0}}{8}} - \frac{f({x_1}) - f(({x_1} + {x_i})/4)}{\frac{{x_1} - {x_0}}{8}}}{\frac{{x_1} - {x_0}}{4}} - \frac{f(({x_1} + {x_i})/2) - f({x_i})}{\frac{{x_1} - {x_0}}{4}}}{\frac{{x_1} - {x_0}}{2}} - \frac{\frac{f({x_i}) - f(({x_i} + {x_0})/2)}{\frac{{x_1} - {x_0}}{4}} - \frac{f(({x_i} + {x_0})/2) - f({x_0})}{\frac{{x_1} - {x_0}}{4}}}{\frac{{x_1} - {x_0}}{2}}}{{x_1} - {x_0}} = \frac{8 (f({x_1}) - 2 f(({x_1} + {x_i})/2) + 2 f(({x_i} + {x_0})/2) - f({x_0}))}{({x_1} - {x_0})^3}
</math>

Lagrange interpolation

One point:

<math>
L_{1}(x_0,x_1) = f(\frac{x_0 + x_1}{2}) = f(x_i)
</math>

Two points:

<math>
L_{2}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_1)}(x - x_1) + \frac{f(x_1)}{(x_1 - x_0)}(x - x_0)
</math>

Three points:

<math>
L_{3}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) + \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x - x_0)(x - x_i)
</math>

Four points:

<math>
L_{4}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) + \frac{f(x_1)}{(x_1 - x_i)(x_1 - x_0)}(x - x_i)(x - x_0)
</math>

Five points:

<math>
L_{5}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) + \frac{f(x_1)}{(x_1 - x_i)(x_1 - x_0)}(x - x_i)(x - x_0)
</math>

Six points:

<math>
L_{6}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) + \frac{f(x_1)}{(x_1 - x_i)(x_1 - x_0)}(x - x_i)(x - x_0)
</math>

Seven points:

<math>
L_{7}(x_0,x_1) = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) + \frac{f(x_1)}{(x_1 - x_i)(x_1 - x_0)}(x - x_i)(x - x_0)
</math>

One point:

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} (L_{1}(x) + h_{1}(x)) dx = \int_{a}^{b} (f(x_i) + 1 h_{1} (x - x_i)) dx
</math>

<math>
\int_{a}^{b} f(x) dx = f(x_i)(x_1 - x_0) + h_{1}\frac{(x_1 - x_i)^2 - (x_0 - x_i)^2}{2}
</math>

<math>
\int_{a}^{b} f(x) dx = f(x_i)(x_1 - x_0)
</math>

<math>
\int_{a}^{b} f(x) dx = f(x_i) \Delta x
</math>

Two points:

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} (L_{2}(x) + h_{2}(x)) dx = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_1)}(x - x_1) + \frac{f(x_1)}{(x_1 - x_0)}(x - x_0) + 2 h_{2} (x - x_i)^2 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{(x_0 - x_1)} \frac{(x_1 - x_1)^2 - (x_0 - x_1)^2}{2} + \frac{f(x_1)}{(x_1 - x_0)} \frac{(x_1 - x_0)^2 - (x_0 - x_0)^2}{2} + 2 h_{2} \frac{(x_1 - x_i)^3 - (x_0 - x_i)^3}{3}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{(x_0 - x_1)} \frac{-(x_0 - x_1)^2}{2} + \frac{f(x_1)}{(x_1 - x_0)} \frac{(x_1 - x_0)^2}{2} + 2 h_{2} \frac{(x_1 - x_0)^3 - (x_0 - x_1)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{-f(x_0)}{2} (x_0 - x_1) + \frac{f(x_1)}{2} (x_1 - x_0) + 2 h_{2} \frac{(x_1 - x_0)^3 + (x_1 - x_0)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{2} (x_1 - x_0) + \frac{f(x_1)}{2} (x_1 - x_0) + 4 h_{2} \frac{(x_1 - x_0)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + f(x_1)}{2} (x_1 - x_0) + h_{2} \frac{(x_1 - x_0)^3}{6}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + f(x_1)}{2} \Delta x + h_{2} \frac{\Delta x^3}{6}
</math>

<math>
\int_{a}^{b} f(x) dx = (f(x_0) + f(x_1)) \frac{\Delta x}{2} + h_{2} \frac{\Delta x^3}{6}
</math>


Three points:

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} (L_{3}(x) + h_{3}(x)) dx = \int_{a}^{b} (L_{3,1}(x) + L_{3,2}(x) + L_{3,3}(x) + h_{3}(x)) dx
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x - x_i)(x - x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x - x_0)(x - x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x - x_0)(x - x_i) \right) dx \\
                     & + \int_{a}^{b} h_{3}(x) (x - x_i)^3 dx
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x^2 - x x_i - x x_1 + x_i x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x^2 - x x_0 - x x_1 + x_0 x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x^2 - x x_0 - x x_i + x_0 x_i) \right) dx \\
                     & + \int_{a}^{b} h_{3}(x) (x - x_i)^3 dx
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x^2 - (x_i + x_1) x + x_i x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1) \right) dx \\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x^2 - (x_0 + x_i) x + x_0 x_i) \right) dx \\
                     & + \int_{a}^{b} h_{3}(x) (x - x_i)^3 dx
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_i + x_1) \frac{x_1^2 - x_0^2}{2} + x_i x_1 (x_1 - x_0)) \\
                     & + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_1) \frac{x_1^2 - x_0^2}{2} + x_0 x_1 (x_1 - x_0)) \\
                     & + \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_i) \frac{x_1^2 - x_0^2}{2} + x_0 x_i (x_1 - x_0)) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (\frac{x_0 + x_1}{2} + x_1) \frac{x_1^2 - x_0^2}{2} + \frac{x_0 + x_1}{2} x_1 (x_1 - x_0)) \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_1) \frac{x_1^2 - x_0^2}{2} + x_0 x_1 (x_1 - x_0)) \\
                    & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + \frac{x_0 + x_1}{2}) \frac{x_1^2 - x_0^2}{2} + x_0 \frac{x_0 + x_1}{2} (x_1 - x_0)) \\
                    & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + 3 x_1}{2} \frac{x_1^2 - x_0^2}{2} + \frac{x_1^2 - x_0^2}{2} x_1) \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2}{2} + x_0 x_1^2 - x_1 x_0^2) \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{3 x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2} + \frac{x_1^2 - x_0^2}{2} x_0) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (\frac{x_0 + 3 x_1}{2} - x_1) \frac{x_1^2 - x_0^2}{2}) \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2 - 2 x_0 x_1^2 + 2 x_1 x_0^2}{2}) \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - (\frac{3 x_0 + x_1}{2} - x_0) \frac{x_1^2 - x_0^2}{2}) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2}) \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{2 x_1^3 - 2 x_0^3 - 3 x_0 x_1^2 + 3 x_0^3 - 3 x_1^3 + 3 x_1 x_0^2 + 6 x_0 x_1^2 - 6 x_1 x_0^2}{6} \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2}) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2}{4}) \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{-x_1^3 + x_0^3 + 3 x_0 x_1^2 - 3 x_1 x_0^2}{6} \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 + x_1^3 - x_0^3 - x_1 x_0^2}{4}) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{4 x_1^3 - 4 x_0^3 - 3 x_0 x_1^2 + 3 x_0^3 - 3 x_1^3 + 3 x_1 x_0^2}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{-x_1^3 + 3 x_0 x_1^2 - 3 x_1 x_0^2 + x_0^3}{6} \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{4 x_1^3 - 4 x_0^3 - 3 x_0 x_1^2 - 3 x_1^3 + 3 x_0^3 + 3 x_1 x_0^2}{12}) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{x_1^3 - x_0^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{6} \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2}{12}) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{6} \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)} \frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{12} \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{(x_1 - x_0)^3}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{6}  \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{12} \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{6}  \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{12} \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{2f(x_0)}{(x_1 - x_0)^2} \frac{(x_1 - x_0)^3}{12} \\
                     & + \frac{4f(x_i)}{(x_1 - x_0)^2} \frac{(x_1 - x_0)^3}{6}  \\
                     & + \frac{2f(x_1)}{(x_1 - x_0)^2} \frac{(x_1 - x_0)^3}{12} \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \frac{f(x_0)}{6}  (x_1 - x_0) \\
                     & + \frac{4f(x_i)}{6} (x_1 - x_0) \\
                     & + \frac{f(x_1)}{6}  (x_1 - x_0) \\
                     & + h_{3} \frac{(x_1 - x_i)^4 - (x_0 - x_i)^4}{4}
\end{align}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{6} (x_1 - x_0) + \frac{4f(x_i)}{6} (x_1 - x_0) + \frac{f(x_1)}{6} (x_1 - x_0) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + 4f(x_i) + f(x_1)}{6} (x_1 - x_0) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + 4f(x_i) + f(x_1)}{6} \Delta x + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = (f(x_0) + 4f(x_i) + f(x_1)) \frac{\Delta x}{6} + 3 h_{3} (x - x_i)^3
</math>

Four points:

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} (L_{4}(x) + h_{4}(x)) dx = \int_{a}^{b} (L_{4,1}(x) + L_{4,2}(x) + L_{4,3}(x) + L_{4,4}(x) + h_{4}(x)) dx
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - {x_i}_{1/3})(x_0 - {x_i}_{2/3})(x_0 - x_1)}(x - {x_i}_{1/3})(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{1/3})}{({x_i}_{1/3} - x_0)({x_i}_{1/3} - {x_i}_{2/3})({x_i}_{1/3} - x_1)}(x - x_0)(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{2/3})}{({x_i}_{2/3} - x_0)({x_i}_{2/3} - {x_i}_{1/3})({x_i}_{2/3} - x_1)}(x - x_0)(x - {x_i}_{1/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - {x_i}_{1/3})(x_1 - {x_i}_{2/3})}(x - x_0)(x - {x_i}_{1/3})(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{\frac{1}{3}(x_0 - x_1) \frac{2}{3}(x_0 - x_1) \frac{1}{1}(x_0 - x_1)}(x - {x_i}_{1/3})(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{1/3})}{\frac{1}{3}(x_1 - x_0) \frac{1}{3}(x_0 - x_1) \frac{2}{3}(x_0 - x_1)}(x - x_0)(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{2/3})}{\frac{2}{3}(x_1 - x_0) \frac{1}{3}(x_1 - x_0) \frac{1}{3}(x_0 - x_1)}(x - x_0)(x - {x_i}_{1/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{\frac{1}{1}(x_1 - x_0) \frac{2}{3}(x_1 - x_0) \frac{1}{3}(x_1 - x_0)}(x - x_0)(x - {x_i}_{1/3})(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{9}{2} \frac{f(x_0)}{(x_0 - x_1) (x_0 - x_1) (x_0 - x_1)}(x - {x_i}_{1/3})(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{27}{2} \frac{f({x_i}_{1/3})}{(x_1 - x_0) (x_0 - x_1) (x_0 - x_1)}(x - x_0)(x - {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{27}{2} \frac{f({x_i}_{2/3})}{(x_1 - x_0) (x_1 - x_0) (x_0 - x_1)}(x - x_0)(x - {x_i}_{1/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{9}{2} \frac{f(x_1)}{(x_1 - x_0) (x_1 - x_0) (x_1 - x_0)}(x - x_0)(x - {x_i}_{1/3})(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( -\frac{9}{2} \frac{f(x_0)}{(x_1 - x_0)^3} (x - {x_i}_{1/3}) (x - {x_i}_{2/3}) (x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{27}{2} \frac{f({x_i}_{1/3})}{(x_1 - x_0)^3} (x - x_0) (x - {x_i}_{2/3}) (x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( -\frac{27}{2} \frac{f({x_i}_{2/3})}{(x_1 - x_0)^3} (x - x_0) (x - {x_i}_{1/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{9}{2} \frac{f(x_1)}{(x_1 - x_0)^3} (x - x_0)(x - {x_i}_{1/3})(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx = & - \frac{9}{2} \frac{f(x_0)}{(x_1 - x_0)^3} \int_{a}^{b} \left( (x - {x_i}_{1/3}) (x - {x_i}_{2/3}) (x - x_1) \right) dx\\
                       & + \frac{27}{2} \frac{f({x_i}_{1/3})}{(x_1 - x_0)^3} \int_{a}^{b} \left( (x - x_0) (x - {x_i}_{2/3}) (x - x_1) \right) dx\\
                       & - \frac{27}{2} \frac{f({x_i}_{2/3})}{(x_1 - x_0)^3} \int_{a}^{b} \left( (x - x_0) (x - {x_i}_{1/3})(x - x_1) \right) dx\\
                       & + \frac{9}{2} \frac{f(x_1)}{(x_1 - x_0)^3} \int_{a}^{b} \left( (x - x_0)(x - {x_i}_{1/3})(x - {x_i}_{2/3}) \right) dx\\
                       & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>


<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - {x_i}_{1/3})(x_0 - {x_i}_{2/3})(x_0 - x_1)}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{1/3})}{({x_i}_{1/3} - x_0)({x_i}_{1/3} - {x_i}_{2/3})({x_i}_{1/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{2/3})}{({x_i}_{2/3} - x_0)({x_i}_{2/3} - {x_i}_{1/3})({x_i}_{2/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{1/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - {x_i}_{1/3})(x_1 - {x_i}_{2/3})}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_0) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>

<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - {x_i}_{1/3})(x_0 - {x_i}_{2/3})(x_0 - x_1)}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{1/3})}{({x_i}_{1/3} - x_0)({x_i}_{1/3} - {x_i}_{2/3})({x_i}_{1/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{2/3})}{({x_i}_{2/3} - x_0)({x_i}_{2/3} - {x_i}_{1/3})({x_i}_{2/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{1/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - {x_i}_{1/3})(x_1 - {x_i}_{2/3})}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_0) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>


<math>
\begin{align}
\int_{a}^{b} f(x) dx & = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - {x_i}_{1/3})(x_0 - {x_i}_{2/3})(x_0 - x_1)}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_1) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{1/3})}{({x_i}_{1/3} - x_0)({x_i}_{1/3} - {x_i}_{2/3})({x_i}_{1/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{2/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f({x_i}_{2/3})}{({x_i}_{2/3} - x_0)({x_i}_{2/3} - {x_i}_{1/3})({x_i}_{2/3} - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1)(x - {x_i}_{1/3}) \right) dx\\
                     & + \int_{a}^{b} \left( \frac{f(x_1)}{(x_1 - x_0)(x_1 - {x_i}_{1/3})(x_1 - {x_i}_{2/3})}(x^2 - ({x_i}_{1/3} + {x_i}_{2/3}) x + {x_i}_{1/3} {x_i}_{2/3})(x - x_0) \right) dx\\
                     & + \int_{a}^{b} 4 h_{4} (x - x_i)^4 dx\\
\end{align}
</math>



<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x^2 - x x_i - x x_1 + x_i x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x^2 - x x_0 - x x_1 + x_0 x_1) + \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x^2 - x x_0 - x x_i + x_0 x_i) + 3 h_{3} (x - x_i)^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx = \int_{a}^{b} \left( \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(x^2 - (x_i + x_1) x + x_i x_1) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(x^2 - (x_0 + x_1) x + x_0 x_1) + \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(x^2 - (x_0 + x_i) x + x_0 x_i) + 3 h_{3} (x - x_i)^3 \right) dx
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{(x_0 - x_i)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_i + x_1) \frac{x_1^2 - x_0^2}{2} + x_i x_1 (x_1 - x_0)) + \frac{f(x_i)}{(x_i - x_0)(x_i - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_1) \frac{x_1^2 - x_0^2}{2} + x_0 x_1 (x_1 - x_0)) + \frac{f(x_1)}{(x_1 - x_0)(x_1 - x_i)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_i) \frac{x_1^2 - x_0^2}{2} + x_0 x_i (x_1 - x_0)) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (\frac{x_0 + x_1}{2} + x_1) \frac{x_1^2 - x_0^2}{2} + \frac{x_0 + x_1}{2} x_1 (x_1 - x_0)) + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + x_1) \frac{x_1^2 - x_0^2}{2} + x_0 x_1 (x_1 - x_0)) + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - (x_0 + \frac{x_0 + x_1}{2}) \frac{x_1^2 - x_0^2}{2} + x_0 \frac{x_0 + x_1}{2} (x_1 - x_0)) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + 3 x_1}{2} \frac{x_1^2 - x_0^2}{2} + \frac{x_1^2 - x_0^2}{2} x_1) + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2}{2} + x_0 x_1^2 - x_1 x_0^2) + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{3 x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2} + \frac{x_1^2 - x_0^2}{2} x_0) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - (\frac{x_0 + 3 x_1}{2} - x_1) \frac{x_1^2 - x_0^2}{2}) + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2 - 2 x_0 x_1^2 + 2 x_1 x_0^2}{2}) + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - (\frac{3 x_0 + x_1}{2} - x_0) \frac{x_1^2 - x_0^2}{2}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2}) + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{2 x_1^3 - 2 x_0^3 - 3 x_0 x_1^2 + 3 x_0^3 - 3 x_1^3 + 3 x_1 x_0^2 + 6 x_0 x_1^2 - 6 x_1 x_0^2}{6} + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 + x_1}{2} \frac{x_1^2 - x_0^2}{2}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 - x_0^3 + x_1^3 - x_1 x_0^2}{4}) + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{-x_1^3 + x_0^3 + 3 x_0 x_1^2 - 3 x_1 x_0^2}{6} + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3}{3} - \frac{x_0 x_1^2 + x_1^3 - x_0^3 - x_1 x_0^2}{4}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{4 x_1^3 - 4 x_0^3 - 3 x_0 x_1^2 + 3 x_0^3 - 3 x_1^3 + 3 x_1 x_0^2}{12} + \frac{4f(x_i)}{(x_1 - x_0)(x_0 - x_1)} \frac{-x_1^3 + 3 x_0 x_1^2 - 3 x_1 x_0^2 + x_0^3}{6} + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{4 x_1^3 - 4 x_0^3 - 3 x_0 x_1^2 - 3 x_1^3 + 3 x_0^3 + 3 x_1 x_0^2}{12}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{x_1^3 - x_0^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2}{12} + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{6} + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - x_0^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2}{12}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{12} + \frac{4f(x_i)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{6} + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)}(\frac{x_1^3 - 3 x_0 x_1^2 + 3 x_1 x_0^2 - x_0^3}{12}) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{2f(x_0)}{(x_0 - x_1)(x_0 - x_1)} \frac{(x_1 - x_0)^3}{12} + \frac{4f(x_i)}{6} (x_1 - x_0) + \frac{2f(x_1)}{(x_1 - x_0)(x_1 - x_0)} \frac{(x_1 - x_0)^3}{12} + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{6} (x_1 - x_0) + \frac{4f(x_i)}{6} (x_1 - x_0) + \frac{f(x_1)}{6} (x_1 - x_0) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + 4f(x_i) + f(x_1)}{6} (x_1 - x_0) + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + 4f(x_i) + f(x_1)}{6} \Delta x + 3 h_{3} (x - x_i)^3
</math>

<math>
\int_{a}^{b} f(x) dx = (f(x_0) + 4f(x_i) + f(x_1)) \frac{\Delta x}{6} + 3 h_{3} (x - x_i)^3
</math>



<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{(x_0 - x_1)} \frac{-(x_0 - x_1)^2}{2} + \frac{f(x_1)}{(x_1 - x_0)} \frac{(x_1 - x_0)^2}{2} + 2 h_{2} \frac{(x_1 - x_0)^3 - (x_0 - x_1)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{-f(x_0)}{2} (x_0 - x_1) + \frac{f(x_1)}{2} (x_1 - x_0) + 2 h_{2} \frac{(x_1 - x_0)^3 + (x_1 - x_0)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0)}{2} (x_1 - x_0) + \frac{f(x_1)}{2} (x_1 - x_0) + 4 h_{2} \frac{(x_1 - x_0)^3}{24}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + f(x_1)}{2} (x_1 - x_0) + h_{2} \frac{(x_1 - x_0)^3}{6}
</math>

<math>
\int_{a}^{b} f(x) dx = \frac{f(x_0) + f(x_1)}{2} \Delta x + h_{2} \frac{\Delta x^3}{6}
</math>


<math>
\int_{a}^{b} f(x) dx = f(x_i)(x_1 - x_0)
</math>

<math>
\int_{a}^{b} f(x) dx = f(x_i) \Delta x
</math>


<math>
\int_{a}^{b} f(x) dx \approx \int_{a}^{b} f(x_i) dx
</math>


S-Sum
<math>
S_{1}(x_0,x_1) = x_i1/2 / 1
</math>

<math>
S_{2}(x_0,x_1) = x_i1/3 + x_i2/3 / 2
</math>

<math>
S_{1}(x_0,x_1) = x_i1/4 + x_i2/4 + x_i3/4 / 3
</math>

Let us consider the integrals

<math>
\int (x - c_1) dx = \frac{x^2}{2} - c_1 x + c_0 = \frac{1 x^2 - 2 c_1 x + 2 c_0}{1 \times 2}
</math>

<math>
\int (x - c_1) (x - c_2) dx = \int (x^2 - (c_1 + c_2) x + c_1 c_2) dx = \frac{x^3}{3} - (c_1 + c_2) \frac{x^2}{2} + c_1 c_2 x + c_0 = \frac{2 x^3 - 3 (c_1 + c_2) x + 6 c_1 c_2 x + 6 c_0}{1 \times 2 \times 3}
</math>

<math>
\int (x - c_1) (x - c_2) (x - c_3) dx = \int (x^3 - (c_1 + c_2 + c_3) x^2 + (c_1 c_2 + c_1 c_3 + c_2 c_3) x + c_1 c_2 c_3) dx = \frac{x^4}{4} - (c_1 + c_2 + c_3) \frac{x^3}{3} + (c_1 c_2 + c_1 c_3 + c_2 c_3) \frac{x^2}{2} + c_1 c_2 c_3 x + c_0 = \frac{2 x^3 - 3 (c_1 + c_2) x + 6 c_1 c_2 x + 6 c_0}{1 \times 2 \times 3}
</math>




An analytical method of calculation of the relative weights of the evaluations of the function <math>f(x)</math> at the points <math>{x_{0}}, {x_{i}}_{({x_{0}},{x_{1}})}, {x_{1}}, ...</math>, consists in evaluating the integral of the function <math>f(x)</math> as the integral of a polynomial approximation.

If one chooses the value of <math>{x_k}</math> as <math>{x_k} = {x_j}</math>, i.e., at the beginning of every subinterval, then the sum of products is equal to:

<math>
\int_{a}^{b} f(x) dx \approx f({x_0} = a) ({x_1} - {x_0}) + f({x_1}) ({x_2} - {x_1}) + ... + f({x_{n-2}}) ({x_{n-1}} - {x_{n-2}}) + f({x_{n-1}}) ({x_n} - {x_{n-1}})
</math>,

If one chooses the value of <math>{x_k}</math> as <math>{x_k} = {x_i}</math>, i.e., at the beginning of every subinterval, then the sum of products is equal to:

<math>
\int_{a}^{b} f(x) dx \approx f({x_0} = a) ({x_1} - {x_0}) + f({x_1}) ({x_2} - {x_1}) + ... + f({x_{n-2}}) ({x_{n-1}} - {x_{n-2}}) + f({x_{n-1}}) ({x_n} - {x_{n-1}})
</math>,

Let us consider the case of equally-spaced points (although this restriction can be easily lifted), according to the definition,

In the case of a semi-continuous function, i.e., a function which is continuous except for a finite numerable set of points in an interval <math>(a,d)</math>, the calculation of the definite integral can be defined as the sum of the definite integrals between the points of discontinuity, i.e.,

<math>\int_{a}^{d} f(x) dx = \int_{a}^{b^{-}} f(x) dx + \int_{b^{+}}^{c^{-}} f(x) dx + \int_{c^{+}}^{d} f(x) dx  \approx \sum_{x = a}^{x = {b^{-}}} f(x) \Delta x + \sum_{x = {b^{+}}}^{x = {c^{-}}} f(x) \Delta x + \sum_{x = {c^{+}}}^{x = d} f(x) \Delta x</math>.

<!--

010 LET X0 = 0.0
020 LET XI = 0.5 
030 LET X1 = 1.0
040 LET Y0 = EXP(X0)
050 LET YI = EXP(XI)
060 LET Y1 = EXP(X1)
070 LET IC = Y1 - Y0
080 LET I0 = Y0 * (X1 - X0)
090 LET II = YI * (X1 - X0)
100 LET I1 = Y1 * (X1 - X0)
110 LET I2 = (Y0 + Y1) / 2 * (X1 - X0)
120 LET I3 = (Y0 + YI + Y1) / 3 * (X1 - X0)
130 LET I4 = (Y0 + 2 * YI + Y1) /  4 * (X1 - X0)
140 LET I6 = (Y0 + 4 * YI + Y1) /  6 * (X1 - X0)
150 LET I8 = (Y0 + 6 * YI + Y1) /  8 * (X1 - X0)
160 LET I9 = (Y0 + 8 * YI + Y1) / 10 * (X1 - X0)
170 LET E0 = (I0 - IC) / IC * 100
180 LET EI = (II - IC) / IC * 100
190 LET E1 = (I1 - IC) / IC * 100
200 LET E2 = (I2 - IC) / IC * 100
210 LET E3 = (I3 - IC) / IC * 100
220 LET E4 = (I4 - IC) / IC * 100
230 LET E6 = (I6 - IC) / IC * 100
240 LET E8 = (I8 - IC) / IC * 100
250 LET E9 = (I9 - IC) / IC * 100
260 PRINT IC
270 PRINT I0
280 PRINT II
290 PRINT I1
300 PRINT I2
310 PRINT I3
320 PRINT I4
330 PRINT I6
340 PRINT I8
350 PRINT I9
360 PRINT E0
370 PRINT EI
380 PRINT E1
390 PRINT E2
400 PRINT E3
410 PRINT E4
420 PRINT E6
430 PRINT E8
440 PRINT E9
450 END

000 ANSWERS

1.71828

1

1.64872

2.71828

1.85914

1.78900

1.75393

1.71886

1.70133

1.69081

-41.8023

-4.04826

58.1977

8.19767

4.11570

2.07471

0.0337172

-0.986777

-1.59908

--

010 LET IC = (EXP(1.0) - EXP(0.0))
020 LET I  = (EXP(0.0) + 4 * EXP(0.25) + 2 * EXP(0.5) + 4 * EXP(0.75) + EXP(1.0)) * 0.5 / 6
030 LET E  = (I - IC) / IC * 100
040 PRINT IC
050 PRINT I
060 PRINT E
070 END

ANSWER

1.71828

1.71832

0.00216456

--

010 LET IC = (EXP(1.0) - EXP(0.0))
020 LET I  = (EXP(0.0) + EXP(0.25) + 6 * EXP(0.5) + EXP(0.75) + EXP(1.0)) * 1.0 / 10
030 LET E  = (I - IC) / IC * 100
040 PRINT IC
050 PRINT I
060 PRINT E
070 END

ANSWER

1.71828

1.70116

-0.99624

--

010 LET IC = (EXP(1.0) - EXP(0.0))
020 LET I  = (EXP(0.0) + 7 * EXP(0.25) + 2 * EXP(0.5) + 7 * EXP(0.75) + EXP(1.0)) * 0.5 / 10
030 LET E  = (I - IC) / IC * 100
040 PRINT IC
050 PRINT I
060 PRINT E
070 END

ANSWER

1.71828

1.70116

-0.99624

--

010 LET X0 = -1.0
020 LET X1 = 1.0
030 LET XI = (X1 + X0) / 2
040 LET Y0 = EXP(X0)
050 LET Y1 = EXP(X1)
060 LET YI = EXP(XI)
070 LET IC = (Y1 - Y0)
080 LET II = YI * (X1 - X0)
090 LET I3 = (Y0 + YI + Y1) / 3 * (X1- X0)
100 LET I4 = (Y0 + 2 * YI + Y1) / 4 * (X1- X0)
110 LET I5 = (Y0 + 3 * YI + Y1) / 5 * (X1- X0)
120 LET I6 = (Y0 + 4 * YI + Y1) / 6 * (X1- X0)
130 LET I7 = (Y0 + 5 * YI + Y1) / 7 * (X1- X0)
140 LET I8 = (Y0 + 6 * YI + Y1) / 8 * (X1- X0)
150 LET I9 = (Y0 + 7 * YI + Y1) / 9 * (X1- X0)
160 LET I10 = (Y0 + 8 * YI + Y1) / 10 * (X1- X0)
170 LET I11 = (Y0 + 9 * YI + Y1) / 11 * (X1- X0)
180 LET I12 = (Y0 + 10 * YI + Y1) / 12 * (X1- X0)
190 LET I13 = (Y0 + 11 * YI + Y1) / 13 * (X1- X0)
200 LET I14 = (Y0 + 12 * YI + Y1) / 14 * (X1- X0)
210 LET I15 = (Y0 + 13 * YI + Y1) / 15 * (X1- X0)
220 LET I16 = (Y0 + 14 * YI + Y1) / 16 * (X1- X0)
230 LET I17 = (Y0 + 15 * YI + Y1) / 17 * (X1- X0)
240 LET I18 = (Y0 + 16 * YI + Y1) / 18 * (X1- X0)
250 LET I19 = (Y0 + 17 * YI + Y1) / 19 * (X1- X0)
260 LET I20 = (Y0 + 18 * YI + Y1) / 20 * (X1- X0)
270 LET I21 = (Y0 + 19 * YI + Y1) / 21 * (X1- X0)
280 LET I22 = (Y0 + 20 * YI + Y1) / 22 * (X1- X0)
290 LET EC = (IC - IC) / IC * 100
300 LET EI = (II - IC) / IC * 100
310 LET E3 = (I3 - IC) / IC * 100
320 LET E4 = (I4 - IC) / IC * 100
330 LET E5 = (I5 - IC) / IC * 100
340 LET E6 = (I6 - IC) / IC * 100
350 LET E7 = (I7 - IC) / IC * 100
360 LET E8 = (I8 - IC) / IC * 100
370 LET E9 = (I9 - IC) / IC * 100
380 LET E10 = (I10 - IC) / IC * 100
390 LET E11 = (I11 - IC) / IC * 100
400 LET E12 = (I12 - IC) / IC * 100
410 LET E13 = (I13 - IC) / IC * 100
420 LET E14 = (I14 - IC) / IC * 100
430 LET E15 = (I15 - IC) / IC * 100
440 LET E16 = (I16 - IC) / IC * 100
450 LET E17 = (I17 - IC) / IC * 100
460 LET E18 = (I18 - IC) / IC * 100
470 LET E19 = (I19 - IC) / IC * 100
480 LET E20 = (I20 - IC) / IC * 100
490 LET E21 = (I21 - IC) / IC * 100
500 LET E22 = (I22 - IC) / IC * 100
510 PRINT IC
520 PRINT II
530 PRINT I3
540 PRINT I4
550 PRINT I5
560 PRINT I6
570 PRINT I7
580 PRINT I8
590 PRINT I9
600 PRINT I10
610 PRINT I11
620 PRINT I12
630 PRINT I13
640 PRINT I14
650 PRINT I15
660 PRINT I16
670 PRINT I17
680 PRINT I18
690 PRINT I19
700 PRINT I20
710 PRINT I21
720 PRINT I22
730 PRINT EC
740 PRINT EI
750 PRINT E3
760 PRINT E4
770 PRINT E5
780 PRINT E6
790 PRINT E7
800 PRINT E8
810 PRINT E9
820 PRINT E10
830 PRINT E11
840 PRINT E12
850 PRINT E13
860 PRINT E14
870 PRINT E15
880 PRINT E16
890 PRINT E17
900 PRINT E18
910 PRINT E19
920 PRINT E20
930 PRINT E21
940 PRINT E22
950 END

ANSWER

2.3504

2

2.72411

2.54308

2.43446

2.36205

2.31033

2.27154

2.24137

2.21723

2.19748

2.18103

2.1671

2.15517

2.14482

2.13577

2.12778

2.12068

2.11433

2.10862

2.10344

2.09874

0

-14.9082

15.8996

8.19767

3.5765

0.495714

-1.70485

-3.35526

-4.63892

-5.66584

-6.50605

-7.20623

-7.79869

-8.30651

-8.74662

-9.13171

-9.47151

-9.77355

-10.0438

-10.287

-10.5071

-10.7071

010 LET X0 = -1.0
020 LET X1 = 1.0
030 LET XI = (X1 + X0) / 2
040 LET Y0 = EXP(X0)
050 LET Y1 = EXP(X1)
060 LET YI = EXP(XI)
070 LET IC = (Y1 - Y0)
080 LET II = YI * (X1 - X0)
090 LET I2 = (Y0 + Y1) / 2  * (X1 - X0)
100 LET I3 = (Y0 + YI + Y1) / 3 * (X1- X0)
110 LET I4 = (Y0 + 2 * YI + Y1) / 4 * (X1- X0)
120 LET I5 = (Y0 + 3 * YI + Y1) / 5 * (X1- X0)
130 LET I6 = (Y0 + 4 * YI + Y1) / 6 * (X1- X0)
140 LET EC = (IC - IC) / IC * 100
150 LET EI = (II - IC) / IC * 100
160 LET E2 = (I2 - IC) / IC * 100
170 LET E3 = (I3 - IC) / IC * 100
180 LET E4 = (I4 - IC) / IC * 100
190 LET E5 = (I5 - IC) / IC * 100
200 LET E6 = (I6 - IC) / IC * 100
210 PRINT IC
220 PRINT II
230 PRINT I2
240 PRINT I3
250 PRINT I4
260 PRINT I5
270 PRINT I6
280 PRINT EC
290 PRINT EI
200 PRINT E2
310 PRINT E3
320 PRINT E4
330 PRINT E5
340 PRINT E6
350 END

2.3504

2

3.08616

2.72411

2.54308

2.43446

2.36205

0

-14.9082

31.3035

15.8996

8.19767

3.5765

0.495714

-->

{{bookcat}}