# Calculus/Integration techniques/Numerical Approximations

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It is often the case, when evaluating definite integrals, that an antiderivative for the integrand cannot be found, or is extremely difficult to find. In some instances, a numerical approximation to the value of the definite value will suffice. The following techniques can be used, and are listed in rough order of ascending complexity.

## Riemann Sum

This comes from the definition of an integral. If we pick n to be finite, then we have:
$\int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i}^{*})\Delta x$
where $x_{i}^{*}$ is any point in the i-th sub-interval $[x_{i-1},x_{i}]$ on $[a,b]$ .

### Right Rectangle

A special case of the Riemann sum, where we let $x_{i}^{*}=x_{i}$ , in other words the point on the far right-side of each sub-interval on, $[a,b]$ . Again if we pick n to be finite, then we have:
$\int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i})\Delta x$

### Left Rectangle

Another special case of the Riemann sum, this time we let $x_{i}^{*}=x_{i-1}$ , which is the point on the far left side of each sub-interval on $[a,b]$ . As always, this is an approximation when $n$ is finite. Thus, we have:
$\int \limits _{a}^{b}f(x)dx\approx \sum _{i=1}^{n}f(x_{i-1})\Delta x$

## Trapezoidal Rule

$\int \limits _{a}^{b}f(x)dx\approx {\frac {b-a}{2n}}\left[f(x_{0})+2\sum _{i=1}^{n-1}{\biglf(x_{i}){\bigr )}+f(x_{n})\right]={\frac {b-a}{2n}}{\bigg{f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots +2f(x_{n-1})+f(x_{n})}{\bigg )}$

## Simpson's Rule

Remember, n must be even,
    $\int \limits _{a}^{b}f(x)dx$ $\approx {\frac {b-a}{6n}}\left[f(x_{0})+\sum _{i=1}^{n-1}\left((3-(-1)^{i})f(x_{i})\right)+f(x_{n})\right]$ $={\frac {b-a}{6n}}{\biggf(x_{0})+4f{\bigl{\tfrac {x_{1}}{2}}{\bigr )}+2f(x_{1})+4f{\bigl{\tfrac {x_{3}}{2}}{\bigr )}+\cdots +4f{\bigl{\tfrac {x_{n-1}}{2}}{\bigr )}+f(x_{n}){\bigg ]}$

## Maclaurin Approximation

A common technique of approximating common trigonometric functions is to use the Taylor-Maclaurin series. Term-by-term integration allows one to easy compute the value of the integral by hand, well up to 5 decimal places of precision, and up to 10 given a factorial table.
For example, using the Maclaurin series of $\sin(x)$, one can easily approximate its integral with a polynomial.
$\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}$
We can then easily integrate each term, taking $(-1)^{n}$ and ${\frac {1}{(2n+1)!}}$ to be constants.
${\displaystyle \sum _{n=0}^{\infty }{\int {\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}}}\implies c_{0}+{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+2}}{(2n+2)!}}}$
We can easily find the constant term by inspecting the known principle integral, $-\cos(x)$, and the new series. This nets us the final equation.
$\int _{0}^{t}{\sin(x)\;\mathrm {d} x}+1={\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n+2}}{(2n+2)!}}}$While this is a rather fast-converging series, converging at $\log _{10}((2x+2)!)$ digits of significance, it is relatively useless, since factorials are expensive to compute.

## Further reading

* Numerical Methods/Numerical Integration
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