Problem $5$: Error analysis in the Riemann Sum Algorithm

When using the left Riemann sum algorithm, one does not know a priori what step size or number of intervals to choose so that the result is accurate to within a desired error tolerance $($the maximum error that can be tolerated$).$ This choice depends on the given function to integrate. For example, if one integrates the constant function, an algorithm using $N=1$ produces the exact result, i$.$e$.,$ no error.

a$)$ Modify the rmlsum file to include a while$-$loop so that the program code terminates when an accuracy based on relative approximate error is achieved. In each iteration of the whileloop, double the number of the intervals $($i$.$e$.,$ the step size $h$ is reduced by one half$)$ in the numerical integration algorithm. The function's input$/$output structure should be:

function $[I,N]=rml{\displaystyle \underset{?}{\overset{?}{}}}N($ fun $,a,b,$ maxtol $)$

where, I is the approximate result of the integration and $N$ is the number of intervals that would yield a result to satisfy the error tolerance criterion. Note that this output value of $N$ is not necessarily the minimum number of intervals to achieve the error tolerance.

Apply this numerical solver to the following integral using maxtol $=1e-5.$

${\displaystyle \underset{0}{\overset{10}{}}}\frac{100}{1+0\mathrm{.}4t^{2\mathrm{.}5}}dt$

One potential difficulty of the above approach is that it might take many iterations to achieve the error tolerance. In general, one would like to set a maximum number of iterations.

b$)$ Modify the above numerical solver so that the while$-$loop checks both the error tolerance and the maximum number of iterations. In other words, the loop terminates when either the error is sufficiently small, or the maximum number of iterations is exceeded. The function structure should be:

function $[I,N]=rml{\displaystyle \underset{?}{\overset{?}{}}}N1($fun$,a,b,$ maxtol, maxitr$)$

Test your program code on the same integral from part $($a$)$ using maxtol $=1e-5$ and maxit $r=19.$ Note that because the number of intervals is doubled in each iteration, $19$ iterations would result in $N=2^{19}=524288.$