# Romberg integration Calculator

## Calculates the integral of the given function f(x) over the interval (a,b) using the Romberg Method.

 $\normal I={\large\int_a^{\hspace{25}b}}f(x)dx=R_{\small n}^{\small\ k}+O$$(\frac{b-a}{2^n})^{\small 2k+2}$$\\[10]\hspace{10} R_{\small 0}^{\small\ 0}\\\hspace{10} R_{\small 1}^{\small\ 0}\hspace{30} R_{\small 1}^{\small\ 1}\\[10]\hspace{10} \cdots\hspace{40} \cdots\hspace{30} \ddots \\[10]\hspace{10} R_{\small n-1}^{\small\ 0}\hspace{15} R_{\small n-1}^{\small\ 1}\hspace{15} \cdots \hspace{10} R_{\small n-1}^{\small\ n-1}\\\hspace{10} R_{\small n}^{\small\ 0}\hspace{30} R_{\small n}^{\small\ 1}\hspace{30} \cdots \hspace{10} R_{\small n}^{\small\ n-1}\hspace{20} R_{\small n}^{\small\ n}\\$
 f(x) a , b maximum step n 2 3 4 5 6 7 8 9 10 11 12 13 14 15  partitions N=2n rule Trapezoidal Midpoint Non-linear substitution (Midpoint)
 6dgt10dgt14dgt18dgt22dgt26dgt30dgt34dgt38dgt42dgt46dgt50dgt
 The Romberg integration Rnn can be rapidly converged with the increase in n.It is calculated by increasing the number of partitions to double from 2 to N=2n.In the case of non-analytic at endpoints of f(x), you can calculate by Midpoint rule.In the case of a periodic function, you can calculate by Non-linear substitution in x. $\normal Romberg\ integration\\[10](1)\ Trapezoidal\ rule\\\hspace{10} R_{\small 0}^{\small\ 0}= {\large\frac{h_0}{2}}\{f(a)+f(b)\}, \hspace{20}h_0=b-a\\\hspace{8} R_{\small n}^{\small\ 0}= {\large\frac{R_{\small n-1}^{\small\ 0}}{2}}+h_n{\large \sum_{\tiny j=1}^{\small 2^{n-1}}}f(a+(2j-1)h_n)\\\hspace{190}h_n={\large\frac{b-a}{2^n}}\\[10](2)\ Midpoint\ rule\\\hspace{10} R_{\small 0}^{\small\ 0}= h_0f(a+{\large\frac{h_0}{2}}),\hspace{20}h_0=b-a\\\hspace{10} R_{\small n}^{\small\ 0}= h_n{\large \sum_{\tiny j=1}^{\small 2^n}}f(a+(j-\frac{1}{2})h_n), \hspace{15}h_n={\large\frac{b-a}{2^n}}\\(3)\ R_{\small n}^{\small\ k}= {\large\frac{4^{\small k} R_{\small n}^{\small\ k-1}-R_{\small n-1}^{\small\ k-1}}{4^{\small k}-1}}\\[10](4)\ Relative\ Error\hspace{30} \epsilon_{\small n}=\left|\frac{R_{\small n}^{\small\ n}-R_{\small n-1}^{\small\ n-1}}{R_{\small n}^{\small\ n}}\right|\\[30]Non-linear\ substitution\ in\ x\\\hspace{25} I={\large\int_a^{\hspace{25}b}}f(x)dx\ ={\large\int_{\small-1}^{\hspace{25}\small1}}f(x)\ \frac{b-a}{2}\ \frac{3(1-u^2)}{2}du\\\hspace{100}x=\frac{b-a}{2}t+\frac{b+a}{2},\hspace{20}t=\frac{u}{2}(3-u^2)$

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